You wish to test the following claim [tex] H_1 [/tex] at a significance level of [tex] \alpha = 0.001 [/tex].

\[
\begin{array}{l}
H_0: \mu_1 = \mu_2 \\
H_1: \mu_1 < \mu_2
\end{array}
\]

You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal.

You obtain a sample of size [tex] n_1 = 18 [/tex] with a mean of [tex] \bar{x}_1 = 53.4 [/tex] and a standard deviation of [tex] s_1 = 12.2 [/tex] from the first population. You obtain a sample of size [tex] n_2 = 19 [/tex] with a mean of [tex] \bar{x}_2 = 59.1 [/tex] and a standard deviation of [tex] s_2 = 10.9 [/tex] from the second population.

a) What is the test statistic for this sample? (Report answer accurate to two decimal places.)
- Test statistic = [tex] \square [/tex]

b) What is the p-value for this sample? (Report answer accurate to four decimal places.)
- [tex] p [/tex]-value = [tex] \square [/tex]

To tackle this hypothesis testing problem, let's go through the steps to find the test statistic and p-value:

### Step-by-Step Solution:

#### a) Calculating the Test Statistic

1. Identify the Given Information:

- Sample sizes: [tex]$ n_1 = 18 $[/tex], [tex]$ n_2 = 19 $[/tex]
- Sample means: [tex]$ \bar{x}_1 = 53.4 $[/tex], [tex]$ \bar{x}_2 = 59.1 $[/tex]
- Sample standard deviations: [tex]$ s_1 = 12.2 $[/tex], [tex]$ s_2 = 10.9 $[/tex]

2. Calculate the Pooled Standard Deviation (Sp):

Since we have equal variances assumption, the pooled standard deviation formula is used:

[tex]\[
S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\][/tex]

Substituting in the values:

[tex]\[
S_p = \sqrt{\frac{(18 - 1) \times (12.2)^2 + (19 - 1) \times (10.9)^2}{18 + 19 - 2}}
\][/tex]

[tex]\[
S_p = \sqrt{\frac{17 \times 148.84 + 18 \times 118.81}{35}}
\][/tex]

[tex]\[
S_p = \sqrt{\frac{2521.28 + 2138.58}{35}}
\][/tex]

[tex]\[
S_p = \sqrt{\frac{4659.86}{35}}
\][/tex]

[tex]\[
S_p \approx 11.51
\][/tex]

3. Calculate the Standard Error (SE) for the Difference in Means:

[tex]\[
SE = S_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}
\][/tex]

[tex]\[
SE = 11.51 \times \sqrt{\frac{1}{18} + \frac{1}{19}}
\][/tex]

[tex]\[
SE \approx 11.51 \times \sqrt{0.0556 + 0.0526}
\][/tex]

[tex]\[
SE \approx 11.51 \times \sqrt{0.1082}
\][/tex]

[tex]\[
SE \approx 11.51 \times 0.329
\][/tex]

[tex]\[
SE \approx 3.79
\][/tex]

4. Compute the Test Statistic (t):

[tex]\[
t = \frac{\bar{x}_1 - \bar{x}_2}{SE}
\][/tex]

[tex]\[
t = \frac{53.4 - 59.1}{3.79}
\][/tex]

[tex]\[
t \approx \frac{-5.7}{3.79}
\][/tex]

[tex]\[
t \approx -1.50
\][/tex]

Therefore, the test statistic is approximately [tex]$-1.50$[/tex].

#### b) Calculating the p-value

1. Calculate the degrees of freedom (df):

[tex]\[
df = n_1 + n_2 - 2 = 18 + 19 - 2 = 35
\][/tex]

2. Find the p-value for the one-tailed t-test:

Using a t-distribution table or software with [tex]$ df = 35 $[/tex] and the calculated test statistic [tex]$ t = -1.50 $[/tex], we determine the p-value.

The p-value is approximately [tex]$ 0.0712 $[/tex].

### Conclusion:

- Test Statistic: [tex]$-1.50$[/tex]
- p-value: [tex]$0.0712$[/tex]

Since the p-value [tex]$0.0712$[/tex] is greater than the significance level [tex]$\alpha = 0.001$[/tex], we do not reject the null hypothesis [tex]$H_0$[/tex]. There isn't enough evidence to support the claim that [tex]$\mu_1 < \mu_2$[/tex] at the [tex]$0.001$[/tex] significance level.