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

`z = -1.53` --- test statistic

`p = 0.1260` --- p value

Conclusion: Fail to reject the null hypothesis.

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 \\e 0` ---- Alternate hypothesis

`\alpha = 0.05`

Solving (a): The test statistic

This is calculated as:

`z = \frac{\bar x_1 - \bar x_2}{\sqrt{(\sigma_1^2)/(n_1) + (\sigma_2^2)/(n_2) }}`

So, we have:

`z = \frac{104 - 106}{\sqrt{(8.4^2)/(80) + (7.6^2)/(70) }}`

`z = \frac{104 - 106}{\sqrt{(70.56)/(80) + (57.76)/(70)}}`

`z = (-2)/(√(0.8820 + 0.8251))`

`z = (-2)/(√(1.7071))`

`z = (-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.