Answer :
We wish to test the claim that the 1988 population’s mean weight is less than the 2012 population’s mean weight. In this problem, let
- Population 1: U.S. Army male personnel weights from 1988, with sample size [tex]$n_1 = 12$[/tex], sample mean [tex]$\overline{x}_1 = 78.35$[/tex], and sample standard deviation [tex]$s_1 = 13.1175$[/tex].
- Population 2: U.S. Army male personnel weights from 2012, with sample size [tex]$n_2 = 15$[/tex], sample mean [tex]$\overline{x}_2 = 87.14$[/tex], and sample standard deviation [tex]$s_2 = 18.0934$[/tex].
Because the claim is that the 1988 mean is less than the 2012 mean, we set up our hypotheses as follows:
[tex]$$
H_0:\ \mu_1 \ge \mu_2 \quad \text{(null hypothesis)}
$$[/tex]
[tex]$$
H_1:\ \mu_1 < \mu_2 \quad \text{(alternative hypothesis)}
$$[/tex]
Because there is no assumption that the population variances are equal, we use Welch’s [tex]$t$[/tex]-test. The test statistic is given by
[tex]$$
t = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}.
$$[/tex]
Substituting the sample estimates, the test statistic is found to be approximately
[tex]$$
t \approx -1.46168.
$$[/tex]
The degrees of freedom are calculated using the Welch–Satterthwaite equation:
[tex]$$
\nu \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}},
$$[/tex]
and with the given numbers, we obtain approximately
[tex]$$
\nu \approx 24.81.
$$[/tex]
A two-tailed test would have resulted in a [tex]$p$[/tex]-value of approximately 0.15638. However, because our alternative hypothesis is one‐tailed (specifically [tex]$H_1: \mu_1 < \mu_2$[/tex]) and the test statistic is negative, the one-tailed [tex]$p$[/tex]-value is half the two-tailed [tex]$p$[/tex]-value. Thus,
[tex]$$
p \approx \frac{0.15638}{2} \approx 0.078.
$$[/tex]
Since we are testing at a significance level of [tex]$\alpha = 0.01$[/tex] and [tex]$p = 0.078 > 0.01$[/tex], we do not have sufficient evidence to reject the null hypothesis. In other words, at the 0.01 significance level, there is not enough evidence to support the claim that the mean weight of the 1988 population is less than that of the 2012 population.
To summarize:
1. The hypotheses are
[tex]$$
H_0:\ \mu_1 \ge \mu_2 \quad \text{and} \quad H_1:\ \mu_1 < \mu_2.
$$[/tex]
2. The one-tailed [tex]$p$[/tex]-value is approximately [tex]$0.078$[/tex].
- Population 1: U.S. Army male personnel weights from 1988, with sample size [tex]$n_1 = 12$[/tex], sample mean [tex]$\overline{x}_1 = 78.35$[/tex], and sample standard deviation [tex]$s_1 = 13.1175$[/tex].
- Population 2: U.S. Army male personnel weights from 2012, with sample size [tex]$n_2 = 15$[/tex], sample mean [tex]$\overline{x}_2 = 87.14$[/tex], and sample standard deviation [tex]$s_2 = 18.0934$[/tex].
Because the claim is that the 1988 mean is less than the 2012 mean, we set up our hypotheses as follows:
[tex]$$
H_0:\ \mu_1 \ge \mu_2 \quad \text{(null hypothesis)}
$$[/tex]
[tex]$$
H_1:\ \mu_1 < \mu_2 \quad \text{(alternative hypothesis)}
$$[/tex]
Because there is no assumption that the population variances are equal, we use Welch’s [tex]$t$[/tex]-test. The test statistic is given by
[tex]$$
t = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}.
$$[/tex]
Substituting the sample estimates, the test statistic is found to be approximately
[tex]$$
t \approx -1.46168.
$$[/tex]
The degrees of freedom are calculated using the Welch–Satterthwaite equation:
[tex]$$
\nu \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}},
$$[/tex]
and with the given numbers, we obtain approximately
[tex]$$
\nu \approx 24.81.
$$[/tex]
A two-tailed test would have resulted in a [tex]$p$[/tex]-value of approximately 0.15638. However, because our alternative hypothesis is one‐tailed (specifically [tex]$H_1: \mu_1 < \mu_2$[/tex]) and the test statistic is negative, the one-tailed [tex]$p$[/tex]-value is half the two-tailed [tex]$p$[/tex]-value. Thus,
[tex]$$
p \approx \frac{0.15638}{2} \approx 0.078.
$$[/tex]
Since we are testing at a significance level of [tex]$\alpha = 0.01$[/tex] and [tex]$p = 0.078 > 0.01$[/tex], we do not have sufficient evidence to reject the null hypothesis. In other words, at the 0.01 significance level, there is not enough evidence to support the claim that the mean weight of the 1988 population is less than that of the 2012 population.
To summarize:
1. The hypotheses are
[tex]$$
H_0:\ \mu_1 \ge \mu_2 \quad \text{and} \quad H_1:\ \mu_1 < \mu_2.
$$[/tex]
2. The one-tailed [tex]$p$[/tex]-value is approximately [tex]$0.078$[/tex].
