As we mentioned on page 378, the following relationship holds between hypothesis test and confidence intervals for one-mean z-procedures: For a two-tailed hypothesis test at the significance level $\alpha$, the null hypothesis role="math" localid="1653038937481" $H_0:\mu ={\mu }_0$will be rejected in favor of the alternative hypothesis $H_a:\mu \ne {\mu }_0$if and only if ${\mu }_0$lies outside the $\left(1-\alpha \right)$-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.

Part (a): Exercise 9.84

Part (b): Exercise9.87

Consider the given question,

The null hypothesis is $H_0:\mu ={\mu }_0$.

The alternative hypothesis is$H_a:\mu \ne {\mu }_0$.

On using the MINITAB procedure,

1. Choose Stat>Basic Statistics>1-Sample Z.
2. In Samples in Column, enter the column of Period.
3. In Standard deviation, enter 3.
4. In Perform hypothesis test, enter the test mean as 23.
5. Check Options, enter Confidence level as 95.
6. Choose not equal is alternative.
7. Click OK in all dialogue boxes.

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

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

It can be concluded that the data provide sufficient evidence to conclude that the mean lactation period of grey seals differs from 23 days at 5% level.

Using the Hypothesis test,

The value of test statistics is -4.2 and the P-value is 0.

The P-value is less than the level of significance that is $P\left(=0\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected at 5% level.

Hence, the data provide sufficient evidence to conclude that the mean lactation period of grey seals differs from 23 days at 5% level.

Thus, both conclusions are same. It means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.

On using the MINITAB procedure,

1. Choose Stat>Basic Statistics>1-Sample Z.
2. In Samples in Column, enter the column of Period.
3. In Standard deviation, enter 3.
4. In Perform hypothesis test, enter the test mean as 23.
5. Check Options, enter Confidence level as 95.
6. Choose not equal is alternative.
7. Click OK in all dialogue boxes.

Hence, from the MINITAB output, the 95% confidence interval is $\left(16.624,18.976\right)$.

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

It can be concluded that the data does not provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia.

Using the Hypothesis test,

The value of test statistics is -4.2 and the P-value is 0.

The P-value is greater than the level of significance that is $P\left(=0.067\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected at 5% level.

Hence, the data does not provide 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. It means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.