7.1

1. The degrees of freedom is n−1.
2. The margin of error is m=t*sn.
3. The alternative hypothesis must be > or <.
4. The units are matched in each treatment.

7.3

1. 2.110.
2. 2.060.
3. 1.708.
4. As the sample size increases, the margin of error decreases. As the confidence level increases, the margin of error increases.

7.5 When the alternative is >, t*=2.539; when the alternative is <, t*=−2.539.

7.7

1. df=12.
2. 2.681<t<3.055.
3. 0.01<P-value<0.02.
4. t=2.78 is significant at the 5% level but not at the 1% level.
5. 0.0167.

7.9

1. m=1.032.
2. m=2.131.
3. m=2.064.

7.11 95% confidence interval: (17.37, 18.95).

7.13

1. The histogram shows that the data are Normally distributed, so t procedures are appropriate.
2. The 95% confidence interval is 36.157 ± 2.603.
3. (33.55, 38.76).
4. Inference from the confidence interval only applies to the mean, not the median. Other tools would need to be used for inference on the median.

7.15

1. Because each participant drank wine and beer in both orders (wine/beer, beer/wine).
2. Order effect estimate : 0.68; standard error : 1.838; P-value=. There is little evidence of an order difference.

7.17

1. H0: μ=10, Ha: μ<10.
2. t=−5.2603, df=33, P-value<0.0005.

7.19

1. The distribution has two peaks, so the distribution is not Normal. The five-number summary is 2.2, 10.95, 28.5, 41.9, 69.3.
2. Maybe. We have a large enough sample to overcome the non-Normal distribution, but we are sampling from a small population.
3. x¯=27.29, s=17.7058, df=39; the 95% confidence interval is (21.63, 32.95).
4. Answers will vary.

7.21 ($639.81, $710.11).

7.23 H0: μ=45, Ha: μ>45. t=5.457. df=49, P-value<0.0005.

7.25

1. H0: μ=0, Ha: μ≠0.
2. x¯=2.73, s=2.8015, t=4.358, df=19, P-value<0.001.

7.27

1. H0: μ=925, Ha: μ>925. t=3.27, df=35, P-value=0.0012.
2. H0: μ=935, Ha: μ>935, t=0.80, df=35, P-value=0.2146.
3. The interval is 931.4 to 945.1, which includes 935 but not 925.

7.29

1. The differences are spread from −0.018 to 0.020. A Normal quantile plot reveals that the data are approximately Normal and, therefore, t methods are appropriate.
2. H0: μ=0, Ha: μ≠0. t=−0.347, df=7, P-value=0.7388.
3. (−0.0117 to 0.0087).
4. The subjects from this sample may be representative of future subjects, but the test results and confidence interval are suspect because this is not a random sample.

7.31 H0: median = 0, Ha: median > 0; P-value=0.0899. In Exercise 7.24, we were able to reject H0; here, we cannot.

7.33 H0: median = 0, Ha: median > 0; P-value=0.0013. Using a t test, we found the same conclusion.

7.35

1. Hypotheses should involve μ1 and μ2.
2. The samples are not independent.
3. We need the P-value to be small to reject H0.
4. Assuming that the researcher computed the t statistic using x¯1−x¯2, a positive value of t does not support Ha.

7.37

1. We cannot reject H0: μ1=μ2 in favor of the two-sided alternative at the 5% level.
2. We can reject H0 in favor of Ha: μ1<μ2 because t*<−1.812.

7.39

1. Both distributions are Normally distributed, except the low-intensity class has a low outlier.
2. H0: μH=μL, Ha: μH≠μL. t=5.30. df=14. P-value<0.001.
3. Because the low-intensity class has an outlier, the t-test is not appropriate.
4. t=6.31. df=13. P-value<0.001. Removing the outlier did not change the results.
5. Because the outlier is not affecting the results, it is probably okay to report both tests.

7.41

1. The t procedure is robust. Because n1+n2≥40, we can use the t procedures on skewed data.
2. H0: μdays=μmonth, Hα: μdays≠μmonth. t=−2.42, 0.01<P-value<0.02.

7.43 H0: μBrown=μBlue, Ha: μBrown>μBlue. t=2.59, 0.005<P-value<0.01.

7.45 Over half of original sample did not respond. Remaining sample could be biased. Might respondents be poorer (value a $10 gift card more)?

7.47

1. The data are not Normally distributed, but because neither distribution is strongly skewed or has outliers, t procedures are still appropriate.
2. For N group: x¯=0.5714, s=0.73, n=14. For S group: x¯=2.1176, s=1.24, n=17.
3. H0: μN=μS, Ha: μN≠μS.
4. t=−4.31, df=13, P-value<0.001.
5. (−2.32, −0.77).

7.49

1. Taking averages on ratings is likely not appropriate.
2. The data are integers, but the samples are large, and the t procedures can be used.
3. Taco Bell: X¯=4.0667, s=1.0308. Chick-fil-A: X¯=4.5683, s=0.6667.
4. H0: μT=μC, Ha: μT≠μC. t=−5.33 df = 164. P-value<0.001.

7.51 You need sample sizes and standard deviations or df and a more accurate P-value. The confidence interval could give us useful information about the magnitude of the difference.

7.53 This is a matched pairs design.

7.55 There could be things that are similar about the next eight employees who need new computers as well as the following eight, which could bias the results.

7.57

1. The north distribution is right-skewed, while the south distribution is left-skewed.
2. The methods of this section seem to be appropriate because the sample sizes are relatively large, and there are no outliers.
3. H0: μN=μS, Ha: μN≠μS.
4. x¯N=23.7, sN=17.5001, x¯S=34.53, and sS=14.2583. t=−2.629, df=29, 0.01<P-value<0.02.
5. (−19.2614, −2.4053).

7.59

1. Using conservative df = 50, CI is (−2.18, 7.28).
2. There is uncertainty in this estimate. 95% confident that true change is somewhere between a loss of 4.18 units and a gain of 7.28 units. Need to sample a larger number of stores in order to reduce uncertainty.

7.61

1. H0: μB=μF, Ha: μB>μF. t=1.654, df=18, 0.05<P-value<0.10.
2. (−0.2434, 2.0434).
3. We need two independent SRSs from Normal populations.

7.63 t=4.17, df=63, P-value<0.001. Confidence interval: (14.57, 41.43). The results are similar.

7.65

1. 2.776.
2. 2.30.
3. The critical value decreased, which will decrease the margin of error.

7.67

1. 25.
2. The margin of error would be smaller when the sample size is 30. The margin of error would be larger if 80% responded.

7.69 No, the confidence interval is for the mean monthly rate, not the individual apartment rates.

7.71

1. n=18.
2. For n=10, the power will be less than 90%.
3. For n=10, the power is 0.80.

7.73 Answers will vary based on the choice of σ. For σ=0.015, power=0.09.

7.75

1. Increase. A larger alpha gives more power.
2. 0.89.

7.77 Using a larger σ for planning the study is advisable because it provides a conservative (safe) estimate of the power.

7.79 x¯=153.5, s=14.708, sx¯=7.354. It would not be appropriate to construct a confidence interval because we cannot consider these four scores to be an SRS.

7.81

2. The plot shows that t* approaches z*=1.96 as the df increases.
3. The plots would be similar, but t* would approach z*=1.645 as the df increases.

7.83

1. Use two independent samples.
2. Use a matched pairs design.
3. Take a single sample of college students and ask them to rate the appeal of the product.

7.85

1. H0: μ=1.5, Ha: μ<1.5; t=−3.67, df=249, P-value≈0.
2. (0.779, 1.283).
3. We have a large sample, so t procedures should be safe.

7.87

1. (−3.008, 1.302).
2. (−1.761, −0.055).
3. The centers of the intervals are the same, but the margin of error for the independent samples interval is much larger.

7.89

1. H0: μ1=μ2 versus Ha: μ1≠μ2; t=−2.32; 0.02<P-value<0.04; fail to reject the null hypothesis at the 1% level.
2. H0: μ1=μ2 versus Ha: μ1<μ2; t=−2.32; 0.01<P-value<0.02; fail to reject the null hypothesis at the 1% level.

7.91

1. Because the same mockingbird responded on each day.
2. 6.9774.
3. H0: μ1=μ4, Ha: μ1≠μ4; t=6.319, df=23, P-value<0.001.
4. t=−0.973, P-value=0.3407.
5. There is a significant difference between Day 1 and Day 4 but not between Day 1 and Day 5.

7.93 A two-sample t procedure was used. We assumed that the data are approximately Normal. H0: μC=μN, Ha: μC>μN; t=0.95, df=89. 0.15<P-value<0.20.

7.95 A two-sample t procedure was used. H0: μC=μN, Ha: μC>μN; t=−0.16, df=89, P-value>0.25.

7.97 x¯=77.76%, s=32.6768%, (64.27% to 91.25%). This seems to support the retailer's claim.

7.99 H0: μ1=μ2, Ha: μ1>μ2; t=3.65, P-value<0.0005. Confidence interval: (0.7714, 2.6086).

7.101

1. H0: μB=μD, Ha: μD>μB; t=2.87, P-value<0.005. Confidence interval: (1.7, 9.7).
2. H0: μB=μS, Ha: μS>μB; t=1.88, P-value<0.05. Confidence interval: (−0.24, 6.7).

7.103 No: What we have is nothing like an SRS of the population of school corporations; we have census data for the state.

7.105

1. H0: μ=0, Ha: μ>0.
2. Left-skewed, x¯=2.5, s=2.8928.
3. t=3.8649, df=19, P-value=0.0005.
4. (1.15, 3.85).

7.107

1. df=80.
2. t*=1.990.
3. | x¯1−x¯2 |≥1.99(10)2/41.
4. 0.519.

7.109

1. H0: μ=0, Ha: μ≠0. t=5.125, df=15, P-value=0.00012.
2. (191.6, 464.4).

7.111 The distributions appear similar. Assessing μ2-μ1. GPA: t=−0.91, df=30, 0.30<P-value<0.40. Confidence interval: (−1.35, 0.52). IQ: t=1.64, 0.10<P-value<0.20 (df=30). Confidence interval: (−1.24, 11.48). These data have that pattern but strength of evidence is relatively weak.