Answer :
We wish to test
[tex]$$
\begin{aligned}
H_0 &: \mu_1 = \mu_2,\\[1mm]
H_a &: \mu_1 < \mu_2,
\end{aligned}
$$[/tex]
using data from two independent samples at a significance level of [tex]$\alpha = 0.01$[/tex].
The procedure is as follows:
1. Compute Sample Statistics.
For each sample, calculate the sample mean, denoted by [tex]$\bar{x}_1$[/tex] and [tex]$\bar{x}_2$[/tex], the sample variance [tex]$s_1^2$[/tex] and [tex]$s_2^2$[/tex], and note the sample sizes [tex]$n_1$[/tex] and [tex]$n_2$[/tex].
2. Welch’s Two-Sample [tex]$t$[/tex]‑Test.
Since we do not assume equal variances, we use Welch’s [tex]$t$[/tex]‑test. The test statistic is given by
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}.
$$[/tex]
3. Degrees of Freedom.
The degrees of freedom for Welch’s test are approximated by the Welch–Satterthwaite equation:
[tex]$$
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}.
$$[/tex]
4. One-Tailed [tex]$p$[/tex]-Value.
The two-sample [tex]$t$[/tex]‑test typically provides a two-tailed [tex]$p$[/tex]‑value. Since our alternative hypothesis is [tex]$H_a: \mu_1 < \mu_2$[/tex], we adjust the two-tailed [tex]$p$[/tex]‑value to obtain a one-tailed [tex]$p$[/tex]‑value. In this case, if the test statistic is negative then the one-tailed [tex]$p$[/tex]‑value is half of the two-tailed value.
After performing the necessary calculations from the sample data, the computed values (rounded as requested) are:
- The test statistic is
[tex]$$
t \approx -0.229.
$$[/tex]
- The corresponding one-tailed [tex]$p$[/tex]‑value is
[tex]$$
p \approx 0.4097.
$$[/tex]
Since the [tex]$p$[/tex]-value is quite high, we do not reject [tex]$H_0$[/tex] at the [tex]$\alpha=0.01$[/tex] level.
Thus, the final answers are:
Test statistic [tex]$=$[/tex] [tex]$-0.229$[/tex]
[tex]$p$[/tex]-value [tex]$=$[/tex] [tex]$0.4097$[/tex]
The [tex]$p$[/tex]-value is greater than [tex]$\alpha = 0.01$[/tex], indicating that there is not enough evidence to support the claim that [tex]$\mu_1 < \mu_2$[/tex].
[tex]$$
\begin{aligned}
H_0 &: \mu_1 = \mu_2,\\[1mm]
H_a &: \mu_1 < \mu_2,
\end{aligned}
$$[/tex]
using data from two independent samples at a significance level of [tex]$\alpha = 0.01$[/tex].
The procedure is as follows:
1. Compute Sample Statistics.
For each sample, calculate the sample mean, denoted by [tex]$\bar{x}_1$[/tex] and [tex]$\bar{x}_2$[/tex], the sample variance [tex]$s_1^2$[/tex] and [tex]$s_2^2$[/tex], and note the sample sizes [tex]$n_1$[/tex] and [tex]$n_2$[/tex].
2. Welch’s Two-Sample [tex]$t$[/tex]‑Test.
Since we do not assume equal variances, we use Welch’s [tex]$t$[/tex]‑test. The test statistic is given by
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}.
$$[/tex]
3. Degrees of Freedom.
The degrees of freedom for Welch’s test are approximated by the Welch–Satterthwaite equation:
[tex]$$
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}.
$$[/tex]
4. One-Tailed [tex]$p$[/tex]-Value.
The two-sample [tex]$t$[/tex]‑test typically provides a two-tailed [tex]$p$[/tex]‑value. Since our alternative hypothesis is [tex]$H_a: \mu_1 < \mu_2$[/tex], we adjust the two-tailed [tex]$p$[/tex]‑value to obtain a one-tailed [tex]$p$[/tex]‑value. In this case, if the test statistic is negative then the one-tailed [tex]$p$[/tex]‑value is half of the two-tailed value.
After performing the necessary calculations from the sample data, the computed values (rounded as requested) are:
- The test statistic is
[tex]$$
t \approx -0.229.
$$[/tex]
- The corresponding one-tailed [tex]$p$[/tex]‑value is
[tex]$$
p \approx 0.4097.
$$[/tex]
Since the [tex]$p$[/tex]-value is quite high, we do not reject [tex]$H_0$[/tex] at the [tex]$\alpha=0.01$[/tex] level.
Thus, the final answers are:
Test statistic [tex]$=$[/tex] [tex]$-0.229$[/tex]
[tex]$p$[/tex]-value [tex]$=$[/tex] [tex]$0.4097$[/tex]
The [tex]$p$[/tex]-value is greater than [tex]$\alpha = 0.01$[/tex], indicating that there is not enough evidence to support the claim that [tex]$\mu_1 < \mu_2$[/tex].
