Large sample inference: Let the test statistic, Z, have a normal distribution whenH0:μ=μ0istrue. Give as much information as you can about the p-value for each of the following situations:(a)Ha:μ > μ0, andZ= 2.0(b)Ha:μ < μ0, andZ=−2.4(c)Ha:μ6=μ0, andZ=−1.6(d)Ha:μ < μ0, andZ=−0.4.(e)Ha:μ < μ0andZ=−3.16(f)Ha:μ > μ0, andZ= 5.0(g)Ha:μ6=μ0, andZ= 2.8

The p-values are:

(a) p-value = 0.023        (e) p-value = 0.001

(b) p-value = 0.008        (f) p-value = 0

(c) p-value = 0.1096       (g) p-value = 0.005

(d) p-value = 0.3446

Step-by-step explanation:

The p-value is well-defined as per the probability, [under the null-hypothesis (H₀)], of attaining a result equivalent to or greater than what was truly observed.

(a)

The alternate hypothesis is:

Hₐ: μ > μ₀

The test statistic value is, Z = 2.0.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.023 implies that the probability of getting the mean value greater than or equal to μ₀ is 0.023.

(b)

The alternate hypothesis is:

Hₐ: μ < μ₀

The test statistic value is, Z = -2.4.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.008 implies that the probability of getting the mean value as extreme as μ₀ is 0.008.

(c)

The alternate hypothesis is:

Hₐ: μ ≠ μ₀

The test statistic value is, Z = -1.6.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.1096 implies that the probability of getting the mean value as extreme as μ₀ is 0.1096.

(d)

The alternate hypothesis is:

Hₐ: μ < μ₀

The test statistic value is, Z = -0.4.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.3446 implies that the probability of getting the mean value as extreme as μ₀ is 0.3446.

(e)

The alternate hypothesis is:

Hₐ: μ < μ₀

The test statistic value is, Z = -3.16.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.001 implies that the probability of getting the mean value as extreme as μ₀ is 0.001.

(f)

The alternate hypothesis is:

Hₐ: μ > μ₀

The test statistic value is, Z = 5.0.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0 implies that the probability of getting the mean value as extreme as μ₀ is 0.

(g)

The alternate hypothesis is:

Hₐ: μ ≠ μ₀

The test statistic value is, Z = 2.8.

Compute the p-value as follows:

*Use a z-table.

The p-value of 0.005 implies that the probability of getting the mean value as extreme as μ₀ is 0.005.