9.5 Review Questions
- Determine whether we should use a two-sample [latex]t[/latex] test or a paired [latex]t[/latex] test in the following applications.
- The data in the following table give the number of young per litter for 24 female cottonmouths in Florida and 44 female cottonmouths in Virginia.
Florida Virginia 8 6 7 5 12 7 7 6 8 7 4 3 12 9 7 4 9 6 [latex]\vdots[/latex] [latex]\vdots[/latex] [latex]\vdots[/latex] [latex]\vdots[/latex] [latex]\vdots[/latex] 5 5 4 5 4 At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, the number of young per litter of cottonmouths in Florida is less than that in Virginia?
- Independent random samples of 10 homes each in Atlantic City and Las Vegas yielded the following data on home prices in thousands of dollars. At the 5% significance level, can you conclude that the mean costs for existing single-family homes differ in Atlantic City and Las Vegas?
Atlantic City Las Vegas 234.0 192.8 226.4 231.5 213.0 256.4 214.7 210.9 [latex]\vdots[/latex] [latex]\vdots[/latex] [latex]\vdots[/latex] [latex]\vdots[/latex] 236.1 301.9 349.4 178.5 - The following table gives data, for a sample of eight years, on the number of days that ice stayed on two lakes in Madison, Wisconsin-Lake Mendota and Lake Monona. At the 10% significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean length of time that ice stays on these two lakes?
Year Mendota Monona 1 119 107 2 115 108 [latex]\vdots[/latex] [latex]\vdots[/latex] [latex]\vdots[/latex] 8 87 91 - The fiber density of 10 samples with varying fiber density was obtained using both an eye-piece method and a TV-screen method. The results, in fibers per square millimeter, are presented in the following table. Test at the 5% significance level whether, on average, the eyepiece method gives a greater fiber-density reading than the TV-screen method.
Sample ID Eyepiece TV Screen 1 182.2 177.8 2 118.5 116.6 [latex]\vdots[/latex] [latex]\vdots[/latex] [latex]\vdots[/latex] 10 85.4 86.6 - Compare the average IQ score of students from U of A and MacEwan. Randomly pick 30 students from each University and obtain their IQ scores.
- Compare weekly earnings of male and female workers. Randomly pick 40 male and 40 female workers and compare their average weekly earnings.
- Compare weekly earnings of male and female workers. Randomly pick 40 households and compare the husbands’ and wives’ average weekly earnings.
- The data in the following table give the number of young per litter for 24 female cottonmouths in Florida and 44 female cottonmouths in Virginia.
- The following table summarizes the operative times of neurosurgeries conducted by a dynamic system (Z-plate) and a static system (ALPS plate), respectively.
Dynamic Static [latex]\bar x_1 = 400[/latex] [latex]\bar x_2 = 480[/latex] [latex]s_1 = 85[/latex] [latex]s_2 = 40[/latex] [latex]n_1 = 60[/latex] [latex]n_2 = 30[/latex] - Test at a 5% significance level whether the dynamic system (Z-plate) reduced the mean operative time relative to the static system (ALPS plate).
- Obtain a confidence interval for the difference in mean operative time between the dynamic and the static systems, [latex]\mu_1-\mu_2[/latex], corresponding to the test in part (a).
- This table shows men and women’s winning times (in minutes) in the New York City Marathon between 1978 and 2006.
Men([latex]y_i[/latex]) Women([latex]x_i[/latex]) Difference([latex]d_i[/latex]) 132.2 152.5 20.3 131.7 147.6 15.9 129.7 145.7 16.0 128.2 145.5 17.3 129.5 147.2 17.7 129.0 147.0 18.0 134.9 149.5 14.6 131.6 148.6 17.0 131.1 148.1 17.0 131.0 150.3 19.3 128.3 148.1 19.8 128.0 145.5 17.5 132.7 150.8 18.1 129.5 147.5 18.0 129.5 144.7 15.2 Men([latex]y_i[/latex]) Women([latex]x_i[/latex]) Difference([latex]d_i[/latex]) 130.1 146.4 16.3 131.4 147.6 16.2 131.0 148.1 17.1 129.9 148.3 18.4 128.2 148.7 20.5 128.8 145.3 16.5 129.2 145.1 15.9 130.2 145.8 15.6 127.7 144.4 16.7 128.1 145.9 17.8 130.5 142.5 12.0 129.5 143.2 13.7 129.5 144.7 15.2 130.0 145.1 15.1 [latex]\bar d = 16.85, s_d = 1.98[/latex] - At the 1% significance level, do the data provide sufficient evidence that there is a difference in winning times between males and females?
- Obtain a confidence interval corresponding to the test in part (a).
- Does the interval in part (b) support the conclusion in part (a)?
- Based on the interval obtained in part (b), can we claim that the winning time of males is at least 15 minutes fast than that of females? How about 20 minutes fast?
Show/Hide Answer
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- The data are two independent samples, use two-sample [latex]t[/latex] test
- The data are two independent samples, use two-sample [latex]t[/latex] test
- The data is a paired sample, use paired [latex]t[/latex] test.
- The data is a paired sample, use paired [latex]t[/latex] test.
- Two independent samples, use two-sample [latex]t[/latex] test.
- Two independent samples, use two-sample [latex]t[/latex] test.
- Paired sample, use paired [latex]t[/latex] test.
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- Check the assumptions:
- We have simple random samples.
- The two samples are independent.
- We have large samples [latex]n_1 = 60>30[/latex] and [latex]n_2 = 30 = 30[/latex].
Steps:
- Hypotheses. [latex]H_0: \mu_1-\mu_2\ge 0[/latex] versus [latex]H_a: \mu_1-\mu_2<0[/latex].
- Significance level [latex]\alpha = 0.05[/latex].
- Test statistic. [latex]t_o = \frac{(\bar x_1-\bar x_2)-\Delta_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} = { \frac{(400-480)- 0}{\sqrt{\frac{85^2}{60}+\frac{40^2}{30}}} = -6.069}, df = \min\{n_1-1, n_2-1\} = \min\{60-1, 30-1\} = 29.[/latex]
- [latex]P\mbox{-value}[/latex]: for a left-tailed test, [latex]P\mbox{-value}[/latex] = [latex]P(t\le t_o) = P(t \le -6.069) = P(t { \ge 6.069})<0.0005<0.05 (\alpha).[/latex]
- Decision: Reject [latex]H_0[/latex], since [latex]P\mbox{-value}\lt\alpha[/latex].
- Conclusion: At the 5% significance level, we have sufficient evidence that the dynamic system (Z-plate) reduced the mean operative time relative to the static system (ALPS plate).
- We should construct a lower-tailed 95% confidence interval for a left-tailed test at the 5% significance level. The upper confidence bound for a 95% lower-tailed interval for [latex]\mu_1-\mu_2[/latex] is
[latex](\bar x_1-\bar x_2) + t_\alpha \times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} = (400-480)+ 1.699 \times \sqrt{\frac{85^2}{60}+\frac{40^2}{30}} = -57.605.[/latex]
The 95% lower-tailed confidence interval is [latex](-\infty, -57.605)[/latex].
Interpretation: we can be 95% confident that the mean operative time of the dynamic system is at least 57.605 minutes shorter than the mean operative time of the static system.
- Check the assumptions:
-
- Check the assumptions:
- We have simple random samples.
- We have large number of pairs [latex]n = 29\approx 30[/latex].
Steps:
- Hypotheses. [latex]H_0: \mu_{\scriptsize M}-\mu_{\scriptsize F} = 0[/latex] versus [latex]H_a: \mu_{\scriptsize M}-\mu_{\scriptsize F}\ne 0[/latex].
- Significance level [latex]\alpha = 0.01[/latex].
- Test statistic. [latex]t_o = \frac{\bar d-{ \delta_0}}{\frac{s_d}{\sqrt{n}}} = \frac{16.85-{ 0}}{\frac{1.98}{\sqrt{29}}} = 45.828, \quad \text{with } df = n-1 = 29-1 = 28.[/latex]
- [latex]P\mbox{-value}[/latex]: for a two-tailed test, [latex]P\mbox{-value}[/latex] = [latex]2P(t\le |t_o|) = 2P(t\ge 45.828){ \bf<} 2\times 0.0005 = 0.001<0.01 (\alpha).[/latex]
- Decision: Reject [latex]H_0[/latex], since [latex]P\mbox{-value}\lt\alpha[/latex].
- Conclusion: At the 1% significance level, we have sufficient evidence that there is a difference in winning times between males and females.
- A two-tailed test at the 1% significance level corresponds to a 99% two-tailed confidence interval. For [latex]df = 28[/latex], [latex]1-\alpha = 0.99 \Longrightarrow \alpha = 0.01 \Longrightarrow \frac{\alpha}{2} = \frac{0.01}{2} = 0.005 \Longrightarrow t_{\alpha/2} = t_{0.005} = 2.763.[/latex] A 99% two-tailed confidence interval for [latex]\mu_d[/latex] is [latex]\bar d\pm t_{\alpha/2} \times \frac{s_d}{\sqrt{n}} = 16.85\pm 2.763 \times \frac{1.98}{\sqrt{29}} = (15.834, 17.866).[/latex]
Interpretation: we can be 99% confident that the difference in the mean winning times between male and female is somewhere between 15.834 minutes and 17.866 minutes. - Yes, because the hypothesized value [latex]\delta_0 = 0[/latex] is outside the entire interval; therefore, we can claim that [latex]\mu_{\scriptsize M}-\mu_{\scriptsize F}\ne 0[/latex] which is the conclusion of the paired [latex]t[/latex] test. Since the entire interval is above 0, we can further claim that [latex]\mu_{\scriptsize M}-\mu_{\scriptsize F}> 0[/latex].
- We can claim that the mean winning time of males is at least 15 minutes fast than that of females, since the entire interval is above 15, we can claim [latex]\mu_{\scriptsize M}-\mu_{\scriptsize F}>15[/latex]. However, we cannot claim that the mean winning time of males is at least 20 minutes faster than that of females since the 20 is inside the interval.
Note: for a right-tailed test at the 1% significance level, it is more precise to construct a 99% upper-tailed interval with a lower bound: [latex]\bar d\pm t_{\alpha} \times \frac{s_d}{\sqrt{n}} = 16.85-2.467 \times \frac{1.98}{\sqrt{29}} = 15.943.[/latex] The 99% upper-tailed interval is [latex](-\infty, 15.943)[/latex].
Interpretation: we can be 99% confident that the winning time of males is at least 15.943 minutes faster than that of females. Since [latex]\delta_0 = 15[/latex] is outside the confidence interval, we can claim that [latex]\mu_{\scriptsize M}-\mu_{\scriptsize F}>15[/latex]; however, [latex]\delta_0 = 20[/latex] is inside the confidence interval, we cannot claim that [latex]\mu_{\scriptsize M}-\mu_{\scriptsize F}>20.[/latex] There is insufficient evidence that the mean winning time of males is at least 20 minutes faster than that of females.
- Check the assumptions:
