Question

For each of the following situations, determine the p-value and make the appropriate conclusion.

a.\({H_0}:\mu \le 25\),\({H_a}:\mu > 25\),\(\alpha = 0.01\),\(z = 2.02\)

b.\({H_0}:\mu \ge 6\),\({H_a}:\mu < 6\),\(\alpha = 0.05\),\(z = - 1.78\)

c.\({H_0}:\mu = 110\),\({H_a}:\mu \ne 110\),\(\alpha = 0.1\),\(z = - 1.93\)

d. \({H_0}:\mu = 10\), \({H_a}:\mu \ne 10\), \(\alpha = 0.05\), \(z = 1.96\)

Short answer

a. The p-value is 0.0217; do not reject the null hypothesis.

b. The p-value is 0.0375; reject the null hypothesis.

c. The p-value is 0.0536; reject the null hypothesis.

d. The p-value is 0.05; do not reject the null hypothesis.

Step-by-step solution

Given information

Four different hypothesis testing problems are provided. The values of the test statistics and significance level for the respective hypothesis are given.

Computing the p-value for test statistic z=2.02

a.

The provided scenario is the one-sided right-tailed test.

The p-value for the right-tailed test is:

\(\begin{aligned}p &= P\left( {Z > 2.02} \right)\\ &= 1 - P\left( {Z \le 2.02} \right)\\ &= 1 - 0.9783\\ &= 0.0217\end{aligned}\).

The z-table is used to obtain the probability of a z-score less than or equal to 2.02.

The p-value is greater than 0.01; therefore, do not reject the null hypothesis.

Computing the p-value for test statistic z=-1.78

b.

The p-value for the left-tailed test is:

\(\begin{aligned}p &= P\left( {Z < - 1.78} \right)\\ &= 0.0375\end{aligned}\).

The value at the intersection of -1.70 and 0.08 is the required probability in the z-table.

The p-value is less than 0.05; therefore, reject the null hypothesis.

Computing the p-value for test statistic z=-1.93

c.

The p-value for the two-tailed test is:

\(\begin{aligned}p &= 2P\left( {Z < - 1.93} \right)\\ &= 2 \times 0.0268\\ &= 0.0536\end{aligned}\).

The value at the intersection of -1.90 and 0.03 is the required probability in the z-table.

The p-value is less than 0.1; therefore, reject the null hypothesis.

Computing the p-value for test statistic z=1.96

d.

The p-value for the two-tailed test is:

\(\begin{aligned}p &= 2P\left( {Z > 1.96} \right)\\ &= 2\left( {1 - P\left( {Z \le 1.96} \right)} \right)\\ &= 2\left( {1 - 0.9750} \right)\\ &= 2 \times 0.0250\\ &= 0.05\end{aligned}\).

The z-table is used to obtain the probability of a z-score less than or equal to 1.96.

The p-value equals the significance level \(\alpha = 0.05\); therefore, do not reject the null hypothesis.