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124-2 Identify the critical value

158-1 use a _____ confidence interval for MU and hence

209-1 so that eight-ounce cups will overflow only 1% of the time

210-1 A company manufactures rope.

720-3 A type I error is always made when:

722-1 The level of significance is (check all that apply):

745-2 Level of confidence is another name for level of significance.

1280-1 What is the calculated value

1285-1 we would reject H(0) if:

1287-2 You will reject H(0) if:

1288-1 The level of significance of this test is approximately:

1290-1 one should _____ the H(0) since the value _____ lies _____

1291-1 "ALPHA = _____, one should _____ the H(0) since the value "

1298-1 A result was said to be statistically significant at the 5% level.

1298-2 The reasoning in rejecting a null hypothesis is __________.

1299-2 The critical value of a test statistic is determined from:

1301-1 significance level of the test is:

1301-2 Which of the following assumptions are needed to

1302-1 "In testing a hypothesis using a statistic Y, a critical region is"

1306-2 Find all values of Z = [XBAR - MU(0)]/[S/SQRT(n)]

1307-1 Which of the following statements do you KNOW is correct?

1308-1 What type of decision is reached when the calculated value of any

1309-1 "ended with a decision of ""reject H(O)""."

1309-2 what is the function of a critical value that is

1312-1 Whether or not causation may be inferred in a research study

1313-3 Your correct interpretation of the outcome is

1323-1 Test the null hypothesis that the new ball

1325-1 Test the appropriate hypothesis at the 5% level.

1336-4 What objection is there to using the rejection region:

1349-3 "If we would reject a null hypothesis at the 5% level, we would also"

1352-1 The decision to use a one-sided or two-sided test is usually made

1352-3 Significance at the ALPHA = .001 level means that the null

1354-2 Testing at a 5% level of significance means that you only have

1489-1 can we conclude that the experimental mean differs

1490-1 Determine a 0.90 confidence interval for the mean reaction time

1493-1 Construct a 95% confidence interval for MU.

1494-1 Do you think that it would be quite unreasonable for

1502-2 Set up a decision rule for H(0): MU => 25

1508-1 "In a sample of 25 physicians, the mean annual income of $47,000"

1567-2 Which of the following assumptions are needed to test

1602-2 "XBAR = 22, one should conclude that:"

1603-1 "If the P-value for your test statistic satisfies P > .25, then:"

1608-1 weight reduction program. After four months the statistics

1609-1 "The meaning of ""testing the hypothesis MU <= 6 versus MU > 6"

1610-3 "Statistical Significance means that, if an experiment were "

1617-1 H(0): MU >= 114 against H(1): MU < 114

1618-1 H(0): MU = 43 against H(1): MU =/= 43

1619-1 "With a computed t-statistic of 2.63, what conclusion should"

1627-1 A new diet for the reduction of cholesterol is introduced.

1628-2 This term she believes that her students are doing sig

1631-1 Past production units of a certain jet engine model showed the mean

1632-1 "Forty-nine American soldiers, observed at random, yield a mean weight"

1635-1 A standard intelligence examination has been given for several years

1646-1 she has collected the final grades of her classes and found

1654-1 "If a statistic is significant at the 5 percent level, then it must be"

1654-2 One can never prove the truth of a statistical (null) hypothesis. One can

1655-1 "If the population mean is known, it makes no sense to test"

1655-2 Since the P-value in a test of hypothesis is based on the specific

1655-3 The descriptive level of significance (P-value) is chosen by

1657-1 A hypothesis accepted at the ALPHA = .20 level of significance

1657-2 A small significance level indicates that the hypothesis

1657-3 If the results of an investigation show that one sleeping tablet

1661-4 "In hypothesis testing, a type I error is"

1662-1 "With which of the following terms is the ""level of significance"" most"

1662-2 better than the old one. The Type I Error is to conclude that:

1664-2 "Usually, one would like the critical region for a test to be _______"

1664-4 "If the number of observations (n) is increased to 2n,"

1665-1 if the population variance (SIGMA**2) is decreased to

1667-2 "If Z(critical) = 2.04, what is the p-value for your test?"

1668-2 For testing the hypothesis MU = 28 against

1669-1 Interpret the quality control procedure described above as a test of

1670-1 What would be the consequences of a Type I and Type II error?

1672-2 When an experimenter selects a particular level of risk (ALPHA) he

1677-1 What type of error might you have made in part a?

1686-3 A Type I error is committed when one accepts the null hypothesis

1688-1 The significance level is computed under the assumption that

1688-2 The risk of type II error does not depend upon the risk of type I

1691-2 "Other things being equal, a small level of significance is desirable."

1693-1 Level of confidence equals (1 - level of significance).

1694-1 "Although we speak of two types of error, in testing any"

1694-2 "Generally, a larger sample size implies a smaller level of"

1702-1 "In this situation, a type II error would be:"

1704-1 Type II error refers to:

1706-2 What is the probability of a Type II error when ALPHA = .05?

2137-1 rejected at the .05 level but not at the .025 level?

2139-1 The value(s) of the test statistic you would use is (are):

2140-1 Sixteen one-acre plots of wheat were harvested.

2143-1 we find a critical value of t equals 2.015. This means that

2143-2 "According to these data, the researcher can reject his H(0) with"

2147-1 "For each of the following sets of information, find and specify"

2151-2 have a higher average height than American males as a whole?

2789-1 the starting point of the region of rejection in terms of XBAR

2790-1 Should the teacher use the new method? Why?

2792-1 What is the smallest ALPHA-level that could be chosen before you

2796-1 The Crapi Cable Company #35 cable has a mean breaking strength of 1800

2799-1 Analyze the data at both levels after setting up appropriate

2800-1 at a .001 significance level. For a sample of 49

2803-1 "Since the observed Z-value is _______ than the table value, we would

2808-1 Test the company's claim using ALPHA = .01.

2811-2 What is the significance probability of the observed result?

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Q: It is desired to test the claim that a steady diet of wolfbane will cause a lycanthrope (werewolf) to lose 10 lbs. over 5 months. A random sample of 49 lycanthropes was taken yielding an average weight loss over 5 months of 12.5 pounds, with S = 7 lbs. Identify the critical value suitable for conducting a two-tail test of the hypothesis at the 2% level. a. 2.06 b. 2.58 c. 1.96 d. 1.65 e. 2.33

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Q: Molybdenum rods produced on a production line are supposed to average 2.2 inches in length. It is desired to check whether the process is in control. Let X = length of such a rod. Assume X is approximately nor- mally distributed with mean = MU and variance = SIGMA**2, where the mean and the variance are unknown. Suppose a sample of n = 400 rods is taken and yields a sample average length of XBAR = 2 inches, and SUM((X - XBAR)**2) = 399. To test H(0): MU = 2.2 vs. H(1): MU =/= 2.2 at level ALPHA = 8%, one would use a _____ confidence interval for MU and hence a table value of _____. a) 92%, 1.67 b) 92%, 1.41 c) 92%, 1.75 d) 96%, 2.06 e) 96%, 1.75

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Q: A soft drink machine can be regulated so that it discharges an average of MU ounces per cup. If the ounces of fill are normally distributed with standard deviation equal to .3 ounces, give the setting for MU so that eight-ounce cups will overflow only 1% of the time.

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Q: A company manufactures rope. From a large number of tests over a long period of time, they have found a mean breaking strength of 300 lbs. and a standard deviation of 24 lbs. Assume that these values are MU and SIGMA. It is believed that by a newly developed process, the mean breaking strength can be increased. (a) Design a decision rule for rejecting the old process with an ALPHA error of 0.01 if it is agreed to test 64 ropes. (b) Under the decision rule adopted in (a), what is the probability of accepting the old process when in fact the new process has increased the mean breaking strength to 310 lbs.? Assume SIGMA is still 24 lbs. Use a diagram to illustrate what you have done, i.e., draw the reference distributions.

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Q: A type I error is always made when: a. the null hypothesis is rejected when it is true b. the null hypothesis is not rejected when it is false c. the research hypothesis is rejected when it is true d. the research hypothesis is not rejected when it is false

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Q: The level of significance is (check all that apply): A. the probability of rejecting the null hypothesis when the null hypothesis is true. B. the magnitude of the sample size. C. symbolized by the greek letter ALPHA. D. none of the above.

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Q: True or False? If False, correct it. Level of confidence is another name for level of significance.

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Q: It is desired to test the claim that a steady diet of wolfbane will cause an 18-year-old lycanthrope werewolf to lose EXACTLY 10 lbs. over 5 months. A random sample of 49 lycanthropes was taken, yielding an average weight loss over 5 months of 12.5 lbs. with S = 7 lbs. Let ALPHA = .02. What is the calculated value suitable for testing the above hypothesis? a. 12.5 b. 7 * 2.5 c. 2.5 d. -2.5 e. .35

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Q: A sample of size 36 is taken from a population with unknown mean MU and standard deviation SIGMA = 3. In a test of H(0): MU = 5 versus H(1): MU =/= 5 at ALPHA = .01, we would reject H(0) if: (a) XBAR - 5 > 1.29 or 5 - XBAR > 1.29 (b) XBAR - 5 > 7.74 or 5 - XBAR > 7.74 (c) XBAR - 5 > 1.29 or 5 - XBAR < 7.74 (d) XBAR - 5 < 1.29 or 5 - XBAR < 1.29 (e) XBAR - 5 < 7.74 or 5 - XBAR < 7.74

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Q: The standard deviation of a large population is 20. To test: H(0): MU = 4 vs. H(A): MU > 4 at level of significance .05, a sample of size 100 will be taken. You will reject H(0) if: (a) XBAR >= 7.3 (d) XBAR >= 7.8 or XBAR <= .2 (b) XBAR >= 7.3 or XBAR <= .8 (e) none of the above (c) XBAR >= 7.8

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Q: To test H(0): MU = 20 vs. H(A): MU =/= 20, a sample of 400 will be taken from a large population, whose standard deviation is 5. H(0) will be rejected if XBAR >= 20.5 or XBAR <= 19.5. The level of significance of this test is approximately: a. .05 b. .02 c. .10 d. .20 e. .15

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Q: Molybdenum rods are produced by a production line setup. It is desir- able to check whether the process is in control. Let X = length of such a rod. Assume X is approximately normally distributed with mean = MU and variance = SIGMA**2, where the mean and variance are unknown. Take n = 400 sample rods, with sample average length XBAR = 2 inches, and SUM((X - XBAR)**2) = 399. In testing H(0): MU = 2.2 vs. H(1): MU =/= 2.2 at level ALPHA = 8%, one should _____ the H(0) since the value _____ lies _____ the confidence interval. a) not reject, 2.2, within b) reject, 2, outside of c) reject, 2.2, outside of d) not reject, 2, within e) either b or c

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Q: Molybdenum rods are produced by a production line setup. It is desired to check whether the process is in control. Let X = length of such a rod. Assume X is approximately normally distributed with mean = MU and variance = SIGMA**2, where the mean and variance are unknown. Take n = 400 sample rods, with sample average length XBAR = 2 inches and SUM((X - XBAR)**2) = 399. If one were testing H(0): MU = 1 vs. H(1): MU =/= 1 at level ALPHA = _____, one should _____ the H(0) since the value 1 lies _____ the confidence interval. a) 16%, not reject (continue), within b) 8%, not reject (continue), within c) 4%, not reject (continue), within d) 4%, not reject (continue), to the left of e) 4%, reject, to the left of

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Q: A result was said to be statistically significant at the 5% level. This means: a. the null hypothesis is probably wrong b. the result would be unexpected if the null hypothesis were true c. the null hypothesis is probably true d. none of the above.

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Q: The reasoning in rejecting a null hypothesis is __________. a. that a significant result usually occurs when the null hypo- thesis is false. b. that a significant result seldom occurs when the null hypo- thesis is true. c. some reason other than (a) or (b). d. both (a) and (b).

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Q: The critical value of a test statistic is determined from: a. calculations from the data. b. calculations based on many actual repetitions of the same experiment. c. the sampling distribution of the statistic assuming H(A). d. the sampling distribution of the statistic assuming H(O).

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Q: Suppose a t-test for the hypothesis that H(O): MU = 0 vs. H(A): MU =/= 0 is carried out and we find t(obs.) = 1.8. The descriptive significance level of the test is: a. the Type I error probability of the test. b. the probability of getting a t-value >= 1.8. c. the probability of getting a t-value >= 1.8 or <= -1.8. d. the Type II error probability of the test. e. none of these.

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Q: Indicate which assumptions are needed to use the sample mean and normal tables to test a hypothesis about a population mean, MU, and known variance, SIGMA**2. Which of the following assumptions are needed to use XBAR, the mean of the data, and normal tables to test a hypothesis about MU? I. the data are a random sample II. the population distribution is normal III. the sample size is large a. I, II, and III b. I and either II or III c. II and III d. only II e. none of the above

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Q: In testing a hypothesis using a statistic Y, a critical region is chosen to meet which of the following conditions: I. the probability of Y falling in the critical region when the null hypothesis is true is ALPHA II. the probability of Y falling in the critical region when the alternative hypothesis is true is greater than it not falling in the critical region. III. the sample size is large a. I, II, and III b. I and II only c. I only d. II only e. none of the above

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Q: Suppose you are going to test H(O): MU = MU(0) H(A): MU =/= MU(0) using ALPHA =.05. Find all values of Z = [XBAR - MU(0)]/[S/SQRT(n)] for which H(O) should be rejected. a) Z < -1.96 b) Z < -1.645 or Z > 1.645 c) -1.96 < Z < 1.96 d) Z < -1.96 or Z > 1.96 e) -1.645 < Z < 1.645

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Q: Suppose a random sample of size 25 is selected from a population with mean MU, the value of which is unknown. The sample statistics are XBAR = 6.4, s = 10. Test H(O): MU = 10 H(A): MU < 10 using ALPHA = .05. Which of the following statements do you KNOW is correct? a) A type 1 error has been committed. b) H(O) is rejected. c) H(O) is not rejected. d) Statements (a) and (b) are correct. e) None of the above.

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Q: What type of decision is reached when the calculated value of any test statistic falls in the critical region when a false null hypothesis is being tested? a) A correct decision. b) Type I error. c) Type II error. d) The type of decision can not be determined from the information given. e) None of the above are correct.

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Q: A home owner claims that the current market value of his house is at least $40,000. Sixty real estate agents were asked independently to estimate the house's value. The hypothesis test that followed ended with a decision of "reject H(O)". Which of the following statements accurately states the conclusion? a) The home owner is right, the house is worth $40,000. b) The home owner is right, the house is worth less than $40,000. c) The home owner is wrong, the house is worth less than $40,000. d) The home owner is wrong, the house is worth more than $40,000. e) The home owner is wrong, he should not sell his home.

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Q: In hypothesis testing, what is the function of a critical value that is taken from the tables? a. It is equal to the calculated statistic from the observed data. b. It is the point where the decision changes from reject to fail to reject. c. It is the center of the distribution of X's. d. It is a point which is 1 standard deviation away from the mean.

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Q: Whether or not causation may be inferred in a research study (a) is indicated by the magnitude of the test statistic employed. (b) is given by the final p-value. (c) must be decided by the investigator. (d) depends upon the sample size employed in the study.

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Q: You have drawn a random sample of size n from a specified population and you test the hypothesis that MU = 50. Suppose that you are unable to reject the hypothesis. Your correct interpretation of the outcome is that a. in the population, MU = 50. b. probability is high that in the population, MU = 50. c. the sample data are not inconsistent with the hypothesis that in the population, MU = 50. d. all of the above are acceptable interpretations.

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Q: The FMA Company has designed a new type of 16 lb. bowling ball. The company knows that the average man who bowls in a scratch league with the company's old ball has a bowling average of 155. The variance of these averages is 100. The company asks a random sample of 100 men bowling in scratch leagues to bowl for five weeks with their new ball. The mean of bowling averages for these men with the new ball is 170. There is no reason to believe the variance is any different with the new ball. Test the null hypothesis that the new ball does not improve a bowler's average at the 5% level of significance.

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Q: A reading coordinator in a large public school system suspects that poor readers may test lower in IQ than children whose reading is satis- factory. He draws a random sample of 30 fifth grade students who are poor readers. Historically fifth grade students in the school system have had an average IQ of 105. The sample of 30 has XBAR = 101.5 and S(XBAR) = 1.42. Test the appropriate hypothesis at the 5% level.

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Q: What objection is there to using the rejection region: [68 - [.125*(SIGMA/SQRT(N))]] < XBAR < [68 + [.125*(SIGMA/SQRT(N))]] in testing the hypothesis H(O): MU = 68? HINT: What would be the level of significance for this test?

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Q: True or False? If False, correct it. If we would reject a null hypothesis at the 5% level, we would also reject it at the 1% level.

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Q: True or False? If False, correct it. The decision to use a one-sided or two-sided test is usually made after the data is analyzed.

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Q: True or false? If false, explain why. Significance at the ALPHA = .001 level means that the null hypo- thesis is almost certainly false.

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Q: True or false? If false, explain why. Testing at a 5% level of significance means that you only have a 5% chance of rejecting the null hypothesis.

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Q: The calculated nitrogen content of pure benzanilide is 7.10%. Five repeat analyses of "representative" samples yielded values of 7.11%, 7.08%, 7.06%, 7.06%, and 7.04%. Using an ALPHA level of size 5%, can we conclude that the experimental mean differs from the expected value? Assume that the measured values are approximately normally distributed.

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Q: The following are the reaction times in seconds of five people to a particular stimulus: 7,8,6,10,9 XBAR = 40/5 = 8 S(X)**2 = [(-1)**2 + (0)**2 + (-2)**2 + (1)**2]/4 = 10/4 = 2.5 a. Do these data present sufficient indication that the mean reac- tion time of all people would be less than ten seconds? Test at the 0.05 level of significance (ALPHA = .05). b. Determine a 0.90 confidence interval for the mean reaction time for all people to this stimulus. c. State any assumptions that must be made in order to do parts (a) and/or (b).

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Q: Suppose in a sample of 25 people, the mean height XBAR was observed to be 70 inches. Suppose also SIGMA = 3. A. Construct a 95% confidence interval for MU. B. Would you reject the hypothesis H(0):MU = 71 versus H(1):MU =/= 71 on the basis of the observations, when testing at level ALPHA = .05? C. Would you reject the hypothesis H(0):MU = 72 versus the alternative H(1):MU =/= 72 on the basis of the observations, when testing at level ALPHA = .05? D. Would you reject the hypothesis H(0):MU = 69 versus the (one-sided) alternative H(1):MU > 69 on the basis of your observations, when testing at level ALPHA = .05?

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Q: A floor manager of a large department store is studying the buying habits of the store's customers. Suppose he assumes that monthly income of these customers is normally distributed with a standard deviation of 500. If he draws a random sample of size N = 100 and obtains a sample mean of YBAR = 800, A 0.95 confidence interval for the true population mean is 702 < MU < 898. Do you think that it would be quite unreasonable for the true population mean to be $600? Explain.

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Q: Suppose that you are concerned with the claim that MU > 25. Assume that you know that SIGMA = 1. Set up a decision rule for H(0): MU => 25 for a sample of size 10 such that the type 1 error rate is 2%.

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Q: In a sample of 25 physicians, the mean annual income of $47,000 with a variance of $360,000. a. Estimate with 95% confidence the mean income for all physicians. b. What assumptions are necessary to make this a valid estimate? Do you feel the assumptions are reasonable in this case? Why or why not? c. Based on your answer in (a), would the null hypothesis that the true mean is $50,000 be continued or rejected at the 5% significance level? Why?

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Q: Which of the following assumptions are needed to test a hypothesis about mean MU in a population with known variance SIGMA**2 using the mean of the data X(1) ... X(N) and normal tables: I. The data are a random sample. II. The population distribution is normal. III. The sample size is large. a. I, II and III d. only II b. I and either II or III e. none of these c. II and III

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Q: In testing H(0): MU <= 20 vs. H(A): MU > 20 when ALPHA = .05, n = 25, S**2 = 16 and XBAR = 22, one should conclude that: a. MU > 20 with 5% chance of error b. MU > 20 with 5 % confidence c. MU = 20 with 95% confidence d. MU = 20 with 95% chance of error

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Q: If the P-value for your test statistic satisfies P > .25, then: (a) you would not reject H(O) (b) you would reject H(O) for ALPHA = .05 (c) you would reject H(O) for ALPHA = .10 (d) your acceptance region has a lower limit of .25 (e) none of these

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Q: Nine men with a genetic condition that causes obesity entered a weight reduction program. After four months the statistics of weight loss were: XBAR = 11.2, S = 9.0. The researcher wants to test the hypothesis: The average four-month weight loss in such a program is <= 6 pounds verses the alternative: > 6 pounds at a 5% significance level. Given the data of our problem, we a. reject the hypothesis. b. reserve judgement about the hypothesis.

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Q: The meaning of "testing the hypothesis MU <= 6 versus MU > 6 at a 5% significance level" is: a. if the population mean MU is > 6, the probability of deciding wrongly is at least 95%. b. if the population mean MU is <= 6, the probability of deciding wrongly is at most 5%. c. if the population mean MU is > 6, the probability of deciding wrongly is at most 5%. d. no matter what the population mean is, the probability of deciding wrongly is at most 5%.

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Q: Statistical Significance means that, if an experiment were replicated over and over again: a) the same results would occur again with certainty b) the same results would probably occur c) the same results would probably not occur d) the same results would certainly not occur again.

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Q: Suppose X(1) ... X(10) is a sample from a normal population with mean = MU and variance = 22.5. The critical region for testing H(0): MU >= 114 against H(1): MU < 114 at significance level .05 is: Reject H(0) if a. XBAR < 112.00 b. XBAR < 111.54 c. XBAR < 111.25 d. XBAR < 111.06 e. none of these

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Q: If X(1),...,X(10) is a sample from a normal population with mean = MU and variance = SIGMA**2, with SIGMA**2 unknown, and the sample standard deviation (biased estimator) s = 7.5, then the critical region for testing H(0): MU = 43 against H(1): MU =/= 43 at ALPHA = .01 is: Reject H(0) if a. XBAR > 50.02 or XBAR < 35.98 b. XBAR > 51.12 or XBAR < 34.88 c. XBAR > 50.56 or XBAR < 35.44 d. XBAR > 50.72 or XBAR < 35.28 e. none of these.

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Q: Blood samples from 40 patients are analyzed for glucose by two different methods. With a computed t-statistic of 2.63, what conclusion should you draw? a. There is a significant difference in methods, p < .005. b. There is a significant difference in methods, p < .01. c. There is a significant difference in methods, p < .05. d. There is no significant difference in the two methods.

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Q: A new diet for the reduction of cholesterol is introduced. In order to test this procedure, nine patients on this new diet had observed choles- terol levels of: patient cholesterol patient cholesterol 1 240 6 220 2 290 7 190 3 220 8 230 4 250 9 200 5 260 XBAR = 210 S**2 = SUM(([X(i) - 210]**2)/8) = 950 S = 30.9 Assume cholesterol levels are normally distributed. This new method of cholesterol reduction was used on a sample from a population with a mean MU cholesterol level of 225. Test the hypo- thesis that the procedure used was effective (ALPHA = .05). a. H(O): _______________. b. H(A): _______________. c. test statistic = _______________. d. critical region = _______________. e. Does the test statistic (c) fall in the critical region (d)? f. Conclusion?

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Q: A teacher has been conducting the same course for many years. As part of her check on her own teaching and on the general effectiveness of the students in her classes from year to year, she has been giving the same pop quiz during the 4th week of classes. Over the years the score has averaged 13.5. This term she believes that her students are doing sig- nificantly different than usual. If she can establish support of her perception she will alter her teaching methods for this class. On the basis of the test scores below should she conclude that there is support for her perception and change her methods? On what basis do you offer your consultation to her? Scores: 20, 19, 6, 4, 3, 10, 13, 14, 16, 17, 10, 11, 8, 9, 11, 20, 18, 6, 4, 13. XBAR = 11.6 S = 5.49

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Q: Past production units of a certain jet engine model showed the mean military thrust to be 7600 pounds. The first ten production units manufactured after a model change yielded military thrusts of 7620, 7680, 7570, 7700, 7650, 7720, 7600, 7540, 7670, and 7630. Is there sufficient evidence (use ALPHA = 0.05) that the model change resulted in a higher average military thrust? Finding: YBAR = 7638 S(Y) = SQRT((583,420,000 - ((76,380)**2/10))/9) = 57.3

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Q: Forty-nine American soldiers, observed at random, yield a mean weight of 160 pounds with a standard deviation (s) of 11 pounds. Are these observations consistent with the assumption that the mean weight of all American soldiers is 170 pounds?

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Q: A standard intelligence examination has been given for several years with an average score of 80 and a standard deviation of 7. If 25 students taught with special emphasis on reading skill, obtain a mean grade of 83 on the examination, is there reason to believe that the special emphasis changes the result on the test? Use ALPHA = .05.

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Q: A teacher has just taken a course in statistics and decides to put her new knowledge to work. During the last ten years she has collected the final grades of her classes and found the mean to be 73.6 with a standard deviation = 9. The final grades for this years class are: 95, 97, 84, 64, 72, 88, 92, 61, 76, 74, 84, 85, 89, 75, 76, 64, 61, 72, 63, 74. The teacher wishes to know if she should consider this years class as significantly different than previous years. The teacher did very well in her statistics class. What would you believe her decision was? Why? XBAR = 77.3; Hint: Do NOT calculate S in this problem.

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Q: True or False? If False, correct it. If a statistic is significant at the 5 percent level, then it must be significant at the 1 percent level also.

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Q: True or False? If False, correct it. One can never prove the truth of a statistical (null) hypothesis. One can only tend to discount it.

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Q: True or False? If False, correct it. If the population mean is known, it makes no sense to test hypotheses concerning the population mean.

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Q: True or False? If False, correct it. Since the P-value in a test of hypothesis is based on the specific observed value of a test statistic, it cannot be used in a two-sided test.

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Q: True or False? If False, correct it. The descriptive level of significance (P-value) is chosen by the investigator before his experiment is conducted.

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Q: True or False? If False, correct it. A hypothesis accepted at the ALPHA = .20 level of significance is probably true.

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Q: True or False? If False, correct it. A small significance level indicates that the hypothesis will probably be continued.

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Q: TRUE OR FALSE. IF FALSE EXPLAIN WHY. If the results of an investigation show that one sleeping tablet works better than another at the 5% level of significance, the conclusion would be similar if tested at the 10% level of significance.

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Q: In hypothesis testing, a type I error is a. failing to reject the null hypothesis when it is false. b. failing to reject the null hypothesis when it is true. c. rejecting the null hypothesis when it is true. d. rejecting the null hypothesis when it is false.

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Q: With which of the following terms is the "level of significance" most closely associated? a. one-tailed test of significance b. two-tailed test of significance c. type 1 error d. type 2 error

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Q: Given the null hypothesis: that a new process is as good as or better than the old one. The Type I Error is to conclude that: (a) the old process is as good or better when it is not (b) the old process is better when it is (c) the old process is better when it is not (d) the new process is as good or better when it is

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Q: Usually, one would like the critical region for a test to be _______ (long or short). The "size" of the critical region is determined by _______ (ALPHA or BETA).

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Q: If the number of observations (n) is increased to 2n, the level of significance (ALPHA) is: a. increased b. unaffected c. decreased

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Q: When conducting a test for the population mean, if the population vari- ance (SIGMA**2) is decreased to [(SIGMA**2)/2], then the level of sig- nificance (ALPHA) is: a. increased b. unaffected c. decreased

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Q: Suppose you are testing H(O): MU <= 21 vs. H(A): MU > 21, from a normal distribution with SIGMA**2 known equal to 28 and n = 13. If Z(critical) = 2.04, what is the p-value for your test? a. .0207 d. .025 b. .4793 e. none of these c. .0414

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Q: A population is known to have a variance of 16. When a sample of size 25 is taken, the sample variance is found to be 14, while the sample mean is 30. For testing the hypothesis MU = 28 against the alternative MU =/= 28 at the 0.10 level, the critical values are: a. +/- 1.711 b. +/- 1.318 c. +/- 1.645 d. +/- 1.28 e. none of these

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Q: Past experience shows that, if a certain machine is adjusted properly, 5 percent of the items turned out by the machine are defective. Each day the first 25 items produced by the machine are inspected for defects. If three or fewer defects are found, production is continued without interruption. If four or more items are found to be defective, produc- tion is interrupted and an engineer is asked to adjust the machine. After adjustments have been made, production is resumed. This proce- dure can be viewed as a test of the hypothesis p = .05 against the alternative p > .05, p being the probability that the machine turns out a defective item. In test terminology, the engineer is asked to make adjustments only when the hypothesis is rejected. Interpret the quality control procedure described above as a test of the indicated hypothesis. A Type I error results in: a. a justified production stoppage to carry out machine adjustments. b. an unnecessary interruption of production. c. the continued production of an excess of defective items. d. the continued production, without interruption, of items that satisfy the accepted standard.

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Q: In assessing the weather prior to leaving our residences on a spring morning, we make an informal test of hypothesis "The weather will be fair today." Using the "best" information available to us, we complete the test and dress accordingly. What would be the consequences of a Type I and Type II error? (1) Type I error: inconvenience in carrying needless rain equipment; Type II error: clothes get soaked. (2) Type I error: clothes get soaked; Type II error: inconvenience in carrying needless rain equipment. (3) Type I error: clothes get soaked; Type II error: no consequence since Type II error cannot be made. (4) Type I error: no consequence since a Type I error cannot be made; Type II error: inconvenience in carrying needless rain equipment.

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Q: When an experimenter selects a particular level of risk (ALPHA) he effectively is setting the probability of rejecting H(0) when in fact it is true. a. true b. false

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Q: The mean weight of adult women in the U.S. is 140 lb. with a standard deviation of 20 lb. You are willing to accept the report of the standard deviation but not the mean. You selected 400 women at random and measured their weight and found the average in this group to be 137 lb. a. Test at ALPHA=.05 the hypothesis that the true weight is 140 lb. b. What type of error might you have made in part a? Do you know the probability of making such an error? If so, what is it? If not, why not?

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Q: True or False? If False, explain why. A Type I error is committed when one accepts the null hypothesis when it is false.

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Q: True or False? If False, correct it. The significance level is computed under the assumption that the alternative hypothesis is true.

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Q: True or False? If False, correct it. The risk of type II error does not depend upon the risk of type I error.

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Q: True or False? If False, correct it. Other things being equal, a small level of significance is desirable.

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Q: True or False? If False, correct it. Level of confidence equals (1 - level of significance).

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Q: True or False? Explain your answer. Although we speak of two types of error, in testing any specified hypothesis we can make only one.

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Q: True or False? If False, correct it. Generally, a larger sample size implies a smaller level of significance.

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Q: Given the null hypothesis: that a process is producing no more than the maximum allowable rate of defective items. In this situation, a type II error would be: (a) to conclude that the process is producing too many defectives when it actually is not (b) to conclude that the process is not producing too many defec- tives when it actually is (c) to conclude that the process is not producing too many defec- tives when it is not (d) to conlcude that the process is producing too many defectives when it is.

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Q: Type II error refers to: a. rejecting the null hypothesis when the alternative is true. b. choosing the wrong decision rule. c. not rejecting the null hypothesis when the alternative is true. d. incorrectly assuming the data are normally distributed. e. none of these.

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Q: What is the probability of a Type II error when ALPHA = .05? (1) .025 (2) .050 (3) .950 (4) .975 (5) Cannot be determined without more information

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Q: 16 observations are taken from a normal population in order to test the null hypothesis H(O): MU = 10 against H(A): MU < 10. A t-statistic is evaluated. In which case would H(O) be rejected at the .05 level but not at the .025 level? a. t = -2.12 b. t = -2.50 c. t = 2.12 d. t = 10 - 2.50 = 7.50 e. t = 10 - 2.12 = 7.88

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Q: A sample of size 26 is taken from a finite Normal population in which the variance is known to be 25. It is desirable to test the hypothe- sis that the mean is greater than 50, with ALPHA = .025. The critical value(s) of the test statistic you would use is (are): (a) -1.96, 1.96 (d) -1.645 (b) 1.96 (e) -2.060, 2.060 (c) 1.645 (f) 1.708

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Q: Sixteen one-acre plots of wheat were harvested. Their average yield was found to be 701 bushels, and their standard deviation was 50 bu. Since last year's crop yielded 680 bu./acre, we wish to test H(0): MU <= 680 against H(A): MU > 680 at ALPHA = .05. Based on the above data we should a. not reject H(0). b. reject H(0).

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Q: Employing ALPHA = 0.05, one-tailed test, for 5 df, we find that the critical value of t equals 2.015. This means that (1) in this t-distribution, 5% of the area lies below t=-2.015. (2) there is a 95% probability of obtaining a t-ratio less than 2.015. (3) if our obtained t = 2.00, we cannot reject H(0) using the .05 level. (4) all of the above. (5) none of the above.

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Q: The birthweight of babies is normally distribution with a mean of 3.8 kg. A researcher suspects that the weight of babies from mothers who smoked a lot during pregnancy will be lower than the population mean. To examine this, he takes a random sample of 26 babies from mothers who smoked a lot. The mean birthweight in this group was 3.48 kg with a standard deviation of .80 kg. (biased estimator of SIGMA**2). He takes as H(0): MU >= 3.80 kg and as H(1): MU < 3.80 kg. According to these data, the researcher can reject his H(0) with a. 2.5 % probability of a type I error b. 5 % probability of a type I error c. 10% probability of a type I error d. 95% probability of a type I error

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Q: For each of the following sets of information, find and specify the appropriate critical region, test the null hypothesis, and draw a conclusion. a) H(0): MU = 16 n = 27 H(A): MU =/= 16 YBAR = 15.0 ALPHA = .02 s**2 = 3.0 b) H(0): MU <= 18.2 n = 250 H(A): MU > 18.2 YBAR = 18.7 ALPHA = .005 s**2 = 3.6 c) H(0): MU >= 113 n = 14 H(A): MU < 113 YBAR = 108 ALPHA = .01 s**2 = 56

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Q: Suppose that the heights of American males are normally distributed with MU=71". A random sample of n=100 university students has XBAR=72.5, S**2 =25.0. Can it be claimed at the ALPHA=.10 level of significance that university students have a higher average height than American males as a whole?

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Q: Suppose a test was taken by 36 students and the variance of the distribution of scores was 100. It is desired to test H(O): MU>=80 against H(1): MU<80, using ALPHA=.05 and z as the test statistic. (Assume the population of test scores is normally distributed.) What ( to the nearest tenth) is the starting point of the region of rejection in terms of XBAR values? a. 76.1 b. 76.7 c. 77.3 d. 77.7 e. None of these.

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Q: A fourth grade teacher wants to try a new teaching method which the authors recommend should only be used with particularly bright children. The authors offer a short test which they feel can be used as a guide to decide whether the method should be used. They believe that the average score for a class on this test should significantly exceed 73 and indicate that the national standard deviation for the test is 10. The teacher's students take the test and exhibit the following scores: 91, 91, 94, 63, 61, 40, 73, 75, 83, 84, 70, 71, 88, 93, 92, 91, 87, 83, 74, 80. Should the teacher use the new method? Why? (Use ALPHA = .05.)

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Q: In a Cancer Experiment, Guinness Beer was given in large quantities to rats (fortunate things]) to see if it inhibited the growth of tumors. From comparative, comprehensive studies it is known that the "typical" tumor should weigh 1.35 grams. These 25 rats had a mean weight of 1.31 grams with a standard deviation of 0.2 grams. a. Has there been a significant inhibiting of growth? Use an ALPHA of 0.10. (Assume tumors are normally distributed.) b. What is the smallest ALPHA-level that could be chosen before you would change your conclusion?

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Q: The Crapi Cable Company #35 cable has a mean breaking strength of 1800 pounds with a standard deviation of 100 pounds. A new material is used which, it is claimed, increases the breaking strength. To test this claim a random sample of 50 cables, manufactured with the new material, is tested. It is found that the sample has a mean breaking strength of 1850 pounds. Test this claim using ALPHA = .01.

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Q: In the past a chemical fertilizer plant has produced an average of 1100 pounds of fertilizer per day. The record for the past year based on 256 operating days shows the following: XBAR = 1060 lbs/day S = 320 lbs/day where XBAR and S have the usual meaning. It is desired to test whether or not the average daily production has dropped significantly over the past year. Suppose that in this kind of operation, the traditionally acceptable level of significance has been .05. But the plant manager, in his report to his bosses, uses level of significance .01. Analyze the data at both levels after setting up appropriate hypotheses, and comment.

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Q: In J.B. Nimble's occupation, he is concerned with the length of candles. Assuming that candle lengths are normally distributed with mean MU and variance 4.0 square inches, he is interested in testing the hypothesis that MU <= 15.0 inches against the alternative that MU > 15.0 inches at a .001 significance level. For a sample of 49 candles, he obtains a sample mean of 15.6 inches. Conclusions?

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Q: It is desired to test the claim that a steady diet of wolfbane will cause a lycanthrope (werewolf) to lose 10 lbs. over 5 months. A random sample of 49 lycanthropes was taken yielding an average weight loss over 5 months of 12.5 pounds, with S = 7 lbs. Use a two-tailed test and let ALPHA = .02. Since the observed Z-value is _______ than the table value, we would ___ H(0). (Regard Z as an adequate approximation to t.) a. smaller, reject b. greater, not reject c. greater, reject d. smaller, not reject

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Q: The Pfft Light Bulb Company claims that the mean life of its 2 watt bulbs is 1300 hours. Suspecting that the claim is too high, Nalph Rader gathered a random sample of 64 bulbs and tested each. He found the average life to be 1295 hours with s = 20 hours. Test the com- pany's claim using ALPHA = .01.

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Q: A study was conducted to see if students living off-campus had a grade point average which differed significantly from the university-wide average GPA of 2.65. Extensive records indicate that the standard deviation of the GPA is 0.3. What is your conclusion, if a random sample of 100 off-campus students had a mean GPA of 2.72? (The researcher feels that it is reasonable to assume that the population of GPA scores is normally distributed.) What is the significance probability of the observed result?

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A: e. 2.33 t(critical,twotail,ALPHA=.02,df=48) == +/- 2.33

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A: c) 92%, 1.75

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A: Z(for ALPHA=.01) = 2.33 Z = (XBAR - MU)/SIGMA 2.33 = (8 - MU)/.3 8 - MU = (2.33) (.3) MU = 8 - (.699) MU = 7.301

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A: a. One tail test at ALPHA = .01, therefore Z = 2.33. Z = (YBAR-MU)/(SIGMA/SQRT(n)) 2.33 = (YBAR-300)/(24/SQRT(64)) YBAR = 307 Decision Rule: If the mean strength of 64 ropes tested is 307 lbs. or more, we reject the hypothesis of no im- provement, i.e., we continue that the new process is better. b. If available, consult file of graphs and diagrams that could not be computerized for reference distributions. Z = (307-310)/(24/SQRT(64)) = 1.00 Area = 0.1587 or 15.87% P(type II error) = 0.1587

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A: a. the null hypothesis is rejected when it is true.

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A: A. the probability that the observed experimental difference is due to chance given the null hypothesis. C. symbolized by the greek letter ALPHA.

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A: False. Level of confidence = 1 - (level of significance).

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A: c. 2.5 t(calculated) = (12.5-10)/(7/SQRT(49)) = 2.5

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A: (a) XBAR - 5 > 1.29 or 5 - XBAR > 1.29 Half Confidence Interval = Z*SIGMA/SQRT(n) = 2.576 * .5 = 1.29 Hence, the critical value at the 1% level is 1.29.

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A: (a) XBAR >= 7.3 Z = (XBAR- MU)/(S/SQRT(n)) 1.645 = (XBAR - 4)/2 XBAR = 7.29

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A: a. .05 Half Confidence Interval = Z * S/SQRT(N) .5 = Z * 5/20 2 = Z Area beyond Z = .023 So for both tails the significance level used = .023 + .023 = .046

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A: c) reject, 2.2, outside of S**2 = 399/399 = 1 S(XBAR) = SQRT(S**2/n) = .05 If you center the confidence interval on the sample mean the confidence interval = 2 +/- (1.75)(.05) = from 1.9125 to 2.0875 which does not contain the hypothesized value, 2.2.

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A: e) 4%, reject, to the left of S**2 = 399/399 = 1; S(XBAR) = SQRT(S**2/n) = .05; C.I. = 2 +/- Z(ALPHA/2) * .05; Z(16%/2) = 1.41, Z(8%/2) = 1.75, Z(4%/2) = 2.05 C.I.(ALPHA=16%) = 2 +/- (1.41)(.05) = from 1.93 to 2.07 C.I.(ALPHA= 8%) = 2 +/- (1.75)(.05) = from 1.91 to 2.09 C.I.(ALPHA= 4%) = 2 +/- (2.05)(.05) = from 1.90 to 2.10 1 is not included in any of the confidence intervals, so H(0) should be rejected in all cases.

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A: b. the result would be unexpected if the null hypothesis were true.

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A: d. both (a) and (b).

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A: d. the sampling distribution of the statistic assuming H(O).

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A: c. the probability of getting a t-value >= 1.8 or <= -1.8.

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A: b. I and either II or III.

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A: b. I and II only

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A: d) Z < -1.96 or Z > 1.96

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A: b) H(O) is rejected. t(calc) = [XBAR - MU]/[s/SQRT(n)] = [6.4 - 10]/[10/5] = [-3.6]/2 = -1.8 t(crit., df=24, ALPHA=.05, one-tailed) = -1.711 Since t(calc) < t(crit), reject H(O). However, since the true value of MU is not known, we do not know if a type 1 error has been committed.

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A: a) A correct decision.

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A: c) The home owner is wrong, the house is worth less than $40,000.

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A: b. It is the point where the decision changes from reject to fail to reject.

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A: (c) must be decided by the investigator.

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A: c. the sample data are not inconsistent with the hypothesis that in the population, MU = 50.

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A: H(O): MU <= 155 H(A): MU > 155 ALPHA = .05 MU = 155 SIGMA = 10 SIGMA(XBAR) = 10/SQRT(100) = 1 Z(calculated) = (XBAR - MU)/SIGMA(XBAR) = (170 - 155)/1 = 15 Z(critical, ALPHA =.05, one-tailed) = 1.645 Since Z(calculated) > Z(critical), reject H(O). Conclude that the new bowling ball does improve a bowler's average at the 5% level.

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A: H(O): MU >= 105 H(A): MU < 105 t(obtained) = (XBAR - MU)/S(XBAR) = (101.5 - 105)/1.42 = -3.5/1.42 = -2.465 t(critical, ALPHA=.05, one-tail, df=29) = -1.699 Since t(obtained) < t(critical), reject H(O). Therefore, the sample evidence is strong enough to suggest that poor readers test signifi- cantly lower in IQ.

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A: The main objection is that your level of confidence would be approximately equal to .10 while the significance level would be approximately equal to .90. This situation is the reverse of most testing situations.

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A: False - A result may be significant at the 5% level, but not at the 1% level.

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A: False, the decision to use a one-sided or two-sided test should be made before the data is collected, so that the experimenter is not influenced by the data.

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A: True, if the results are significant at .001 level, it is very unlikely (one in one thousand) that the null hypothesis is true.

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A: False, this statement is accurate only if the null hypothesis is true.

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A: H(O): MU = 7.10 H(A): MU =/= 7.10 YBAR = 7.07 S(Y) = 0.0265 t = (YBAR - MU)/S(YBAR) = (7.07 - 7.10)/(0.0265/SQRT(5)) = 2.53 t(critical, ALPHA=.05, df=4) = +/- 2.776 Since the calculated value of t is not in the critical region, continue H(O) that the nitrogen content has a true value of 7.10%, i.e., the 0.03% difference is ascribable to random error. or YBAR +/- t*(S(Y)/SQRT(n)) YBAR +/- 2.776*(0.0265)/(SQRT(5)) P(7.037 <= MU <= 7.103) = 0.95 Continue H(O) that the nitrogen content has a true value of 7.10% at 95% level since 7.10 lies within the 95% confidence interval.

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A: a. H(O): MU = 10 H(A): MU < 10 t(calculated) = (XBAR - MU)/(S(X)/SQRT(n) = (8 - 10)/SQRT(2.5/5) = -2/.707 = -2.8 t(critical, one-tail, ALPHA = .05, df = 4) = -2.13 Decision: reject H(O) and conclude that this data does present sufficient indication, with confidence level = .95, that the mean reaction time of all people would be less than ten seconds. b. C.I. = XBAR +/- t(ALPHA = .10)*S(XBAR) = 8 +/- (2.13 * .707) = 8 +/- 1.51 6.49 < MU < 9.51 c. 1.) The sample of n=5 observations is a simple random sample of times from the population or phenomena being studied.

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A: A. XBAR +/- Z*SIGMA/SQRT(n) = 70 +/- (1.96*3)/SQRT(25) = 66.824 to 71.176 B. Not Reject H(0) C. Reject H(0) D. H(0): MU = 69 vs. H(1): MU > 69. Decision Rule: If Z obtained > Z critical then reject H(0). Z (Calculated) = (XBAR - MU)/(SIGMA(XBAR)) = (70-69)/.6 = 1.667 Z (Critical) = 1.645 1.667 > 1.645 => Reject H(0)

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A: Yes, based on the above confidence interval, we would reject the hypothesis that MU = 700 (at ALPHA = .05).

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A: H(0): MU > 25 H(1): MU <= 25 With ALPHA = .02, a one-tailed Z = 2.05 Critical value = MU - (Z) (SIGMA(XBAR)) = 25 - (2.05)(1/SQRT(10)) = 24.35 Decision rule: If the sample mean is greater than or equal to 24.35, then the claim that MU > 25 is to be continued.

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A: a. XBAR - t*SQRT((S**2)/n) <= MU <= XBAR + t*SQRT((S**2)/n) 47000 - 2.06*SQRT(360000/25) <= MU <= 47000 + 2.06*SQRT(360000/25) 46750 <= MU <= 47250 b. Physician incomes are normally distributed. This is not likely to be a valid assumption - incomes are usually skewed. c. The null hypothesis would be rejected since the hypothesized value of $50,000 is not included in the confidence interval found in (a.).

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A: b. I and either II or III

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A: a. MU > 20 with 5% chance of error t = (22-20)/(4/5) = 2.5 Critical value for t is 1.711, with df = 24 Reject H(0) and conclude H(A) is true with .05 chance of error.

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A: (a) You would not reject H(O). The probability of the data under H(O) is greater than .25, therefore you would not reject H(O).

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A: b. reserve judgement about the hypothesis. S(XBAR) = S/SQRT(n) = 9.0/SQRT(9) = 3.0 t(calc) = [XBAR - MU]/[S(XBAR)] = [11.2 - 6.0]/[3.0] = 1.73 t(crit, df=8, ALPHA=.05, one-tail) = 1.86 Since t(calc) < t(crit), continue (or do not reject) H(O).

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A: b. if the population mean MU is <= 6, the probability of deciding wrongly is at most 5%.

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A: b) the same results would probably occur again. Statistical tests are always performed at some given probability level which gives the probability of occurrence of same results.

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A: b. XBAR < 111.54 SIGMA(XBAR) = SQRT(22.5/10) = 1.5 Critical point = 114 - (1.645)(1.5) = 111.5325

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A: b. XBAR > 51.12 or XBAR < 34.88 S/SQRT(n-1) = 7.5/3 = 2.5 Critical points = 43 +/- t(df=9, ALPHA=.01/2)*(2.5) = 43 +/- (3.25)(2.5)

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A: b. There is a significant difference in methods, p < .01.

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A: a. H(O): MU = 225 b. H(A): MU < 225 c. t(calculated) = (210 - 225)/(30.9/SQRT(9)) = (-15)/(10.3) = -1.456 d. t(critical, df = 8, ALPHA = .05, onetail) = -1.86 e. No f. The result of this sample does not provide strong enough evidence to support the claim that cholesterol level was reduced.

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A: Using ALPHA = .05 and two-tailed t-test with df = 19, we test H(0): MU = 13.5 H(1): MU =/= 13.5 t(calculated) = (11.6 - 13.5)/(5.49/SQRT(20)) = -1.55 t(critical) = +/- 2.093 We conclude that, at the .05 ALPHA level, she should continue H(0), since the mean score of this class is not significantly different from the usual. The basis for this conclusion rests on the assumption that the scores have a normal distribution.

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A: Using ALPHA = .05 and a one-tailed t-test we test: H(0): MU <= 7600 H(A): MU > 7600 t = (YBAR - MU)/(S(Y)/SQRT(N)) t = (7638 - 7600)/(57.3/SQRT(10)) t = 2.097 t(critical) = 1.83 Since t(calculated) is larger than t(critical) for a one-sided test at ALPHA = .05, reject the null hypothesis. At the 95% confidence level, the sample evidence indicates a detectable increase.

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A: Null hypothesis: MU = 170 Alternative: MU =/= 170 ALPHA = .05 Reject if t is not between -2.01 and 2.01. t = (160 - 170)/(11/SQRT(49)) = -6.36 Data from sample is not consistent with the assumption.

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A: H(O): MU = 80 H(A): MU =/= 80 SIGMA(XBAR) = SIGMA/SQRT(n) = 7/5 = 1.4 Using a Confidence Interval to test our hypothesis: XBAR(CRIT) = MU +/- Z(CRIT)*SIGMA(XBAR) = 80 +/- 1.96*1.4 = 77.256 to 82.744 Since 83 > 82.744 we reject H(O) and conclude that the result on the test is significantly different from 80.

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A: MU = 73.6 SIGMA = 9 Test the hypothesis that this years class is a sample of a population with MU = 73.6 and standard deviation = 9. H(0): MU = 73.6 H(1): MU =/= 73.6 SIGMA(XBAR) = 9/SQRT(20) = 2.01 Z(calculated) = (77.3 - 73.6)/2.01 = 3.7/2.01 = l.84 Z(critical) = 1.96 with ALPHA = .05 At the 5% significance level, Z(calculated) is less than Z(critical) so based on this evidence I would conclude that this years class is not different from previous years classes.

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A: False. A statistic significant at the 5% level is not necessarily significant at the 1% level, although the latter may be true.

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A: True. In statistical inference there is always some probability, how- ever small, of drawing the wrong conclusion. The fact that a hypothesis is consistent with a set of data does not mean that it is correct; whereas, if it is not consistent with the data set it may be incorrect.

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A: True, the purpose of a hypothesis test is to draw a conclusion about a population based on a sample. If the population is known, there is nothing to hypothesize.

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A: False, if the assumed distribution is symmetric, the P-value can be used in a two-sided test by doubling it.

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A: False, the descriptive level of significance (P-value) is dependent on the sample. The ALPHA level is chosen before the experiment is conducted.

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A: False. The failure to reject does not imply the null hypothesis is true.

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A: False. A small significance level merely indicates that the probability of a type I error is small.

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A: True, 95% confidence automatically implies more than 90% confidence.

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A: c. rejecting the null hypothesis when it is true.

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A: c. type 1 error

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A: (c) the old process is better when it is not. Type I Error = rejecting null hypothesis when it is true = rejecting [newer is as good or better than old] when true = continuing [old better than new] when it is not = Prob(getting old betternew process is better than old one)

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A: Usually, one would like the critical region for a test to be SHORT. The "size" of the critical region is determined by ALPHA.

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A: b. unaffected

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A: b. unaffected.

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A: a. .0207 Area beyond Z corresponding to Z value of 2.04 = .0207.

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A: c. +/- 1.645

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A: b. an unnecessary interruption of production.

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A: (1) Type I error: inconvenience in carrying needless rain equipment; Type II error: clothes get soaked.

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A: a. true

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A: a. H(0): MU = 140 H(A): MU =/= 140 ALPHA = .05 Test statistic: Z = (XBAR-MU)/(SIGMA/SQRT(n)) Assumptions: Random sample. Sampling distribution of mean is normal. Critical region: Z > 1.96 or Z < -1.96 Z = (137-140)/(20/SQRT(400)) = -3.00 Reject H(0) Conclude true mean is not equal to 140. b. Type I error: rejection of true H(0). Probability is .05 (ALPHA).

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A: False, a Type I error is committed when one rejects the null hypothesis when it is true.

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A: False - The significance level is computed under the assumption that the null hypothesis is true.

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A: False - The risk of type II error depends upon the risk of type I error. As the risk of one type increases, the risk of the other type decreases.

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A: True. Level of significance is the probability of Type I error. We would always like to have it small.

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A: True. Level of confidence = (1 - ALPHA).

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A: True. Reject H(0) Don't Reject H(0) ________________________________________________ NO ERROR H(0) True TYPE1ERROR 1 - ALPHA ________________________________________________ H(0) False NO ERROR TYPE2ERROR 1 - BETA ________________________________________________ Therefore, either we can make a Type I error or Type II error depending on whether our hypothesis is true or false, but it is impossible to make both simultaneously.

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A: False, the level of significance is usually chosen before the measurements are collected and is not dependent on the sample size.

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A: (b) to conclude that the process is not producing too many defec- tives when it actually is. Type II error = P(do not reject H(O) when H(O) is false) H(O) = process producing no more than k defectives H(O) false means process producing more than k de- fectives.

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A: c. not rejecting the null hypothesis when the alternative is true.

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A: (5) Cannot be determined without more information.

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A: a. t = -2.12 With df = 16 - 1 = 15, t(critical, ALPHA = .05) = -1.753 t(critical, ALPHA = .025) = -2.131

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A: (b) 1.96 This is a one-sided test and, since SIGMA is known, Z(ALPHA = .025) = 1.96.

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A: a. not reject H(0). t(calculated) = (701 - 680)/(50/SQRT(16)) = 21/(50/SQRT(16)) = 1.68 t(critical, one tail, ALPHA = .05, df = 15) = 1.753 Therefore t(calculated) < t(critical) and we should not reject H(0) at this significance level.

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A: (4) all of the above.

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A: b. 5% probability of a type I error t = [XBAR - MU]/[S(XBAR)] = [3.48-3.80]/[[.80)/[SQRT(25)]] = -2.00, df = 25, one-tailed test

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A: a) t(critical, df=26, ALPHA=.02, two-tail) = +/-2.479 critical region: t's < -2.479 and t's > +2.479 t(calculated) = (15-16)/SQRT(3/27) = -3 t(calculated) falls in the critical region, so based on this sample evidence you should reject H(0). b) t(critical, df=249, ALPHA=.005, one-tail) == Z(critical, ALPHA=.005, onetail) = +2.576 critical region: Z's > +2.576 Z(calculated) = (18.7-18.2)/(SQRT(3.6/250)) = 4.166 Z(calculated) falls in the critical region so based on this sample evidence you should reject H(0). c) t(critical, df=13, ALPHA=.01, one-tail) = -2.65 critical region: t's < -2.65 t(calculated) = (108-113)/(SQRT(56/14)) = -2.5 t(calculated) does not fall in the critical region so based on this sample evidence do not reject (continue) H(0).

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A: Using a one-tail t-test, we test: H(0): MU<=71 H(1): MU>71 t(calculated) =(XBAR-MU)/(S/SQRT(n)) =1.5/(5/SQRT(100)) =3 t(ALPHA=.10,df=99)==1.296Back to review this question

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A: c. 77.3 Z(crit., ALPHA=.05, one-tail) = 1.645 rejection region = MU - Z(crit.) * SIGMA/SQRT(n) = 80 - (1.645 * 10/SQRT(36)) = 77.258Back to review this question

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A: H(0): MU <= 73 SIGMA = 10 H(1): MU > 73 XBAR = 79.2 Z = (79.25 - 73)/(10/SQRT(20)) = 2.79 Z(critical) = 1.645; with ALPHA = .05 and using a onetail decision rule. Reject H(0) and conclude that the average score for this class is significantly higher than 73. Therefore, she should use the new method.Back to review this question

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A: a. H(O): Guinness Beer does not affect tumors. H(A): Guinness Beer inhibits the growth of tumors. i.e., H(O): MU => 1.35 H(A): MU < 1.35 t(calc.) = [XBAR-MU]/[S/SQRT(n)] with (n-1) df = [1.31-1.35]/[0.2/SQRT(25)] = [-0.04]/[0.04] = -1.00 t(crit., df=24, ALPHA=.10, one-tail) = -1.318 Since t(calc.) > t(crit.), continue (do not reject) H(O). It would thus appear that Guinness Beer does not affect the growth of tumors. b. In part (a) we continued H(O), now we must reject H(O) for some level of ALPHA. Keeping same degrees of freedom we notice: ALPHA t(crit.) ----- -------- 0.05 -1.711 0.10 -1.318 0.15 -1.059 0.20 -0.857 0.25 -0.685 Since t(calc.) = -1.000, the test is significant for ALPHA=0.20, but not significant for ALPHA=0.15. The approximate significance level obtained by interpolation is about 0.16.Back to review this question

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A: Hypothetical population: All Crapi #35 cables made with the new material. Sample: The 50 cables randomly selected. H(O): MU <= 1800. The mean breaking strength of the new cable is 1800 lb. H(A): MU > 1800. The mean breaking strength of the new cable is more than 1800 lb. MU(XBAR) = 1800 by H(O) SIGMA(XBAR) = SIGMA/SQRT(n) = 100/SQRT(50) = 14.142 XBAR(crit) = MU(XBAR) + Z(crit)*SIGMA(XBAR) = 1800 + (2.33)*(14.142) = 1832.951 Since the sample mean breaking strength is 1850, which is greater than 1832.51, we must reject H(O) and conclude that the mean breaking strength of the new cable is significantly more than 1800 lb.Back to review this question

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A: H(O): MU => 1100 H(A): MU < 1100 Since n = 256, use Z to approximate t. S(XBAR) = 320/SQRT(256) = 320/16 = 20 Z(calculated) = (1060 - 1100)/20 = -40/20 = -2 Z(critical, ALPHA=.05, one-tailed) = 1.645 Z(critical, ALPHA=.01, one-tailed) = 2.33 Therefore, H(0) is rejected at ALPHA=.05 but not at ALPHA=.01. It appears that the manager is trying to pull a fast one on his bosses by using ALPHA=.01 and saying production has not dropped. However, if the traditional level of significance is used, ALPHA=.05, there is evidence that indicates a drop in production.Back to review this question

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A: XBAR = 15.6 SIGMA**2 = 4 n = 49 H(O): MU <= 15.0 H(A): MU > 15.0 ALPHA = .001 Critical Region: Z > 3.090 Z(calculated) = (XBAR - MU)/(SIGMA/SQRT(n)) = 2.1 Conclusion: Mean candle lengths may be 15 inches.Back to review this question

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A: c. greater, reject Z(calculated) = (12.5-10)/(7/SQRT(49)) = 2.5/1 = 2.5 Z(critical,ALPHA=.02,twotail) = +/- 2.33Back to review this question

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A: Hypothetical population: All Pfft 2 watt bulbs. Sample: The 64 randomly selected bulbs. H(O): MU => 1300. The mean life of 2 watt bulbs is 1300 hours. H(A): MU < 1300. The mean life of 2 watt bulbs is less than 1300 hours. MU(XBAR) = 1300 by H(O) S(XBAR) = S/SQRT(n) = 20/8 = 2.5 XBAR(crit) = MU(XBAR) + Z(crit)*S(XBAR) = 1300 - 2.33*2.5 = 1294.18 Since 1295 is not less than 1294.18, we cannot reject H(O). There is not enough evidence to conclude that the mean life of the 2 watt bulbs is significantly less than 1300 hours.Back to review this question

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A: MU = 2.65 SIGMA = 0.3 N = 100 XBAR = 2.72 H(O): MU(off-campus) = 2.65 H(A): MU(off-campus) =/= 2.65 Z(calculated) = (XBAR - MU)/(SIGMA/SQRT(100)) = 2.33 P(Z > 2.33 or Z < -2.33) = .02 Based on this sample evidence, it appears that the null hypothesis is incorrect and that off-campus students do have a different GPA.Back to review this question

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Item is still being reviewed Multiple Choice TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 3 GeneralBack to this chapter's Contents

Based upon item submitted by S. Selvin - UC Berkeley Multiple Choice TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 3 GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice TYPE1ERROR ONETAIL/Z CONCEPT STATISTICS ZTEST PARAMETRIC T= 2 Computation Comprehension D= 2 General ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Multiple Choice TWOTAIL/Z TYPE1ERROR ZTEST PARAMETRIC STATISTICS CONCEPT T= 2 Comprehension D= 2 General ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Multiple Choice TYPE1ERROR CONCEPT STATISTICS T= 5 Comprehension D= 3 Natural Sciences Business GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice TYPE1ERROR TYPE2ERROR CONCEPT STATISTICS T= 2 Comprehension Application D= 4 GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 3 GeneralBack to this chapter's Contents

Item is still being reviewed Short Answer ***Calculus Necessary*** TWOTAIL/Z TYPE1ERROR ZTEST PARAMETRIC STATISTICS CONCEPT T=10 Application Comprehension D= 4 General Biological Sciences ***Multiple Parts*** ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed True/False TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 4 GeneralBack to this chapter's Contents

Based upon item submitted by W. J. Hall - Univ. of Rochester True/False TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 2 GeneralBack to this chapter's Contents

Based upon item submitted by W. J. Hall - Univ. of Rochester True/False TYPE1ERROR TYPE2ERROR CONCEPT STATISTICS T= 5 Comprehension D= 3 GeneralBack to this chapter's Contents

Item is still being reviewed True/False TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 3 GeneralBack to this chapter's Contents

Item is still being reviewed True/False TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 2 GeneralBack to this chapter's Contents

Item is still being reviewed True/False TYPE1ERROR TYPE2ERROR CONCEPT STATISTICS T= 2 Comprehension D= 4 GeneralBack to this chapter's Contents

Based upon item submitted by J. L. Mickey -UCLA True/False TYPE1ERROR CONCEPT STATISTICS T= 2 Comprehension D= 2 GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice TYPE2ERROR CONCEPT STATISTICS T= 2 Comprehension D= 2 GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice TYPE2ERROR CONCEPT STATISTICS T= 2 Comprehension D= 3 GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice TYPE2ERROR CONCEPT STATISTICS T= 2 Comprehension D= 2 GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice ONETAIL/T TTEST PARAMETRIC STATISTICS T= 2 Comprehension D= 3 General ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Multiple Choice ONETAIL/T TTEST PARAMETRIC STATISTICS T= 2 Computation D= 2 GeneralBack to this chapter's Contents

Based upon item submitted by F. J. Samaniego - UC Davis Multiple Choice ONETAIL/T TTEST PARAMETRIC STATISTICS T= 5 Application D= 3 General Biological Sciences ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Multiple Choice ONETAIL/T TTEST PARAMETRIC STATISTICS T= 2 Comprehension D= 3 GeneralBack to this chapter's Contents

Item is still being reviewed Multiple Choice ONETAIL/T TTEST PARAMETRIC STATISTICS T= 2 Computation D= 3 Biological Sciences General ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Numerical Answer ONETAIL/T TWOTAIL/T TTEST PARAMETRIC STATISTICS T=10 Computation Application D= 5 General ***Calculator Necessary*** ***Multiple Parts*** ***Statistical Table Necessary***Back to this chapter's Contents

Based upon item submitted by W. J. Hall - Univ. of Rochester Short Answer ONETAIL/T STANDERROROFMEAN TESTOFSIGNIFICAN I650I TTEST PARAMETRIC STATISTICS DESCRSTAT/P CONCEPT T= 5 Application D= 5 Biological Sciences General General ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Multiple Choice ONETAIL/Z ZTEST PARAMETRIC STATISTICS T= 5 Application D= 4 General Education ***Calculator Necessary*** ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Numerical Answer ONETAIL/Z STANDERROROFMEAN I650I ZTEST PARAMETRIC STATISTICS DESCRSTAT/P T=10 Application D= 4 Education ***Calculator Necessary*** ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Numerical Answer ONETAIL/Z ZTEST PARAMETRIC STATISTICS T=10 Application D= 4 General Biological Sciences ***Multiple Parts*** ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Short Answer ONETAIL/Z ZTEST PARAMETRIC STATISTICS T=10 Computation D= 2 General Business ***Calculator Necessary*** ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Short Answer ONETAIL/Z COMMONPITFALLS ZTEST PARAMETRIC STATISTICS MISCELLANEOUS T=10 Application D= 4 Business Natural Sciences General ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Short Answer ONETAIL/Z MEAN ZTEST PARAMETRIC STATISTICS DESCRSTAT/P T=10 Application D= 4 GeneralBack to this chapter's Contents

Based upon item submitted by F. J. Samaniego - UC Davis Multiple Choice TWOTAIL/Z ZTEST PARAMETRIC STATISTICS T= 5 Comprehension Computation D= 3 General ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Short Answer TWOTAIL/Z ZTEST PARAMETRIC STATISTICS T=10 Computation D= 2 General Business ***Statistical Table Necessary***Back to this chapter's Contents

Item is still being reviewed Short Answer TWOTAIL/Z MEAN ZTEST PARAMETRIC STATISTICS DESCRSTAT/P T=10 Application D= 4 General EducationBack to this chapter's Contents

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