Question

In: Statistics and Probability

Consider the following hypothesis test. H0: μ ≤ 12 Ha: μ > 12 A sample of...

Consider the following hypothesis test.

H0: μ ≤ 12

Ha: μ > 12

A sample of 25 provided a sample mean x = 14 and a sample standard deviation s = 4.45.

Compute the value of the test statistic. (Round your answer to three decimal places.)

Use the t distribution table to compute a range for the p-value.

p-value > 0.2000.

100 < p-value < 0.200    

0.050 < p-value < 0.1000

.025 < p-value < 0.0500

.010 < p-value < 0.025

p-value < 0.010

At α = 0.05, what is your conclusion?

Reject H0. There is sufficient evidence to conclude that μ > 12.

Do not reject H0. There is insufficient evidence to conclude that μ > 12.

Reject H0. There is insufficient evidence to conclude that μ > 12.

Do not reject H0. There is sufficient evidence to conclude that μ > 12.

What is the rejection rule using the critical value? (If the test is one-tailed, enter NONE for the unused tail. Round your answer to three decimal places.)

test statistic≤ ?

test statistic≥ ?

What is your conclusion?

Reject H0. There is sufficient evidence to conclude that μ > 12.

Do not reject H0. There is insufficient evidence to conclude that μ > 12.

Reject H0. There is insufficient evidence to conclude that μ > 12.

Do not reject H0. There is sufficient evidence to conclude that μ > 12.

Solutions

Expert Solution

Test statistics

The value of the test statistic is 2.247

Degrees of Freedom df = n-1 = 25-1 = 24

P-value corresponding to t = 2.247 and df = 24 for a right tailed test is obtained using p-value calculator. Screenshot below:

P-value = 0.0171

Thus, correct answer is: .010 < p-value < 0.025

Since p-value = 0.0171 < α = 0.05, we reject null hypothesis H0. There is sufficient evidence to conclude that μ > 12.

Critical t value corresponding to df = 24 and α = 0.05 for a right-tailed test is obtained using critical t value calculator. screenshot below:

Critical t-value = 1.711

Since it is a right tailed test so rejection region is

test statistic ≤ NONE

test statistic ≥ 1.711

In this case test statistic = 2.247 ≥ Critical t value = 1.711, so we reject H0. There is sufficient evidence to conclude that μ > 12.


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