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forsale [732]
3 years ago
15

Consider the following hypothesis test. : : The following results are for two independent samples taken from two populations. Ex

cel 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 -
Mathematics
1 answer:
I am Lyosha [343]3 years ago
4 0

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.

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