9.95 Two-Tailed Hypothesis Tests and CIs. 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 $H_a:\mu \ne {\mu }_0$if and only if ${\mu }_0$ lies outside the $(1-\alpha )$-level confidence interval for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean $z$-interval (Procedure 8.1 on page 322 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.84
b. Exercise 9.87

To 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 $9.84$.

The null hypothesis is indicated as follows:
$H_0:\mu ={\mu }_0$
The alternative hypothesis is indicated as follows:
$H_0:\mu \ne {\mu }_0$
Calculate the confidence interval by using MINITAB.
The output of MINITAB will be:
One-Sample Z: PERIODS

The $95\%$confidence interval from the MINITAB output is $(18.064,21.207)$.

The population mean $(=23)$ is not in the middle of the range.
As a result, the null hypothesis is rejected at the $5\%$ level.
The findings offered enough information to conclude that the average lactation time of grey seals is longer than $23$days at the $5\%$ level.
Hypothesis test for exercise $9.84$ as follows:
The test statistic's value is $-4.20$, and the $P$-value is $0$.
The $P$-value is lower than the level of significance in this case.
in other words, $P(=0)<\alpha (=0.05)$
At a $5\%$ level, the null hypothesis is rejected.
The data was sufficient to infer that the average lactation time of grey seals differed from $23$days at a $5\%$ level.
As a result, both conclusions are the same. In other words the conclusion for the confidence interval and the hypotheses test are the same.

To 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 $9.87$.

The null hypothesis is indicated as follows:
$H_0:\mu ={\mu }_0$
The alternative hypothesis is indicated as follows:
$H_0:\mu \ne {\mu }_0$

Calculate the confidence interval by using MINITAB.
The output of MINITAB will be:
One-Sample Z:

The $95\%$confidence interval for the MINITAB output is $(16.624,18.976)$.

The population mean $(16.7)$ is in the middle of the range.
As a result, at the $5\%$ level, the null hypothesis is not rejected.
The data does not support the conclusion that the average length of incarceration for motor vehicle theft offenders in Sydney differs from the national average in Australia.
Hypothesis test for exercise $9.87$ as follows:
The test statistic's value is $1.83$, and the P-value is $0.067$.
The P-value is higher than the level of significance in this case.
$P(=0.067)>\alpha (=0.05)$, for example.
At the $5\%$ level, the null hypothesis is not rejected.
The data does not support the conclusion that the average length of incarceration for motor vehicle theft offenders in Sydney differs from the national average in Australia.
As a result, both conclusions are the same. In other words the conclusion for the confidence interval and the hypotheses test are the same.