Q. 9.96 - Elementary Statistics

Question

9.96 Left-Tailed Hypothesis Tests and Cls. In Exercise 8.105 on page 335, we introduced one-sided one-mean z-intervals. The following relationship holds between hypothesis tests and confidence intervals for one-mean $z$-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 $(1-\alpha )$-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.
a. Exercise $9.85$
b. Exercise $9.86$

Step-by-Step Solution

Answer

(a) The null hypothesis $H_0:\mu =18\mathrm{mg}$ is rejected in support of the alternative hypothesis $H_a:\mu <18\mathrm{mg}$.

(b) The null hypothesis $H_0:\mu =55$ years is not rejected in support of the alternative hypothesis $H_a:\mu <55$ years.

Part (a) Step 1: Given information

To 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 $9.85$.

Part (a) Step 2: Explanation

Since, the population standard deviation as $\sigma =4.2\mathrm{mg}$

The null hypothesis is determined as follows:
$H_0:\mu ={\mu }_0$

$H_0:\mu =18\mathrm{mg}$

The alternative hypothesis is determined as follows:
$H_a:\mu <{\mu }_0$

$H_a:\mu <18\mathrm{mg}$

Then the significance level $=1\%$ i.e. $\alpha =0.01$

Sample size,$n=45$.
Sample mean, $\overline{x}=16.68$
Calculate the population mean's $100(1-\alpha ) \%$ upper confidence bound (one sided one mean z-interval) using MINITAB.
The population mean's $99\%$ upper confidence bound is $16.137mg$.
The null hypothesis mean $\mu =18 mg$ is higher than the upper confidence bound obtained.
It shows that we have rejected the null hypothesis $H_0:\mu =18\mathrm{mg}$ in favor of the alternative hypothesis $H_a:\mu <18\mathrm{mg}$ at the $1\%$ level of significance.
As a result, the null hypothesis $H_0:\mu =18\mathrm{mg}$ is rejected in support of the alternative hypothesis $H_a:\mu <18\mathrm{mg}$.

Part (b) Step 1: Given information

To 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 $9.86$.

Part (b) Step 2: Explanation

Since, the population standard deviation $\sigma =6.8$ years .

The null hypothesis is determined as follows:

$H_0:\mu ={\mu }_0$

$H_0:\mu =55$ years.

And the alternative hypothesis is determined as follows: 

$H_a:\mu <{\mu }_0$

$H_a:\mu <55$ years.

And the significance level $=1\%$ i.e. $\alpha =0.01$.
Then the sample size, $n=21$.
The sample mean, $\overline{x}=52.5$
Calculate the population mean's $100(1-\alpha )$ upper confidence bound (one sided one mean z-interval) using MINITAB.
The population mean has a $99\%$ upper confidence bound of $55.95$ years.
The null hypothesis mean $H_0:\mu =55$ years is lower than the upper confidence bound obtained.
It shows that we have not rejected the null hypothesis $H_0:\mu =55$ years in support of the alternative hypothesis $H_a:\mu =55$ years at the $1$ percent level of significance.
As a result, the null hypothesis $H_0:\mu =55$ years is not rejected in support of the alternative hypothesis $H_a:\mu <55$ years.