Certainly! Let's step through each part of the hypothesis test with the given information:
### Given Data:
- Sample sizes:
[tex]\( n_1 = 40 \)[/tex]
[tex]\( n_2 = 45 \)[/tex]
- Sample means:
[tex]\( \bar{x}_1 = 25.1 \)[/tex]
[tex]\( \bar{x}_2 = 23.0 \)[/tex]
- Population standard deviations:
[tex]\( \sigma_1 = 4.8 \)[/tex]
[tex]\( \sigma_2 = 6.0 \)[/tex]
- Significance level:
[tex]\( \alpha = 0.05 \)[/tex]
### Hypothesis:
[tex]\[
\begin{aligned}
H_0: \mu_1 - \mu_2 \leq 0 \\
H_a: \mu_1 - \mu_2 > 0
\end{aligned}
\][/tex]
### a. Calculation of the Test Statistic (z-value):
The test statistic (z-value) for two independent samples can be calculated using the formula:
[tex]\[
z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\left(\frac{\sigma_1^2}{n_1}\right) + \left(\frac{\sigma_2^2}{n_2}\right)}}
\][/tex]
From the given data:
[tex]\[
\sigma_1 = 4.8, \quad n_1 = 40, \quad \sigma_2 = 6.0, \quad n_2 = 45
\][/tex]
First, calculate the standard error (SE):
[tex]\[
SE = \sqrt{\left(\frac{\sigma_1^2}{n_1}\right) + \left(\frac{\sigma_2^2}{n_2}\right)} = \sqrt{\left(\frac{4.8^2}{40}\right) + \left(\frac{6.0^2}{45}\right)}
\][/tex]
Then, calculate:
[tex]\[
z = \frac{25.1 - 23.0}{SE}
\][/tex]
Using this formula, we obtain:
[tex]\[
z \approx 1.79
\][/tex]
So, the value of the test statistic is:
[tex]\[
\boxed{1.79}
\][/tex]
### b. Calculation of the p-value:
The p-value can be determined by finding the area to the right of the z-value in the standard normal distribution.
For [tex]\( z = 1.79 \)[/tex], the p-value is the area to the right of [tex]\( z = 1.79 \)[/tex].
The p-value comes out to be approximately:
[tex]\[
\boxed{0.0367}
\][/tex]
### c. Hypothesis Testing Conclusion:
With [tex]\(\alpha = 0.05\)[/tex]:
- We compare the p-value to the significance level [tex]\(\alpha\)[/tex]:
- If [tex]\( p \text{-value} < \alpha \)[/tex], reject [tex]\( H_0 \)[/tex].
- If [tex]\( p \text{-value} \geq \alpha \)[/tex], do not reject [tex]\( H_0 \)[/tex].
In this case:
[tex]\[
p \text{-value} = 0.0367 < 0.05
\][/tex]
Since the p-value is less than the significance level [tex]\( \alpha \)[/tex], we reject the null hypothesis [tex]\( H_0 \)[/tex].
Thus, our conclusion with [tex]\( \alpha = 0.05 \)[/tex] is:
[tex]\[
\boxed{\text{Reject } H_0}
\][/tex]
