Chapter 9, Problem R9.3RE

(a)

To determine

To find: the standardized test statistic and P-value in each setting and make an appropriate conclusion for part (a)

Answer to Problem R9.3RE

t= -1.311

  $0.10=2\left(0.05\right)<P<2\left(0.10\right)=0.20$

Or P=0.19622

There is no enough convincing proof that the true mean height of this year’s female graduates from the large high school differs from the national average.

Explanation of Solution

Given:

  $\begin{array}{l}\alpha =0.05\\ n=48\\ \overline{x}=63.5\\ s=3.7\end{array}$

Claim is that the mean is different from 64.2

Calculation:

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is that the population mean is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

  $\begin{array}{l}{H}_0:\mu =64.2\\ H_1:\mu \ne 64.2\end{array}$

The P-value is the probability of getting the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. $df=n-1=48-1=47$ . The test is two-tailed (due to the $\ne$ in the alternative hypothesis $H_1$ ). Although the table is not having $df=47$ , so using $df=40$ instead.

  $0.10=2\left(0.05\right)<P<2\left(0.10\right)=0.20$

Command Ti83/84- calculator: 2*tcdf (-1E99, -1.311, 47) which would return a P-value of 0.19622. Note: it could replace -1E99 by any other very small negative number.

If the P-value is lesser than the significance level $\alpha$ , then reject the null hypothesis

  $P>0.05\Rightarrow \text{Fail to reject }{H}_0$

There is no enough convincing proof that the true mean height of this year’s female graduates from the large high school differs from the national average.

(b)

To determine

To find: the standardized test statistic and P-value in each setting and make an appropriate conclusion for part (b)

Answer to Problem R9.3RE

Z=2.07

P=0.0192

There is enough convincing proof that the true proportion of students in their school who have played in the rain is greater than 0.25.

Explanation of Solution

Given:

  $\begin{array}{l}\alpha =0.05\\ n=48\\ x=28\end{array}$

Formula used:

  $\begin{array}{c}\widehat{p}=\frac{x}{n}\\ z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\end{array}$

Calculation:

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis statement is that the population proportion is equal to the value given in the claim. If the null hypothesis is the claim, then the alternative hypothesis statement is the opposite of the null hypothesis.

  $\begin{array}{c}{H}_0:p=0.25\\ H_1:p>0.25\end{array}$

The sample proportion is

  $\widehat{p}=\frac{x}{n}=\frac{28}{80}=0.35$

The test-statistic is

  $\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.35-0.25}{\sqrt{\frac{0.25\left(1-0.25\right)}{80}}}\\ =2.07\end{array}$

The P-value is the probability of getting the value of the test statistic or a value more extreme, when the null hypothesis is true. Find the P-value using the normal probability table $\begin{array}{c}P=P\left(Z>2.07\right)\\ =1-P\left(Z<2.07\right)\\ =1-0.9808\\ =0.0192\end{array}$

If the P-value is lesser than the significance level $\alpha$ , then reject the null hypothesis:

  $P<0.05\Rightarrow \text{Reject }{H}_0$

There is enough convincing proof that the true proportion of students in their school who have played in the rain is greater than 0.25.