Practice Questions for Business Statistics

Warning: This web page document is quite long and has many (intra)connecting links. Do NOT click on any links until the entire document has been loaded by your web browser.

Return to Brian Schott @ GSU

Return to the list of chapters

Chapter: Analysis of Variance Questions

Contents:

98-2 P (X < c) = .95?

105-2 the probability of obtaining an F-ratio exceeding

1278-1 The appropriate statistical test for this comparison is:

1430-1 chicken soup consumption between sexes statistically significant?

1536-1 Set 95% confidence limits for the difference between means

1612-1 F > 9.48773 and the computed value of F from the data is .86.

1614-2 and your P-value satisfies .01 < P < .05.

1650-1 What useful information can you supply future investigators when

1687-4 "In ANOVA, if we wish to investigate the difference among five"

1872-3 Suggest some factors that might influence variability in weights

1873-2 Data containing less variability than would be expected by chance

1895-1 You are to develop plans for a follow-up study.

1917-1 s**2 = (Sum of Squares)/(n-1)

1953-2 The degrees of freedom associated with a error (or within)

1953-3 the degrees of freedom associated with within sum of squares depend

2102-3 which of the following is an estimate of the variance of

2118-1 MEAN SQUARE

2548-1 "If known variation is large compared to unknown variation,"

2550-1 sees that his MS within groups is larger than his MS among groups.

2551-1 What decision would be made regarding H(0):

2552-1 Then we know that the experimenter used

2553-1 Items i. thru iii. are based on a teaching experiment involving

2567-1 what is the number of degrees of freedom for

2570-1 How many degrees of freedom do we have for the within groups sum of

2577-2 F = 3.5 so hypothesis is

2582-1 A fisheries researcher wishes to conclude that there is a difference in

2584-2 "If the sample means for each of the treatment groups were identical,"

2586-1 exceeding that reported in the F table

2588-1 -0.10 to 1.5. Which one of the following statements is true?

2591-1 Samples from 3 classes were given an identical math test.

2593-1 "If you reject, which shelving policies are different? "

2598-1 "the hypothesis that ""mothers are more likely to be success"

2601-1 """The wealthier a person, the more likely he will be relatively"

2603-1 hypothesis that male students are more assertive than female students

2609-1 Complete the ANOVA Table and test if all routes are equally

2612-1 What hypothesis is tested using this F ratio?

2702-1 Complete the following analysis of variance table:

2705-1 what is the value of (corrected) total sums of squares?

2706-1 what is the sum of squares for between groups?

2709-1 Consider each of the pairs of methods.

2719-4 indicates how a particular sum-of-squares value is calculated.

2723-4 "In a one-way ANOVA, the between mean square plus the within mean square"

2771-2 Are you satisfied with this conclusion? Why or why not?

2779-1 but use of an LSD is much less exposed to Type II errors.

2781-2 Which of the following pairs of sample means should NOT be judged

Return to the list of chapters

Return to Brian Schott @ GSU

Questions:

98-2

    Q:  If X has an F distribution with df=4,5, what is the value of c  where
        P (X < c) = .95?

Back to this chapter's Contents

Look at the answer


105-2

    Q:  True or False?  If False, correct it.

        In a one-way classification ANOVA, when the null hypothesis is false,
        the probability of obtaining an F-ratio exceeding that reported in the
        F table at the .05 level of significance is greater than .05.

Back to this chapter's Contents

Look at the answer


1278-1

    Q:  In a study, subjects are randomly assigned to one of three groups:
        control, experimental A, or experimental B.  After treatment, achieve-
        ment test scores for the three groups are compared.  The appropriate
        statistical test for this comparison is:

        a.  the correlation coefficient
        b.  chi square
        c.  the t-test
        d.  the analysis of variance

Back to this chapter's Contents

Look at the answer


1430-1

    Q:  A survey of 114 men and 126 women produced the result that  the  mean
        amount  of chicken soup consumed by the men in a month's time was .67
        liters, compared with a mean of .54 liters for the women.   The  var-
        iance  of  the  chicken  soup consumption for the men was 25% greater
        than that for  the  women.     A  95%  confidence  interval  for  the
        difference  between  the means (men's mean - women's) was found to be
        -.07 to +.33 liters.  At the .05 level, is  the  difference  in  mean
        chicken soup consumption between sexes statistically significant?

        a)  no
        b)  yes
        c)  Can't tell from the data given

Back to this chapter's Contents

Look at the answer


1536-1

    Q:  The report for an investigation includes:

        Means for weight loss by dieters
        Cream puff diet                 0
        Frogurt diet                    5
        Beer and raw eggs diet          7
        Saltines and kind words diet    8

        HSD(.05)                        2

        Set 95% confidence limits for the difference between means for:
        a.  Cream puff diet vs. beer and raw eggs
        b.  Beer and raw eggs vs. saltines and kind words

Back to this chapter's Contents

Look at the answer


1612-1

    Q:  Suppose the critical region for a certain test of hypothesis is of the
        form F > 9.48773 and the computed  value  of F from the data  is  .86.
        (F refers to an F statistic.)  Then:

        a)  H(O) should be rejected.
        b)  H(A) is two-tailed.
        c)  The significance level is given by the area to the right of 9.48773
            under the appropriate F distribution.
        d)  None of these.

Back to this chapter's Contents

Look at the answer


1614-2

    Q:  Suppose H(O): MU(1) = MU(2) and your P-value satisfies .01 < P < .05.
        Which of the following conclusions can be drawn?

        (a)  Fail to Reject H(O) because P is small.
        (b)  Reject H(O) because P is small.
        (c)  Fail to Reject H(O) because P is large.
        (d)  Reject H(O) because P is large.
        (e)  Fail to Reject H(O) at ALPHA = .01.

Back to this chapter's Contents

Look at the answer


1650-1

    Q:  A carefully designed experiment has just been concluded.  Execution of
        the experiment was flawless.  Unfortunately use of ALPHA (.05) indic-
        ated no significant differences among treatments.

        What useful information can you supply future investigators when you
        report on this experiement?

        Indicate agreement with yes or disagreement with no for each of the fol-
        lowing items.

        a.  No useful information.  These are negative results and there is
            nothing useful to report.
        b.  The estimated variance (and its df) can be useful to future invest-
            igators.
        c.  A careful description of experimental conditions and treatments may
            be useful as an indication of circumstances where responses are
            about the same.
        d.  Significant differences can be reported by changing the Type I
            error rate from .05 to .10, .20 or whatever is needed to declare
            significance.

Back to this chapter's Contents

Look at the answer


1687-4

    Q:  True or False?  If False, correct it.

        In ANOVA, if we wish to investigate the difference among five
        means, it is good statistical procedure to perform a t-test on
        each pair of means.

Back to this chapter's Contents

Look at the answer


1872-3

    Q:  Suggest some factors that might influence variability in weights
        measured on a group of men.

Back to this chapter's Contents

Look at the answer


1873-2

    Q:  True or False?  If False, correct it.

        Data containing less variability than would be expected by chance
        are always preferable.

Back to this chapter's Contents

Look at the answer


1895-1

    Q:  A report on an investigation includes the following information
        related to the influence of several mouth washes to length of
        time that breath remains "great".

        Analysis of Variance

        Source                   df       SS          Mean Square

        Treatments                 3         500      166.67
        Error                    196      19,600      100
        Total                    199      20,100

        Means (hours that breath remained "great")

        Whiskey      12
        Brand X      11
        Water         9
        Brand L       8
            F(critical) (.05) = 3.92 with 196 df

        You are to develop plans for a follow-up study.  In particular,
        you are to re-examine the difference between Brand X and Brand L.
        In looking at methods for estimating number of replicates needed
        you find that you need values for -

                1.  the size of the difference to be detected
                2.  the anticipated standard deviation
                3.  the anticipated variance

        On the basis of the above report what values will you use for
        each of the above 3 items?  Why?

Back to this chapter's Contents

Look at the answer


1917-1

    Q:  Given the following observed number of pigs for 8 litters, the numerator
        of the formula for s**2 is called the corrected sum of squares as illus-
        trated.

            s**2 = (Sum of Squares)/(n-1)

        Find the sum of squares.

        X(1) =  9      X(5) = 10
        X(2) =  6      X(6) =  7
        X(3) = 14      X(7) =  8
        X(4) =  9      X(8) =  9

        a)  5      b)  SQRT(5)      c)  40      d)  80

Back to this chapter's Contents

Look at the answer


1953-2

    Q:  True or False?  If False, correct it.

        The degrees of freedom associated with a error (or within) term
        in ANOVA depends only on sample size.

Back to this chapter's Contents

Look at the answer


1953-3

    Q:  True or false? IF FALSE, CORRECT IT.

        In ANOVA the degrees of freedom associated with within sum of squares
        depend on how many independent sample values were used and how many
        other independent parameters have been estimated.

Back to this chapter's Contents

Look at the answer


2102-3

    Q:  In a simple analysis of variance problem, which of the following is an
        estimate  of  the  variance  of  individual  measurements (after the
        various effects have been accounted for)? (MS means SS/df so each of 
        answers is a Mean Square.)

             a.  MS(between)
             b.  MS(within)
             c.  MS(total)
             d.  none of the above

Back to this chapter's Contents

Look at the answer


2118-1

    Q:  Define the following term and give an example of its use.
        Your example should not be one given in class or in a handout.

        MEAN SQUARE

Back to this chapter's Contents

Look at the answer


2548-1

    Q:  The total variation in response, assuming no bias,  is  due  to  error
        (unexplained  variation)  plus  differences  due to treatments  (known
        variation).  If known variation is large compared to unknown variation,
        which of the following conclusions is the best?

        a)  There is no difference in response due to treatments.
        b)  There is a difference in response due to treatments.
        c)  The treatments are not comparable.
        d)  The cause of the response is due to something other than treatments.

Back to this chapter's Contents

Look at the answer


2550-1

    Q:  An investigator randomly assigns 30 college students into three equal
        size study groups (early morning, afternoon, late night) to determine
        if the period of the day at which people study has an effect on their
        retention.   The  students  live  in a controlled environment for one
        week, on the third day of which the experimental treatment (study  of
        predetermined   material)  is  administered.   The  seventh  day  the
        investigator tests for retention, and in computing  his  analysis  he
        sees  that  his  MS within groups is larger than his MS among groups.
        What is the indication of this result?

        a.  An error in calculation was made.
        b.  There was more than the expected variability between groups.
        c.  There was more variability between subjects within the same
            group than there was between groups.
        d.  That there should have been additional controls in the experi-
            ment.

Back to this chapter's Contents

Look at the answer


2551-1

    Q:          Source             SS        df
                ------             --        --
                Between            30.5       4
                Within
                Total             165.0      99

            What decision would be made regarding H(0):  population means are
            equal?
            (a)  Reject H(0) at the .05 level
            (b)  Fail to reject H(0) at the .01 level
            (c)  Insufficient information is given to answer

Back to this chapter's Contents

Look at the answer


2552-1

    Q:  In reading a scientific article you encounter the following table:

                                 Analysis of Variance
        ------------------------------------------------------------------------
        Source                  SS            df            MS             F
        ------------------------------------------------------------------------
        Between samples        722.7           4           180.68         15.3**
        Within samples         473.3          40            11.83
        ------------------------------------------------------------------------
            Total             1196.0          44

        Further reading indicates that all sample sizes are equal.  Then we know
        that the experimenter used
            (a)  4 samples of size 10.
            (b)  5 samples of size 10.
            (c)  4 samples of size 9.
            (d)  5 samples of size 9.
            (e)  None of these

Back to this chapter's Contents

Look at the answer


2553-1

    Q:  NOTE:  Items i. thru iii. are based on a teaching experiment involving
               four elementary statistics classes.  Below are scores for 24
               students who took the same final examination.

        Statistics          Statistics          Statistics           Statistics
         Cookbook           with Humor          Made Useful        in Story Form
        ------------------------------------------------------------------------
            78                  51                  64                   54
            78                  57                  54                   61
            79                  64                  61                   79
            70                  75                  66                   69
            83                  42                  57                   69
            74                  83                  71                   65
        ------------------------------------------------------------------------
           462                 372                 373                  397

        Suppose further that calculations yield SS(total) = 800, and
            SS(between) = 300.

        i.  What is the observed value of the statistic one computes to test
            H(0):  MU(1) = MU(2) = MU(3) = MU(4) against H(1):  not all 4 means
            are equal?
            (a)  2.5       (b)  5.2        (c)  4.0        (d)  11.1
            (e)  None of these

        ii.  If ALPHA = .01, then the critical value of the statistic is
            (a)  3.10      (b)  4.94       (c)  8.66       (d)  26.7
            (e)  None of these

        iii.  Suppose that the observed value of the statistic in Item i. is
            5.6 while the critical value in Item ii. is 7.21.  With only this
            information, which of the following conclusions is most logical?
            (a)  The four populations do not all have the same mean.
            (b)  The four populations have the same mean.
            (c)  The four populations do not all have the same mean.
                 Statistics Cookbook and Statistics in Story Form produce
                 higher means than the other two books.
            (d)  Statistics Cookbook has the highest population mean.
            (e)  The four population means may be different but these samples
                 fail to demonstrate any difference.

Back to this chapter's Contents

Look at the answer


2567-1

    Q:  An experiment was conducted as a oneway random ANOVA design yielding K
        sample means, each based on n scores.  If the between and within mean
        squares are represented by S(m)**2 and S(p)**2, respectively, what is
        the number of degrees of freedom for S(m)**2?

        a.  n - 1
        b.  K - 1
        c.  n - K
        d.  (n - 1)(K - 1)
        e.  none of the above

Back to this chapter's Contents

Look at the answer


2570-1

    Q:  Suppose we have five independent groups (or samples), each of size 9.
        How  many  degrees of freedom do we have for the within groups sum of
        squares in the ANOVA assuming a single factor experiment?

        1.  4
        2.  41
        3.  44
        4.  45
        5.  Not given.

Back to this chapter's Contents

Look at the answer


2577-2

    Q:  Samples of size 5 are taken from 3 populations and the following
        analysis of variance table found.  Test the hypothesis that the
        three populations have the same means.

        Source             d.f.        M.S.        F
        ------             ----        ----        -
        Between means                  350
        Within samples                 100
        Total

        a.  F = 3.5 so hypothesis is rejected at 5% level
        b.  F = 3.5 so hypothesis is not rejected at 5% level
        c.  F = 3.5 so hypothesis is not rejected at 10% level
        d.  F = 3.5 so hypothesis is rejected at 1% level
        e.  Do not have enough information to perform test.

Back to this chapter's Contents

Look at the answer


2582-1

    Q:  A fisheries researcher wishes to conclude that there is a difference in
        mean weights of three species of fish caught in  a large lake near Lin-
        coln, Nebraska.  The data are as follows:   (Use ALPHA = .05.)

                                 SPECIES
                                 -------
            X                       Y                       Z
            1.5                     1.5                     6.0
            4.0   SUM(X) = 13       1.0   SUM(Y) = 9        4.5   SUM(Z) = 20.5
            4.5                     4.5                     4.5
            3.0                     2.0                     5.5

        ANOVA Table (incomplete) :

        Source of Variation        SS        df        MS        F
        Between Groups            17.04       2       8.52     
        Within Groups             14.19       9       1.58
        TOTAL                     31.23      11

           i)  The null hypothesis is:

              a)  H(O):  BETA = 0
              b)  H(O):  MU = 0
              c)  H(O):  MU(X) = MU(Y) = MU(Z)
              d)  H(O):  BETA(X) = BETA(Y)= BETA(Z)

         ii)  The test statistic is:

              a)  t(calc) = 2.52
              b)  t(calc) = 3.09
              c)  F(calc) = 1.20
              d)  F(calc) = 5.40

        iii)  The critical value is:

              a)  t(.05,9) = 2.262
              b)  t(.10,9) = 1.833
              c)  F(.05,2,9) = 4.26
              d)  F(2.5,2.9) = 5.71

         iv)  What is your conclusion?

              a)  Reject H(O) because F(calc) > F(crit), (at least 1 pair
                  has different means).
              b)  Reject H(O) because ]t(calc)] > t(crit), (all means are
                  different).
              c)  Fail to reject H(O) because F(calc) < F(crit), (insuffi-
                  cient evidence that means are different).
              d)  Fail to reject H(O) because ]t(calc)] < t(crit), (means
                  are equal).

Back to this chapter's Contents

Look at the answer


2584-2

    Q:  If the sample means for each of the treatment groups were  identical,
        what would be the observed value of the ANOVA F-ratio?

        a.  1.0
        b.  0.0
        c.  A value between 0.0 and 1.0
        d.  A negative value
        e.  Infinite

Back to this chapter's Contents

Look at the answer


2586-1

    Q:  Assuming a single factor experiment is being analyzed  and  that  the
        null hypothesis being tested by the F-test in the ANOVA is false, the
        probability of obtaining a mean-square ratio exceeding that  reported
        in the F table as the 95th percentile is

        a.  greater than .05.
        b.  less than .05.
        c.  equal to .05.

Back to this chapter's Contents

Look at the answer


2588-1

    Q:  A one-way classification analysis of variance is performed on experi-
        mental data for which there were 10 subjects in each of two groups.
        The .95 confidence interval around the difference YBAR(1) - YBAR(2) is
        -0.10 to 1.5.  Which one of the following statements is true?
            (a)  The F ratio obtained in the analysis of variance was less
                 than 4.41
            (b)  The F ratio obtained in the analysis of variance was greater
                 than 8.28
            (c)  The true difference MU(1) - MU(2) must lie between -0.10 and
                 1.50.
            (d)  The best estimate of MU(1) - MU(2) possible from the results
                 is 1.50.

Back to this chapter's Contents

Look at the answer


2591-1

    Q:  Samples from 3 classes were given an identical math test. The  scores
        are  given  as  follows. Test the hypothesis that the performances of
        the 3 classes are equal. (ALPHA = .05)

                     Class A     Class B      Class C
                     -------     -------     -------
                        5          10           7
                        4           2           7
                        6           1           6
                       10           5           5
                        4           2           7
                        7           7           8
                        3           8          10
                        9           2           1
                       ---         ---        ---
               Total   48          37          51

        Source of variation        SS         df       MS         F
        Between classes            13.583      2      6.7915    
        Within classes (error)    220.75      21     10.5
        Total                     234.333     23

Back to this chapter's Contents

Look at the answer


2593-1

    Q:  To test the hypothesis that shelf placement influences sales, a
        marketing researcher has collected data on sales in a random sample
        of 15 comparable supermarkets with 3 different shelving policies
        for an identical brand of soup.  The data is weekly sales figures
        (in tens of cans).  Perform the appropriate test at the 5% level.
        If you reject, which shelving policies are different? 

                          bottom shelf    middle shelf    top shelf
                              sales           sales         sales
                          ------------    ------------    ---------
                               10              25            10
                                5              20            10
                               10              25            20
                               10              30            20
                               15              50            40

        MEANS FOR SHELF
        TREATMENT     MEAN(SALES)
        MIDDLE         30
        TOP            20
        BOTTOM         10

         ANALYSIS OF VARIANCE (incomplete) :
        ---------------------
        SOURCE OF VARIATION         DF          SS      MEAN SQUARE    F(CALC.)
        SHELF                        2      1000.0000   500.00000       
        EXPERIMENTAL ERROR          12      1200.0000   100.00000
        TOTAL                       14      2200.0000

         HSD FOR ABOVE MEANS IS 16.86 at CONF.LEVEL 95
        (Note:  HSD means Tukey's Honest Significant Difference.)

Back to this chapter's Contents

Look at the answer


2598-1

    Q:  A reseaecher assigns each of his interviewers a list of 7 families,
        drawn randomly from a region, to be interviewed.  Each interviewer
        is instructed to administer a successful parenting scale (SPS) to
        each parent in his sample.  The SPS scores, Y(i), are defined as
        ranging from 0 (no parenting skills deemed successful) to 100
        (successful parenting skills consistently and skillfully applied).

        An interviewer returns with data for both parents.  Use this data to
        test, using classical analysis of variance, at the 90% level of con-
        fidence, the hypothesis that "mothers are more likely to be success-
        ful parents". 

            Mothers    Fathers
             Y(i)       Y(j)
              68         63
              72         48
              48         30
              54         52
              83         55
              92         41
              87         57

         MBAR = 72.00
        FBAR = 49.43

        ANOVA Table (incomplete) :
        Source of Variation          SS          df          MS      F(calc)
        Between Groups             1783.143       1       
        Within Groups              2391.714      12         199.310
        Total                      4174.857      13

Back to this chapter's Contents

Look at the answer


2601-1

    Q:  Use one-way analysis of variance, with an F test, to test the hypothesis
        that "The wealthier a person, the more likely he will be relatively
        politically conservative," at the 90% level of confidence.  Note that,
        for purposes of research, the researcher operationally defined "wealthy"
        as those with an annual income of $50,000, while "poor" subjects re-
        ceived less than $5,000/year.  Note, too, that the "political conser-
        vatism" scale used produced scores of 0 for "extremely liberal", and 100
        for "extremely conservative".  Sample data:

                             X, Income Category
                             Wealthy       Poor
                            +---------+--------+
                            | Y   Y   |    Y   |
                            |    90   |   50   |
                            |    80   |   60   |
                            |    70   |   40   |
                            |    60   |   50   |
                            |    90   |   30   |
                            +---------+--------+
        ANOVA Table (incomplete) :
        Source of variation |   SS   |   df   |   MS   |   F(calc.)
        --------------------+--------+--------+--------+-----------
        Between groups      |   2560 |    1   |        |   
        Within groups       |   1200 |    8   |        |
        Corrected total     |   3760 |    9   |        |

Back to this chapter's Contents

Look at the answer


2603-1

    Q:  A sociologist conducted a study of assertion by having one of her top
        students, after appropriate training, note the number of assertive acts
        performed in a day by each of 10 randomly selected coeds, producing the
        following sample of data, in acts per day:  [5,3,10,6,4,9,5,5,7,5].
        Another sociologist wonders whether the male and female students at
        Bedrock indeed differ in assertiveness and, by a similar procedure,
        gathers the following data for male students, in acts per day:  [8,3,5,
        8,12,10,7,7,9,7].  Use a one-way analysis of variance to test the
        hypothesis that male students are more assertive than female students
        at Bedrock College, at the 90% level of confidence.
        ANOVA table (incomplete) :
        Source of variation  |   SS   |   df   |   MS   |  F(calc.)
        ---------------------+--------+--------+--------+----------
        Between Groups       |   14.45|    1   |        |  
        Within Groups        |   99.30|   18   |        |
        Total                |  113.75|   19   |        |

Back to this chapter's Contents

Look at the answer


2609-1

    Q:  Mr. Martin can drive to work along four different routes, and
        the following are the number of minutes in which he timed him-
        self on five different occasions for each route:

                Route 1    Route 2    Route 3    Route 4
                ________________________________________
                  22         25         26         26
                  26         27         29         28
                  25         28         33         27
                  25         26         30         30
                  31         29         33         30
                ________________________________________
        T.(j) =  129        135        151        141

        ANOVA Table  (incomplete) 
        Source       df       SS       M.Sq.      F ratio
        Routes       .3     52.8      
         Error       16    100.4       
         Total       19    153.2
        Complete the ANOVA Table and test if all routes are equally
        fast (ALPHA = 5%).

Back to this chapter's Contents

Look at the answer


2612-1

    Q:  An imaginary study has been conducted on the effects of three brands of
        laxatives on regularity of TV actresses where each brand was tested by
        one actress belonging to each of 10 age groups.  Results obtained
        included:  F= (brand M.Sq.)/(Error M.Sq.) = 2.1 with 2 and 18 df.

        a.  What hypothesis is tested using this F ratio?
        b.  Interpret these results using a significance level of 5%.

Back to this chapter's Contents

Look at the answer


2702-1

    Q:  Samples of size 11 are taken from each of 5 populations.  Complete
        the following analysis of variance table:

        Source          S.S.    d.f.    M.S.    F
        ------          ----    ----    ----    -

        Between means   1000     a       c      e
        Within samples  5000     b       d
        Total           6000

        a.  a = 4  b = 44  c = 250  d = 113.6  e = 2.2
        b.  a = 4  b = 44  c = 250  d = 113.6  e = 0.2
        c.  a = 5  b = 55  c = 200  d =  90.9  e = 0.2
        d.  a = 5  b = 50  c = 200  d = 100    e = 2.0
        e.  a = 4  b = 50  c = 250  d = 100    e = 2.5

Back to this chapter's Contents

Look at the answer


2705-1

    Q:  In a single factor experiment with four levels  if  the  mean  square
        (between)=25,  mean  square(within)=10, n(1)=n(2)=n(3)=8 and n(4)=10,
        what is the value of (corrected) total sums of squares?

        a.  435
        b.  786
        c.  1221
        d.  Insufficient information
        e.  Sufficient information but correct value is not given

Back to this chapter's Contents

Look at the answer


2706-1

    Q:  In  the ANOVA for a single factor experiment with four levels all n's
        equal 5 and YBAR(1)=22, YBAR(2)=24, YBAR(3)=20, and YBAR(4)=18.  What
        is the sum of squares for between groups?

        a.  25.00
        b.  33.33
        c.  100.00
        d.  Cannot be determined without more data

Back to this chapter's Contents

Look at the answer


2709-1

    Q:  Nine children were randomly split into three groups of three each.  In
        a spelling unit, individuals in one group were criticized each time they
        misspelled a word.  The individuals in another group were praised each
        time they correctly spelled a word, while the individuals in the third
        group were neither praised or criticized.  At the end of the unit, each
        child was given ten words to spell with the following results (number
        correct is given for each child):

            Praised    Neutral    Criticized
               8          3            9
               9          2           10
               7          5            7
        MEANS FOR RESPONSE
        TREATMENT    MEAN(WORDS CORRECT)
        CRITICIZED   8.66667
        PRAISED      8
        NEUTRAL      3.33333

        ANALYSIS OF VARIANCE:
        --------------------
        SOURCE OF VARIATION        DF          SS        MEAN SQUARE
        RESPONSE                    2        50.6667     25.33333
        EXPERIMENTAL ERROR          6        11.3333      1.88889
        TOTAL                       8        62.0000

         PROBABILITY LEVEL FOR COMPARING MEANS = 0.05
        HSD FOR ABOVE MEANS IS 3.444 AT PROB.LEVEL 0.05

         Is there any evidence for significant differences among the methods?
        Consider each of the pairs of methods.  Are there significant differ-
        ences within any of the pairs?

Back to this chapter's Contents

Look at the answer


2719-4

    Q:  NOTE:  Items i. thru iv. are based on the following situation.  Each
               item indicates how a particular sum-of-squares value is
               calculated.  Use the following code in responding to each item.

               (a)  Between
               (b)  Within
               (c)  Total (corrected)
               (d)  Either (a) or (b)
               (e)  None of these

        i.    Squaring the deviations of each score from its sample (or group)
              mean.
        ii.   Squaring the deviations of each score from the overall mean.
        iii.  Squaring the deviations of each sample mean from the overall mean.
        iv.   Squaring the deviations of one sample mean from all other sample
              means.

Back to this chapter's Contents

Look at the answer


2723-4

    Q:  True or False?  If False, correct it.

        In a one-way ANOVA, the between mean square plus the within mean square
        must equal the total mean square if computations are correct.

Back to this chapter's Contents

Look at the answer


2771-2

    Q:  A physiologist reports an investigation of potential plant hormones.  He
        reports the following averages for lengths of 20 stem segments treated.

        Compound X      1.18
        Compound Y      1.17
        Compound Z      1.15
        Control         1.14

        The conclusion is that there are no treatment differences.  Are you
        satisfied with this conclusion?  Why or why not?

Back to this chapter's Contents

Look at the answer


2779-1

    Q:  When various methods for comparing means are discussed two methods
        that are often cited as extremes are -
        a.  LSD for which the Type I error rate is ALPHA per comparison
        b.  Tukey's HSD, for which the Type I error is ALPHA per experiment
            regardless of the number of comparisons made.

        If there are many comparisons to be made in an experiment, use of
        an LSD is much more exposed to Type I error than use of Tukey's
        method, but use of an LSD is much less exposed to Type II errors.

        In each of the items below indicate whether or not you would use
        an LSD to compare means.  Explain your choice.

        a.  An experiment involves 10 treatments.  If you commit a Type I
            error, there will be little immediate loss.  If you commit a
            Type II error there will be somewhat greater losses.

        b.  An experiment involves 2 treatments.  If you commit a Type I
            error, there will be appreciable loss.  If you commit a Type
            II error, there will be little loss.

        c.  An experiment involves 20 treatments.  If you commit a Type I
            error, there could be widespread injury (e.g. loss of life).
            If you commit a Type II error, there will be bothersome
            losses (loss of investment in developing treatments), but
            less loss than from Type I error.

Back to this chapter's Contents

Look at the answer


2781-2

    Q:  A F-ratio signficant at the .05 level was found in a one-factor ANOVA
        in which there are 3 levels and n=25.  A set of simultaneous confidence
        intervals with confidence coefficient .95 is established around
        differences between pairs of means using Tukey's HSD method.  The
        following intervals are obtained:

            YBAR(1) - YBAR(2):  (-7.50, -7.00)
            YBAR(1) - YBAR(3):  (8.50, 8.75)
            YBAR(2) - YBAR(3):  (-.80, 1.50)

        Which of the following pairs of sample means should NOT be judged to
        be significantly different?
            (a)  YBAR(1) - YBAR(2)
            (b)  YBAR(1) - YBAR(3)
            (c)  YBAR(2) - YBAR(3)
            (d)  None of the above

Back to this chapter's Contents

Look at the answer


Return to the list of chapters

Return to Brian Schott @ GSU

Answers:

98-2

    A:  F(95th percentile, df=4,5) = 5.19

Back to review this question

Look at this question's identification

Back to this chapter's Contents


105-2

    A:  True

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1278-1

    A:  d.  the analysis of variance

            Of the four tests given, the analysis of variance is the only
            exact test that can be performed.  The t-test does not compare
            more than two means and chi square is not exact.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1430-1

    A:  a)  no

            The confidence interval includes zero.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1536-1

    A:  a.  (7-0) +/- HSD
            7 +/- 2
            Interval is from 5 to 9 (significant difference)
        b.  (8-7) +/- 2
            Interval is from -1 to +3 (not a significant difference)

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1612-1

    A:  c)  The significance level is given by the area to the right of 9.48773
            under the appropriate F distribution.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1614-2

    A:  (e)  Fail to Reject H(O) at ALPHA = .01.

             Since P > .01 you cannot reject H(O) at ALPHA = .01.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1650-1

    A:  a.  no
        b.  yes, a variance estimate is needed in order to estimate sample size
            required when someone else investigates this phenomenon in the
            future.
        c.  yes, if circumstances that produce no response aren't reported, the
            reports on record sometimes will give the misleading impression that
            there generally is a response.
        d.  no, Type I error rate is to be chosen before the investigation
            and is not to be changed in order to declare significance.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1687-4

    A:  False, good statistical procedure would use an F-test and Tukey's
        test to avoid the problem of increased type I error in the
        multiple comparisons.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1872-3

    A:  Factors that might influence variability in weights include:

            variability in height, age, level of activity,
            diet, and random error.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1873-2

    A:  False - When data contains less variability than would be expected by
        chance, it is time to be suspicious of the techniques or procedures
        used in collecting the data. But business managers are typically con-
        servative and prefer data with less variability.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1895-1

    A:  1.  The estimated difference between Brands X and L in this trial was 3.
            Since a follow up study has been commissioned, it would seem that a
            difference at least this small is to be detected.

        2.  The mean square for experimental error gives us an estimate of
            the variance and an estimate of the standard deviation would be
            the square root of that, so SQRT(100) = 10.

        3.  As explained above, it is 100.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1917-1

    A:  c)  40

            SUM(X) = 72
            XBAR = 72/8 = 9

            Corrected Sum of Squares = (0**2) + (-3**2) + (5**2) + (0**2) +
                                       (1**2) + (-2**2) + (-1**2) + (0**2)
                                     = 9 + 25 + 1 + 4 + 1
                                     = 40

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1953-2

    A:  False.  It also depends on how many independent parameters (or treatment
        level means) are being estimated.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


1953-3

    A:  True.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2102-3

    A:  b.  MS(within)

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2118-1

    A:  Definition:  A sum of squares divided by its degrees of freedom.

        Example:     If an ANOVA table includes:
                        Source                df      SS
                        Experimental Error    12      3600,
                     the mean square for experimental error is
                     300 = 3600/12.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2548-1

    A:  b)  There is a difference in response due to treatments.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2550-1

    A:  c.  There was more variability between subjects within the same
            group than there was between groups.

        This is precisely the null hypothesis of the F-test in this analysis.
        Since the MS(within) is larger than MS(among), this F-test will  have
        a  calculated  value  less  than  one.   Based on this result we will
        undoubtedly not reject the null hypothesis.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2551-1

    A:      (a)  Reject H(0) at the .05 level

            Completed ANOVA table:

            Source            SS       df       MS
            ------            --       --       --
            Between           30.5      4       7.625
            Within           134.5     95       1.416
            Total            165.0     99

            F(calc.) = 7.625 / 1.416
                     = 5.385

            F(crit., ALPHA=.05, df=4,95, one-tail) = 2.47
            F(crit., ALPHA=.05, df=4,95, one-tail) = 3.59


            H(0) would be rejected at both ALPHA levels

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2552-1

    A:  (d)  5 samples of size 9.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2553-1

    A:  i.  (c)  4.0

            ANOVA Table:

            Source            SS        df        MS       F
            ------            --        --        --       -
            Between samples  300         3       100       4.00
            Within samples   500        20        25
            Total (corrected)800        23

        ii.  (b)  4.94

            F(crit., ALPHA=.01, df=3,20, one-tail) = 4.94

        iii.  (e)  The four population means may be different but these samples
            fail to demonstrate any difference since F(obs.) < F(crit.).

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2567-1

    A:  b.  K - 1

              Total Categories = K
            Degrees of Freedom = K - 1

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2570-1

    A:  5.  Not given.

            df(within) = (total # of observations) - 1 - df(between)
                       = (5*9) - 1 - 4
                       = 45-5
                       = 40

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2577-2

    A:  b.  F = 3.5 so hypothesis is continued at 5% level.

            Completed ANOVA table:

            Source            d.f.      M.S.       F
            ------            ----      ----       -
            Between samples    2        350       3.5
            Within samples    12        100
            Total             14

            F(critical, df = 2, 12, ALPHA = .05) = 3.89

            Therefore the hypothesis should not be rejected.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2582-1

    A:    i)  c)  H(O):  MU(X) = MU(Y) = MU(Z)
         ii)  d)  F(calc) = 5.40
        iii)  c)  F(.05,2,9) = 4.26
         iv)  a)  Reject H(O) because F(calc) > F(crit), (at least 1 pair
                  has different means).
        ANOVA Table:

        Source of Variation        SS        df        MS        F
        Between Groups            17.04       2       8.52      5.40
        Within Groups             14.19       9       1.58
        TOTAL                     31.23      11

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2584-2

    A:  b.  0.0  because Sum of Squares between would be zero and the F-ratio
            is:

                 F(calc.) = [SS(between)/df(between)]/[SS(within)/df(within)]

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2586-1

    A:  a.  greater than .05.

            The null hypothesis being tested is:

               H(O):  MU(1) = M(2) = ... = MU(K)

            where K is the number of experimental groups or samples.  But if
            this is false, then the K population means are not equal and the
            between groups mean square will be greater than the within groups
            mean square.  So the sampling distribution of this ratio will be
            centered further to the right than the F-distribution associated
            with the null hypothesis.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2588-1

    A:  (a)  The F ratio obtained in the analysis of variance was less than 4.41

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2591-1

    A:  H(O):  MU(1)  =  MU(2)  =  MU(3)
        H(A):  MU(1) =/= MU(2) =/= MU(3)

        F(calculated) = MS for classes/MS for error
                      = 6.7915/10.5
                      = .646

        F(critical, df=2, 21, ALPHA =.05) = 3.47

        Since F(calculated) < F(critical) do not reject H(O). Therefore
        performances of the 3 classes are not found to be different.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2593-1

    A:  PROBABILITY LEVEL FOR COMPARING MEANS = .05
        F(critical, df=2,12, ALPHA=.05) = 3.88

        Therefore, reject the null hypothesis that shelving policy does not
        influence sales.

        Based on the HSD given above, it appears that there is a significant
        difference between the middle shelf and the bottom shelf.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2598-1

    A:  n(M) = n(F) = 7

        H(O):  MU(females)  =  MU(males)
        H(A):  MU(females) =/= MU(males)

        ANOVA Table:

        Source of Variation          SS          df          MS      F(calc)
        Between Groups             1783.143       1        1783.143   8.947
        Within Groups              2391.714      12         199.310
        Total                      4174.857      13

        F(crit., ALPHA=.10, one-tail, df=1,12) = 3.18

        Since F(calc) > F(crit), H(O) should be rejected.  Thus, based on this
        sample evidence, it appears that the means for mothers and fathers are
        different and the indication from the sample means is that mothers are
        more likely to be successful parents.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2601-1

    A:  H(0):  MU(Wealthy)  =  MU(Poor)
        H(A):  MU(Wealthy) =/= MU(Poor)

        ANOVA Table:
        Source of variation |   SS   |   df   |   MS   |   F(calc.)
        --------------------+--------+--------+--------+-----------
        Between groups      |   2560 |    1   |  2560  |    17.1
        Within groups       |   1200 |    8   |   150  |
        Total               |   3760 |    9   |        |

        F(crit., df=1,8, ALPHA=.10) = 3.46

        Since F(calc.) > F(crit.) reject H(0) and based on the fact that wealthy
        people have a higher mean income assume that the wealthier a person, the
        more likely he will be relatively politically conservative.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2603-1

    A:  H(0):  MU(Males)  =  MU(Females)
        H(A):  MU(Males) =/= MU(Females)

        ANOVA table:
        Source of variation  |   SS   |   df   |   MS   |  F(calc.)
        ---------------------+--------+--------+--------+----------
        Between Groups       |   14.45|    1   |  14.45 |   2.62
        Within Groups        |   99.30|   18   |   5.52 |
        Total                |  113.75|   19   |        |

        F(crit., df=1,18, ALPHA=.10) = 3.01

        Since F(calc.) < F(crit.), do not reject H(0).  Therefore, conclude that
        there is no difference between men and women at Bedrock College.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2609-1

    A:  H(0): The means for the routes are all equal.
        H(A): The means for the routes are not all equal.

        ANOVA Table
        Source       df       SS       M.Sq.      F ratio
        Routes       .3     52.8       17.6        2.805
         Error       16    100.4        6.275
         Total       19    153.2

        F(ALPHA = .05, df = 3,16) = 3.24

        Since the F ratio is less than 3.24, we continue the null
        hypothesis that all routes are equally fast.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2612-1

    A:  a.  The null hypothesis tested is that all brand effects equal zero,
            H(O):  MU(1) = MU(2) = MU(3).
        b.  F(calculated) = 2.1
            F(critical, df = 2.18, ALPHA = .05) = 3.55

        Since the calculated F value is smaller than the critical F value, we
        do not reject the null hypothesis that all brand effects are equal.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2702-1

    A:  e.  a = 4  b = 50  c = 250  d = 100    e = 2.5

            a = number of treatments - 1 = 5 - 1 = 4

            b = (a+1)*(number of replications - 1) = (4+1)*(11-1) = 50

            c = 1000/a = 250

            d = 5000/b = 100

            e = c/d = 250/100 = 2.5

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2705-1

    A:  e.  Sufficient information but correct value is not given

            Known parts of ANOVA table are:

             Source of Variation          df         SS         MS
             -------------------          --         --         --
             Between (Treatment)           3                    25
             Within                       30                    10
             (Corrected) Total            33

             Therefore between Sum of Squares is:

                 SS(B) = mean square(between) * df(between)
                       = (25*3)
                       = 75

             Within Sum of Squares is:

                 SS(W) = mean square(within) * df(within)
                       = (10*30)
                       = 300

             Corrected Total SS is:

                 SS(T) = SS(B) + SS(W)
                       = 75+300
                       = 375

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2706-1

    A:  c.  100.00

            Overall Mean = (22+24+20+18)/4
                         = 84/4
                         = 21

            SS(between) = [5*((22-21)**2)]+[5*((24-21)**2)]+[5*((20-21)**2)]
                          +[5*((18-21)**2)]
                        = [5*1]+[5*9]+[5*1]+[5*9]
                        = 5 + 45 + 5 + 45
                        = 100.00

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2709-1

    A:  F(calculated) = (25.3/1.9)
                      = 13.40

        --------------------------------------------------------------
        H(O):  No difference between treatments
        H(A):  Not all treatment means are the same

        Critical Region:  F(df=2,6) > 5.14

        Reject H(O), and conclude that amount and/or kind of praise
        makes a difference in spelling ability.

        Using the HSD and the treatment means given above, we see that
        "criticizing vs. neutral" are significantly different and "prais-
        ing vs. neutral" are significantly different.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2719-4

    A:  i.    (b)  Within
        ii.   (c)  Total (corrected)
        iii.  (a)  Between
        iv.   (e)  None of these

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2723-4

    A:  False, in a one-way ANOVA, the between SS plus the within SS must equal
        the total SS if computations are correct.

        SS(TOTAL) = SS(Between) + SS(Within)

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2771-2

    A:  On the basis of the information given I am not satisfied with this con-
        clusion.  No indication of an estimate for the background variation has
        been given, nor any indication of a basis for comparing these means,
        such as an HSD with a specified significance level.
        
        Due to the lack of information reported, there is no basis for making
        a conclusion one way or another.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2779-1

    A:  a.  I would use an LSD because I want to protect against Type II
            error, not Type I errors.

        b.  I could use either method in this case because when there
            are only two treatments, the two methods are identical.

        c.  I would not use an LSD in this case, I would use Tukey's test.
            Tukey's method protects us from Type I errors when more than
            one comparison is to be made.

Back to review this question

Look at this question's identification

Back to this chapter's Contents


2781-2

    A:  (c)  YBAR(2) - YBAR(3)

Back to review this question

Look at this question's identification

Back to this chapter's Contents


Return to the list of chapters

Return to Brian Schott @ GSU

Identification:

98-2

Item is still being reviewed
        Multiple Choice
FDISTRIBUTION
        PROBDISTRIBUTION  PROBABILITY

T= 2    Comprehension
D= 2    General             Education
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


105-2

Based upon item submitted by R. Pruzek - SUNY at Albany
        True/False
FDISTRIBUTION     TESTING           FTEST
        PROBDISTRIBUTION  PROBABILITY       CONCEPT
        STATISTICS        ANOVA             PARAMETRIC

T= 2    Comprehension
D= 3    General

Back to review this question

Back to this chapter's Contents


1278-1

Based upon item submitted by R. Shavelson - UCLA
        Multiple Choice
ONEWAY/CR         TESTING
        DESIGN            ANOVA             SIMPLE/COR
        COMPLETELYRANDOM  PARAMETRIC        STATISTICS
        CONCEPT           CORRELATION/P

T= 5    Comprehension
D= 5    General

Back to review this question

Back to this chapter's Contents


1430-1

Based upon item submitted by R. L. Stout & R. M. Paolino - Brown
        Multiple Choice
SIMPLE/CI         TESTOFSIGNIFICAN
        TESTING           CONFIDENCEINTERV  ESTIMATION
        CONCEPT           STATISTICS

T= 5    Comprehension
D= 5    General             Biological Sciences Sociology

Back to review this question

Back to this chapter's Contents


1536-1

Based upon item submitted by J. Warren - UNH
        Numerical Answer
OTHER/CI          LSD
        I650C             CONFIDENCEINTERV  ESTIMATION
        CONCEPT           STATISTICS        APOSTERIORI
        MULTIPLECOMPARIS  PARAMETRIC

T= 5    Computation     Comprehension
D= 2    General
                ***Multiple Parts***

Back to review this question

Back to this chapter's Contents


1612-1

Based upon item submitted by D. Kleinbaum - Univ of North Carolina
        Multiple Choice
TESTOFSIGNIFICAN  FTEST
        FDISTRIBUTION     CONCEPT           STATISTICS
        ANOVA             PARAMETRIC        PROBDISTRIBUTION
        PROBABILITY

T= 2    Comprehension
D= 5    General

Back to review this question

Back to this chapter's Contents


1614-2

Based upon item submitted by D. Kleinbaum - Univ of North Carolina
        Multiple Choice
TESTOFSIGNIFICAN
        CONCEPT           STATISTICS

T= 2    Comprehension
D= 3    General

Back to review this question

Back to this chapter's Contents


1650-1

Based upon item submitted by J. Warren - UNH
        Short Answer
HSD               TESTOFSIGNIFICAN
        TYPICALSUMMARY    I650C             APOSTERIORI
        MULTIPLECOMPARIS  PARAMETRIC        STATISTICS
        CONCEPT

T= 5    Comprehension
D= 4    General
                ***Multiple Parts***

Back to review this question

Back to this chapter's Contents


1687-4

Item is still being reviewed
        True/False
TTEST             FTEST             TYPE1ERROR
        NEWMANKEULS       DUNCANS           TUKEYSHSD
        PARAMETRIC        STATISTICS        ANOVA
        CONCEPT           APOSTERIORI       MULTIPLECOMPARIS

T= 5    Comprehension
D= 2    General

Back to review this question

Back to this chapter's Contents


1872-3

Item is still being reviewed
        Short Answer
VARIABILITY/P
        DESCRSTAT/P       PARAMETRIC        STATISTICS

T= 5    Comprehension
D= 3    General             Biological Sciences

Back to review this question

Back to this chapter's Contents


1873-2

Based upon item submitted by W. J. Hall - Univ. of Rochester
        True/False
VARIABILITY/P     RANDOMVARIATION
        DESCRSTAT/P       PARAMETRIC        STATISTICS
        SAMPLINGERROR     SAMPLING

T= 5    Comprehension
D= 4    General

Back to review this question

Back to this chapter's Contents


1895-1

Item is still being reviewed
        Short Answer
VARIANCE          STANDARDDEVIATIO  OTHER/AN
        I650C             DESCRSTAT/P       PARAMETRIC
        STATISTICS        ANOVA

T=10    Comprehension
D= 4    General
                ***Multiple Parts***

Back to review this question

Back to this chapter's Contents


1917-1

Item is still being reviewed
        Multiple Choice
VARIANCE          SUMSOFSQUARES
        DESCRSTAT/P       PARAMETRIC        STATISTICS
        ANOVA

T= 2    Computation
D= 2    General

Back to review this question

Back to this chapter's Contents


1953-2

Based upon item submitted by J. Warren - UNH
        True/False
VARIANCE          DEGREESOFFREEDOM
        I650I             DESCRSTAT/P       PARAMETRIC
        STATISTICS        ANOVA

T= 2    Comprehension
D= 3    General

Back to review this question

Back to this chapter's Contents


1953-3

Based upon item submitted by J. Warren - UNH
        True/False
VARIANCE          DEGREESOFFREEDOM
        SAMPLESIZE        I650I             DESCRSTAT/P
        PARAMETRIC        STATISTICS        ANOVA
        SAMPLING

T= 2    Comprehension
D= 2    General

Back to review this question

Back to this chapter's Contents


2102-3

Item is still being reviewed
        Multiple Choice
VARIANCE/OTHER    SUMSOFSQUARES
        COMPLETELYRANDOM  DESCRSTAT/P       PARAMETRIC
        STATISTICS        ANOVA

T= 2    Comprehension
D= 4    General

Back to review this question

Back to this chapter's Contents


2118-1

Based upon item submitted by J. Warren - UNH
        Definition
VARIANCE/OTHER    EXPERDESIGN/TERM
        DESCRSTAT/P       PARAMETRIC        STATISTICS
        I650C             ANOVA

T= 5    Comprehension
D= 3    General

Back to review this question

Back to this chapter's Contents


2548-1

Item is still being reviewed
        Multiple Choice
FTEST             OTHER/AN
        ANOVA             PARAMETRIC        STATISTICS

T= 2    Comprehension
D= 3    General

Back to review this question

Back to this chapter's Contents


2550-1

Item is still being reviewed
        Multiple Choice
OTHER/AN          FTEST
        ONEWAY/CR         ANOVA             PARAMETRIC
        STATISTICS        COMPLETELYRANDOM

T= 2    Comprehension
D= 2    General             Education

Back to review this question

Back to this chapter's Contents


2551-1

Item is still being reviewed
        Multiple Choice
OTHER/AN          SUMSOFSQUARES     ONEWAY/CR
        ANOVA             PARAMETRIC        STATISTICS
        COMPLETELYRANDOM

T= 3    Application
D= 3    General
                ***Calculator Necessary***

Back to review this question

Back to this chapter's Contents


2552-1

Item is still being reviewed
        Multiple Choice
ONEWAY/CR         OTHER/AN
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS

T= 2    Comprehension   Computation
D= 3    General

Back to review this question

Back to this chapter's Contents


2553-1

Item is still being reviewed
        Multiple Choice
OTHER/AN          ONEWAY/CR         FTEST
        ANOVA             PARAMETRIC        STATISTICS
        COMPLETELYRANDOM

T= 6    Application     Comprehension
D= 4    General             Education
                ***Multiple Parts***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2567-1

Based upon item submitted by W. J. Hall - Univ. of Rochester
        Multiple Choice
DEGREESOFFREEDOM  ONEWAY/CR
        DESIGN            ANOVA             PARAMETRIC
        STATISTICS        COMPLETELYRANDOM  CONCEPT

T= 2    Comprehension
D= 2    General

Back to review this question

Back to this chapter's Contents


2570-1

Item is still being reviewed
        Multiple Choice
DEGREESOFFREEDOM
        ANOVA             PARAMETRIC        STATISTICS

T= 2    Computation
D= 4    General             Education

Back to review this question

Back to this chapter's Contents


2577-2

Item is still being reviewed
        Multiple Choice
FTEST             ONEWAY/CR
        DEGREESOFFREEDOM  ANOVA             PARAMETRIC
        STATISTICS        COMPLETELYRANDOM

T= 5    Computation
D= 5    General
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2582-1

Item is still being reviewed
        Multiple Choice
FTEST             ONEWAY/CR
        ANOVA             PARAMETRIC        STATISTICS
        COMPLETELYRANDOM

T= 5    Application
D= 4    Biological Sciences
                ***Multiple Parts***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2584-2

Item is still being reviewed
        Multiple Choice
FTEST
        ANOVA             PARAMETRIC        STATISTICS

T= 2    Comprehension
D= 3    General             Education

Back to review this question

Back to this chapter's Contents


2586-1

Item is still being reviewed
        Multiple Choice
FTEST
        ANOVA             PARAMETRIC        STATISTICS

T= 2    Comprehension
D= 2    General             Education

Back to review this question

Back to this chapter's Contents


2588-1

Item is still being reviewed
        Multiple Choice
FTEST             ONEWAY/CR
        ANOVA             PARAMETRIC        STATISTICS
        COMPLETELYRANDOM

T= 2    Comprehension   Computation
D= 3    General

Back to review this question

Back to this chapter's Contents


2591-1

Item is still being reviewed
        Numerical Answer
ONEWAY/CR         FTEST
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS

T= 5    Computation
D= 5    General
                ***Calculator Necessary***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2593-1

Item is still being reviewed
        Numerical Answer
FTEST             ONEWAY/CR         LSD
        ANOVA             PARAMETRIC        STATISTICS
        COMPLETELYRANDOM  APOSTERIORI       MULTIPLECOMPARIS

T= 6    Application
D= 4    General             Business
                ***Calculator Necessary***
                ***Multiple Parts***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2598-1

Item is still being reviewed
        Numerical Answer
ONEWAY/CR         FTEST
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS

T= 5    Computation
D= 3    General             Social Sciences     Sociology
                ***Calculator Necessary***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2601-1

Item is still being reviewed
        Numerical Answer
ONEWAY/CR         FTEST
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS

T= 5    Computation
D= 3    Social Sciences     Sociology           General
                ***Calculator Necessary***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2603-1

Item is still being reviewed
        Numerical Answer
ONEWAY/CR         FTEST
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS

T= 5	    Computation
D= 3    Sociology           General
                ***Calculator Necessary***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2609-1

Based upon item submitted by A. Bugbee - UNH
        Short Answer
ONEWAY/CR         FTEST
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS

T= 5    Computation     Comprehension   Application
D= 3    General
                ***Calculator Necessary***
                ***Multiple Parts***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2612-1

Based upon item submitted by J. Warren - UNH
        Short Answer
FTEST             FRATIO
        I650C             ANOVA             PARAMETRIC
        STATISTICS        APRIORI           MULTIPLECOMPARIS

T= 5    Computation     Comprehension
D= 3    General
                ***Multiple Parts***
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2702-1

Item is still being reviewed
        Multiple Choice
ONEWAY/CR
        FTEST             DEGREESOFFREEDOM  SUMSOFSQUARES
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS

T= 5    Computation
D= 5    General

Back to review this question

Back to this chapter's Contents


2705-1

Item is still being reviewed
        Multiple Choice
SUMSOFSQUARES     ONEWAY/CR
        ANOVA             PARAMETRIC        STATISTICS
        COMPLETELYRANDOM

T= 5    Comprehension   Computation
D= 5    General             Education

Back to review this question

Back to this chapter's Contents


2706-1

Item is still being reviewed
        Multiple Choice
SUMSOFSQUARES     ONEWAY/CR
        ANOVA             PARAMETRIC        STATISTICS
        COMPLETELYRANDOM

T= 5    Comprehension   Computation
D= 6    General             Education

Back to review this question

Back to this chapter's Contents


2709-1

Based upon item submitted by D. Zahn - Florida State U.
        Essay
ONEWAY/CR         HSD
        COMPLETELYRANDOM  ANOVA             PARAMETRIC
        STATISTICS        APOSTERIORI       MULTIPLECOMPARIS

T= 5    Application
D= 5    General             Education
                ***Statistical Table Necessary***

Back to review this question

Back to this chapter's Contents


2719-4

Item is still being reviewed
        Multiple Choice
SUMSOFSQUARES
        ANOVA             PARAMETRIC        STATISTICS

T= 5    Comprehension
D= 3    General
                ***Multiple Parts***

Back to review this question

Back to this chapter's Contents


2723-4

Based upon item submitted by R. Pruzek - SUNY at Albany
        True/False
SUMSOFSQUARES
        ANOVA             PARAMETRIC        STATISTICS

T= 2    Comprehension
D= 2    General

Back to review this question

Back to this chapter's Contents


2771-2

Item is still being reviewed
        Short Answer
MULTIPLECOMPARIS
        TESTOFSIGNIFICAN  I650C             PARAMETRIC
        STATISTICS        CONCEPT

T= 5    Comprehension
D= 3    General             Biological Sciences

Back to review this question

Back to this chapter's Contents


2779-1

Based upon item submitted by J. Warren - UNH
        Short Answer
HSD               TUKEYSHSD
        TYPE1ERROR        TYPE2ERROR        I650C
        APOSTERIORI       MULTIPLECOMPARIS  PARAMETRIC
        STATISTICS        CONCEPT

T=10    Comprehension
D= 4    General
                ***Multiple Parts***

Back to review this question

Back to this chapter's Contents


2781-2

Item is still being reviewed
        Multiple Choice
TUKEYSHSD
        APOSTERIORI       MULTIPLECOMPARIS  PARAMETRIC
        STATISTICS

T= 2    Computation
D= 3    General

Back to review this question

Back to this chapter's Contents


Return to the list of chapters

Return to Brian Schott @ GSU