Left-Tailed Hypothesis Tests and CIs. 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 α, the null hypothesis H0:μ=μ0 will be rejected in favor of the alternative hypothesis $H_o:\mu <{\mu }_0$ if and only if μ0 is greater than or equal to the (1−α)-level upper confidence bounif for μ. In each case, illustrate the preceding relationship by obtaininy the appropriate upper confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise9.117
b. Exercise9.118

To determine the hypothesis test's conclusion.

Given that, the expressions are $H_0:\mu ={\mu }_0$, $H_0:\mu <{\mu }_0$

Calculation:

Confidence interval:

To determine the hypothesis test's conclusion.

$0.255$Given that, $H_0:\mu ={\mu }_0$, $H_0:\mu <{\mu }_0$

Calculation:

Confidence interval,

MINITAB is used to calculate the $90\%$confidence interval.

Output for MINITAB is,
$2.087$ is the$90$percent upper bound.
The population mean is $2$, which is lower than the $90\%$ upper bound. As a result, at the ten percent level, the null hypothesis is not rejected.
Hypothesis test:
The P - value for issue $9118E$is . The significance level is bigger than the P-value.
As a result, the null hypothesis is not ruled out. As a result, the data does not support the conclusion that the average gasoline tank capacity of all dirt bikes is less than $2$litres.
As a result, both conclusions are the same. That is, the conclusion for the confidence interval and the hypotheses test are the same.