Question

Consider the following hypothesis test. : : The following results are for two independent samples taken from two populations. Excel File: data10-03.xlsx Enter negative values as negative numbers. a. What is the value of the test statistic? (to 2 decimals) b. What is the -value? (to 4 decimals) c. With , what is your hypothesis testing conclusion? - Select your answer -

Answer 1

Answer:

$z = -1.53$ --- test statistic

$p = 0.1260$ --- p value

Conclusion: Fail to reject the null hypothesis.

Step-by-step explanation:

Given

$n_1 = 80$     $\bar x_1= 104$   $\sigma_1 = 8.4$

$n_2 = 70$    $\bar x_2 = 106$    $\sigma_2 = 7.6$

$H_o: \mu_1 - \mu_2 = 0$ --- Null hypothesis

$H_a: \mu_1 - \mu_2 \ne 0$ ---- Alternate hypothesis

$\alpha = 0.05$

Solving (a): The test statistic

This is calculated as:

$z = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} }}$

So, we have:

$z = \frac{104 - 106}{\sqrt{\frac{8.4^2}{80} + \frac{7.6^2}{70} }}$

$z = \frac{104 - 106}{\sqrt{\frac{70.56}{80} + \frac{57.76}{70}}}$

$z = \frac{-2}{\sqrt{0.8820 + 0.8251}}$

$z = \frac{-2}{\sqrt{1.7071}}$

$z = \frac{-2}{1.3066}$

$z = -1.53$

Solving (b): The p value

This is calculated as:

$p = 2 * P(Z < z)$

So, we have:

$p = 2 * P(Z < -1.53)$

Look up the z probability in the z score table. So, the expression becomes

$p = 2 * 0.0630$

$p = 0.1260$

Solving (c): With $\alpha = 0.05$, what is the conclusion based on the p value

We have:

$\alpha = 0.05$

In (b), we have:

$p = 0.1260$

By comparison:

$p > \alpha$

i.e.

$0.1260 > 0.05$

So, we fail to reject the null hypothesis.