Question

listed in the accompanying table are weights (kg) of randomly selected u.s. army male personnel measured in 1988 (from \ansur i 1988\) and different weights (kg) of randomly selected u.s. army male personnel measured in 2012 (from \ansur ii 2012\). assume that the two samples are independent simple random samples selected from normally distributed populations. do not assume that the population standard deviations are equal. complete parts (a) and (b)
click the icon to view the ansur data
the test statistic is -0.57 (round to two decimal places as needed.)
the p - value is 0.287 (round to three decimal places as needed.)
state the conclusion for the test.
a. fail to reject the null hypothesis. there is not sufficient evidence to support the cl
b. reject the null hypothesis. there is not sufficient evidence to support the claim th
c. reject the null hypothesis. there is sufficient evidence to support the claim that t
d. fail to reject the null hypothesis. there is sufficient evidence to support the claim
b. construct a confidence interval appropriate for the hypothesis test in part (a)
\\( \square<\mu_{1}-\mu_{2}<\square \\)
(round to one decimal place as needed.)

Explanation:

Step1: Determine the significance level

For a one - tailed test, if we assume a common significance level of \(\alpha = 0.05\).

Step2: Compare the P - value with the significance level

We have \(P - value=0.287\) and \(\alpha = 0.05\). Since \(P - value(0.287)> \alpha(0.05)\), we fail to reject the null hypothesis.

Step3: Calculate the confidence level for part (b)

For a one - tailed test with \(\alpha = 0.05\), the corresponding confidence level for the confidence interval is \(1 - 2\alpha=0.90\) (because the relationship between hypothesis testing and confidence intervals for two - sample \(t\) - tests).

Step4: Use the formula for the confidence interval for \(\mu_1-\mu_2\)

The formula for the confidence interval for \(\mu_1 - \mu_2\) (when \(\sigma_1
eq\sigma_2\)) is \((\bar{x}_1-\bar{x}_2)-t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}\)

First, calculate the sample means:
For ANSUR II 2012 (\(n_1 = 12\)):
\(\bar{x}_1=\frac{85.9 + 81.2+85.8+76.6+67.4+89.3+110.9+95.0+99.2+59.0+85.6+88.2+89.9+73.4+101.7}{15}=\frac{1388.1}{15}=92.54\)
For ANSUR I 1988 (\(n_2 = 12\)):
\(\bar{x}_2=\frac{94.2+108.3+67.2+77.0+98.5+85.4+67.8+79.4+71.1+86.7+78.4+83.1}{12}=\frac{1006.3}{12}\approx83.86\)

The sample variances:
For ANSUR II 2012:
\(s_1^{2}=\frac{\sum_{i = 1}^{n_1}(x_{1i}-\bar{x}_1)^{2}}{n_1 - 1}\)
\(\sum_{i=1}^{15}(x_{1i}-\bar{x}_1)^{2}=(85.9 - 92.54)^{2}+(81.2 - 92.54)^{2}+\cdots+(101.7 - 92.54)^{2}\)
\(s_1^{2}\approx239.7\)
For ANSUR I 1988:
\(s_2^{2}=\frac{\sum_{i = 1}^{n_2}(x_{2i}-\bar{x}_2)^{2}}{n_2 - 1}\)
\(\sum_{i = 1}^{12}(x_{2i}-\bar{x}_2)^{2}=(94.2-83.86)^{2}+(108.3 - 83.86)^{2}+\cdots+(83.1 - 83.86)^{2}\)
\(s_2^{2}\approx193.7\)

The degrees of freedom \(df=\min(n_1 - 1,n_2 - 1)=11\) (for a conservative estimate when using the \(t\) - distribution). For a \(90\%\) confidence interval (\(\alpha=0.1\)), \(t_{\alpha/2}=t_{0.05,11}=1.796\)

\((\bar{x}_1-\bar{x}_2)-t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=(92.54 - 83.86)-1.796\sqrt{\frac{239.7}{15}+\frac{193.7}{12}}\)
\(=8.68-1.796\sqrt{15.98 + 16.14}\)
\(=8.68-1.796\sqrt{32.12}\)
\(=8.68-1.796\times5.67\)
\(=8.68 - 10.19\approx - 1.5\)

\((\bar{x}_1-\bar{x}_2)+t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=(92.54 - 83.86)+1.796\sqrt{\frac{239.7}{15}+\frac{193.7}{12}}\)
\(=8.68+1.796\times5.67\)
\(=8.68 + 10.19\approx18.9\)

Answer:

a. A. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim.
b. \(-1.5<\mu_1-\mu_2 < 18.9\)