In Exercise 8.146 on page 345, we introduced one-sided one-mean t-intervals. The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a left-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 $H_a:\mu <{\mu }_0$if and only if ${\mu }_0$is greater than or equal to the $\left(1-\alpha \right)$- level upper confidence bound for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.117

Part (b): Exercise 9.118

On computing the 95%confidence interval,

1. Choose Stat>Basic Statistics>1-Sample t.
2. In Summarized data, enter the sample size 187 and mean 0.64.
3. In Standard deviation, enter the value 0.15.
4. In Perform hypothesis test, enter the test mean as 0.9.
5. Check Options, enter Confidence level as 95.
6. Choose less than in alternative.
7. Click OK in all dialogue boxes.

Test of mu$=0.9vs<0.9$

Hence, the population mean is 0.9 which lies above 95% upper bound. Therefore, the null hypothesis is rejected at 5% level.

Test of mu$=2vs<2$

Hence, from the MINITAB output, the 90% upper bound is 2.087.

Here, the population mean is 2, which lies below the 90% upper bound. Therefore, the null hypothesis is not rejected at 10% level.

Using the Hypothesis test,

From Problem 9-118E, the P-value is 0.255. The P-value is greater than the level of significance.

Therefore, the null hypothesis is not rejected. The data does not provide sufficient evidence to conclude that the mean fuel tank capacity of all dirt bikes is less than2 gallons.

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