1) p. 641, problem 5

Hypothesized proportions

1    0.033333

2    0.066667

3    0.100000

4    0.133333

5    0.166667

6    0.166667

7    0.133333

8    0.100000

9    0.066667

10   0.033333

 

H0: pi = (5.5-|i - 5.5|)/30

Ha: at least one pi differs from this value.

 

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: obs counts

 

Using category names in location

 

                          Test            Contribution

Category  Observed  Proportion  Expected     to Chi-Sq

1                4    0.033333    6.6667       1.06667

2               15    0.066667   13.3333       0.20833

3               23    0.100000   20.0000       0.45000

4               25    0.133333   26.6667       0.10417

5               38    0.166667   33.3333       0.65333

6               31    0.166667   33.3333       0.16333

7               32    0.133333   26.6667       1.06667

8               14    0.100000   20.0000       1.80000

9               10    0.066667   13.3333       0.83333

10               8    0.033333    6.6667       0.26667

 

  N  DF  Chi-Sq  P-Value

200   9  6.6125    0.677

 

Since the p-value > .10, we fail to reject H0.

 

There is no evidence that the data is not consistent with the previously determined proportions.

 

2) p. 662, problem 32

First, looking at the data through segmented bar graphs:

We see that in the sample conservatives tend to use marijuana less frequently (5% of conservatives in frequently category) and “other” more frequently (28% of others in frequently category). To see if these differences are statistically significant, we can run a chi-square test.

 

H0: no relationship between marijuana usage and political views in the population.

Ha: there is a relationship. 

 

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

 

        never  rarely  frequently  Total

    1     479     173         119    771

       494.38  151.46      125.17

        0.478   3.064       0.304

 

    2     214      47          15    276

       176.98   54.22       44.81

        7.746   0.961      19.828

 

    3     172      45          85    302

       193.65   59.33       49.03

        2.420   3.459      26.394

 

Total     865     265         219   1349

 

Chi-Sq = 64.654, DF = 4, P-Value = 0.000

 

From Minitab, all expected cell counts exceed 5 (smallest = 44.81).

 

The chi-square value is 64.654 with df=4 and p-value =.000. The p-value is < .01, so we reject the null hypothesis.

 

There is strong evidence of a relationship between marijuana usage and political views for this population of high school and college students.  Since this was an observational study, no causal conclusions can be drawn.

 

The largest contributions to the chi-square sum come from “other, frequently” and “conservative, frequently” supporting what we observed in the segmented bar graphs.

 

3) p. 666, problem 47

(a) Since we want to compare proportions in three groups, we will use a chi-square test.

If we assume these were three independent samples (soccer players, nonsoccer athletes, controls), the best way to state the null hypothesis is

Descriptively, it appears that soccer players do get more concussions (about 50%) compared to 29% and 15% in the other groups.

 

H0: p1 = p2 = p3 (the population proportions with at least one concussion are the same)

Ha: at least one pi differs

 

Chi-Square Test: soccer, nonsoccer, control

 

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

 

       soccer  nonsoccer  control  Total

    1      45         28        8     81

        30.71      32.40    17.89

        6.647      0.598    5.465

 

    2      46         68       45    159

        60.29      63.60    35.11

        3.386      0.304    2.784

 

Total      91         96       53    240

 

Chi-Sq = 19.184, DF = 2, P-Value = 0.000

 

Since the p-value is small (.000 < .001), we reject the null hypothesis and conclude that at least one population has a different proportion with concussions.

 

(b) This shows a moderate negative relationship indicating (as long as the relationship is linear) that an increase in soccer exposure is associated with a lower score on the immediate memory recall test.

 

Other it didn’t tell us to, we can check whether this sample proportion is statistically significant.

H0: r = 0 (no association in the population)

Ha: r ≠ 0 (there is an association)

 

t = r = -.220(91-2)/(1-.2202) = -2.13

Comparing this to the t table with 89 degrees of freedom, we can round down to df = 60 and see that our two-sided p-value would be between 2(.01) =.02 and 2(.025) = .05.  This indicates that the p-value is small enough to reject the null hypothesis.  We have moderate evidence (below .05 but above .01) of an association between memory performance and how much soccer is played for this population of soccer players.

 

(c) Let m1 = population mean test score for soccer players and m2 for the nonsoccer players.

Using a two-sample t­-test in Minitab:

Two-Sample T-Test and CI

 

Sample   N   Mean  StDev  SE Mean

1       26  37.50   9.13      1.8

2       56   39.6   10.2      1.4

 

 

Difference = mu (1) - mu (2)

Estimate for difference:  -2.13000

95% CI for difference:  (-6.63998, 2.37998)

T-Test of difference = 0 (vs not =): T-Value = -0.95  P-Value = 0.348  DF = 54

 

With a large p-value (.348>.1), we do not have evidence that the mean oral word-association test score differs between soccer players and nonsoccer athletes.

 

(d) Now we want to run an ANOVA to compare the three population means

            H0: m1 = m2 = m­3 (the three populations have the same amount of concussions on average)

            Ha: at least one population has a different amount of concussions on average

 

We can calculate MSE by pooling the variances:

            = .5244

 

We can calculate MSTr by comparing the sample means to the overall mean

            overall mean = [92(.3) + 97(.49) + 53(.19)]/237 = .359

            SSTr = 91(.3-.359)2 + 96(.49-.359)2 + 52(.19-.359)2 = 3.47

So MSTr = 3.47/(3-1) = 1.73

 

F = 1.73/.5244 = 3.30

 

The degrees of freedom are 2 and 237.  Using Table A.9, with df = 2, 200, the p-value is between .01 and .05. So we have moderate evidence that that the average number of concussions differs in at least one of these populations.

 

4) p. 708, problem 5

 

5) p. 708, problem 7

Need mean of ’s and mean of s’s: 12.95+ 3(.526)/(.940sqrt(5)) gives LCL=12.20, UCL=13.70  Every one of the 22  values is well within these limits, so the process appears to be in control with respect to location.

 

6) p. 708, problem 12 No points fall outside 2-sigma limits and only two points fall outside the 1-sigma limits. There are no runs of eight on the same side of the center line.

 

7) p. 717, problem 19 (a) pbar=578/(100*25) = .231

(b) .231 + 3sqrt(.231(.769)/100) gives LCL=.105 and UCL=.357

39/100 exceeds UCL generating an out-of-control signal.

 

8) p. 717, problem 23 =4.08, 4.08 + 3 sqrt(4.08) gives LCL=-2.0 or 0 and UCL=10.14.  No values exceed 10.1 so the process is in control.