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 \(9.87\)

The null hypothesis.

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

Alternative hypothesis

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

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.

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.