Question

Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and T test statistic. Also determine if the null hypothesis would be rejected at alpha = 0.05(a) HA : u > u0, N = 11, T = 1 :91(b) HA: < u0, N = 17, T = -3:45(c) HA: u ? u0 , N = 7, T = 0.83(d) HA: u > u0 , N = 28, T = 2.13

Answer 1

Answer:

(a) p-value = 0.043. Null hypothesis is rejected.

(b) p-value = 0.001. Null hypothesis is rejected.

(c) p-value = 0.444. Null hypothesis is not rejected.

(d) p-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a t-test for single mean is used.

The significance level for the test is α = 0.05.

The decision rule is:

If the p - value is less than the significance level then the null hypothesis will be rejected. And if the p-value is more than the value of α then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

Hₐ: μ > μ₀

The sample size is, n = 11.

The test statistic value is, t = 1.91 ≈ 1.90.

The degrees of freedom is, (n - 1) = 11 - 1 = 10.

Use a t-table t compute the p-value.

For the test statistic value of 1.90 and degrees of freedom 10 the p-value is:

The p-value is:

P (t₁₀ > 1.91) = 0.043.

The p-value = 0.043 < α = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

Hₐ: μ < μ₀

The sample size is, n = 17.

The test statistic value is, t = -3.45 ≈ 3.50.

The degrees of freedom is, (n - 1) = 17 - 1 = 16.

Use a t-table t compute the p-value.

For the test statistic value of -3.50 and degrees of freedom 16 the p-value is:

The p-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The p-value = 0.001 < α = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

Hₐ: μ ≠ μ₀

The sample size is, n = 7.

The test statistic value is, t = 0.83 ≈ 0.82.

The degrees of freedom is, (n - 1) = 7 - 1 = 6.

Use a t-table t compute the p-value.

For the test statistic value of 0.82 and degrees of freedom 6 the p-value is:

The p-value is:

P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.

The p-value = 0.444 > α = 0.05.

The null hypothesis is not rejected at 5% level of significance.

(d)

The alternate hypothesis is:

Hₐ: μ > μ₀

The sample size is, n = 28.

The test statistic value is, t = 2.13 ≈ 2.12.

The degrees of freedom is, (n - 1) = 28 - 1 = 27.

Use a t-table t compute the p-value.

For the test statistic value of 0.82 and degrees of freedom 6 the p-value is:

The p-value is:

P (t₂₇ > 2.12) = 0.022.

The p-value = 0.444 > α = 0.05.

The null hypothesis is rejected at 5% level of significance.