9.128 Two-Tailed Hypothesis Tests and CIs. The following relationship holds between hypothesis tests and confidence intervals for one-mean $t$-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 $H_2:\mu >{\mu }_0$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 $t$-interval (Procedure 8.2 on page 338 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.113
b. Exercise 9.116

To illustrate the preceding relationship by obtaining the appropriate one-mean $t$-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise $9.113$.

Let, the $90\%$confidence interval is $(3.872,5.648)$.
The population's average is $4.55$, which is in the middle of the range.
As a result, the null hypotheses are not rejected at the $10\%$ level.
Therefore, the data does not provide adequate information to determine that the typical person's daily television viewing habits changed from $2005$ to last year.
The Hypothesis test as follows:
The null hypothesis as follows:

${\mathrm{H}}_0:\mu =4.55$
In $2005$, the average American watched $4.55$hours of television every day on average.
The alternative hypothesis as follows:
${\mathrm{H}}_0:\mu \ne 4.55$

In $2005$, the average American watched $4.55$ hours of television every day.
Calculate the test statistic's value.
The test statistic is worth $0.41$, while the $P$value is worth $0.687$.
If $P=\alpha$, the null hypothesis must be rejected.
The degree of importance is larger than the $P$- value of $0$.
$P(=0.687)<\alpha (=0.10)$
So, the null hypothesis is not rejected at the $10\%$ level.
As a result, the data does not provide adequate information to determine that the typical person's daily television viewing habits changed from 2005 to last year.
As a result, both conclusions are the same. That is, the conclusion for the confidence interval and the hypotheses test are the same.

To illustrate the preceding relationship by obtaining the appropriate one-mean $t$-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise $9.116$.

Let, the $90\%$confidence interval is $(1,777.9.2,067.6)$.
In between lower and higher limits, the population mean does not lie.
So, the null hypotheses are rejected at the $5\%$ level.
As a result, the statistics provide adequate evidence to establish that the northeast's mean annual expenditure on clothes and services in $2012$deviated from the national average of $\$1,736$.
Hypothesis test is calculated as follows:
The null hypothesis is calculated as follows:
${\mathrm{H}}_0:\mu =\$1,736$
The northeast's typical annual expenditure on clothes and services for consumer units in $2012$ was $\$1,736$, which is similar to the national mean.
The alternative hypothesis is calculated as follows:
${\mathrm{H}}_0:\mu \ne \$1,736$

The northeast's mean annual expenditure on clothes and services for consumer units was $\$1,736$in $2012$, compared to the national average of $\$1,736$.
The value of test statistic is $2.66$and $P$- value is $0.014$.
Rejection rule:
If $P\le \alpha$, then reject the null hypothesis.
Here, the $P$- value is $0.014$ is less than the level of significance is
$P(=0.0137)<\alpha (=0.05)$
Therefore, the null hypothesis is rejected at $5\%$ level.
As a result, the statistics provide adequate evidence to infer that the northeast's average annual expenditure on clothes and services in $2012$deviated from the national average of $\$1,736$.
Hence, both conclusions are the same. That is, the conclusion for the confidence interval and the hypotheses test are the same.