Question

As we mentioned on page \(378\), the following relationship holds between hypothesis tests and confidence intervals for one mean \(z-\)procedures: For a two-tailed hypothesis test at the significance level \(\alpha\), the null hypothesis \(H_{0}:\mu =\mu_{0}\) will be rejected in favor of the alternative hypothesis \(H_{a}:\mu \neq \mu_{0}\) if and only if \(\mu_{0}\) lies outside the \((1-\infty)\) level confidence interval for \(\mu\). In each case, illustrate the preceding relationship by obtaining the appropriate one-mean \(z-\)interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise \(9.84\)

b. Exercise \(9.87\)

Short answer

Part a. Both conclusions are same. i.e. the conclusion for confidence interval is same as the conclusion for hypotheses test.

Part b. Both conclusions are same. i.e. the conclusion for confidence interval is same as the conclusion for hypotheses test.

Step-by-step solution

Part a. Step 1. Given information

The null hypothesis.

\(H_{0}:\mu =\mu_{0}\)

Alternative hypothesis

\(H_{0}:\mu \neq \mu_{0}\)

Part a. Step 2. Calculation

Calculate the confidence interval by using MINITAB.

MINITAB output:

One-Sample Z: PERIODS

From the MINITAB output, the \(95%\) confidence interval is \((18.064, 21.207\)

The population mean \((=23)\) does not lie between lower and upper limit. Therefore, the null hypothesis is rejected at \(5%\) level.

The data provided sufficient evidence to conclude that the mean lactation period of grey seals differs from \(23\) days at \(5%\) level.

Hypothesis test

Problem \(9.84E\)

The value of test statistic is \(-4.20\) and \(P-\)value is \(0\).

Here, the \(P-\)value is less than the level of significance. i.e. \(P(=0)<\alpha (=0.5)\)

The null hypothesis is rejected at \(5%\) level.

The data provided sufficient evidence to conclude that the mean lactation period of grey seals differs from \(23\) days at \(5%\) level.

Thus, both conclusions are same. i.e. the conclusion for confidence interval is same as the conclusion for hypotheses test.

Part b. Step 1. Calculation

Calculate the confidence interval by using MINITAB.

MINITAB output:

One-Sample Z:

From the MINITAB output, the \(95%\) confidence interval is \((16.624, 18.976)\)

The population means \((16.7)\) lies between lower and upper limit. Therefore, the null hypothesis is not rejected at \(5%\) level.

The data does not provided sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia.

Hypothesis test

Problem \(9.87E\)

The value of test statistic is \(1.83\) and \(P-\)value is \(0.067\).

Here, the \(P-\)value is greater than the level of significance. i.e. \(P(=0.067)>\alpha (=0.05)\)

The null hypothesis is not rejected at \(5%\) level.

The data does not provided sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia.

Thus, both conclusions are same. i.e. the conclusion for confidence interval is same as the conclusion for hypotheses test.