Chapter 9.3, Problem 69E

(a)

To determine

To Explain: the sample result gives some evidence for carrying out the test are met.

Answer to Problem 69E

Sample mean 62.8 differs from 64

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:\mu =64\\ H_1:\mu \ne 64\\ \alpha =0.05\\ n=25\\ \overline{x}=62.8\\ s=5.36\end{array}$

It is observed that the sample mean 62.8 differs from 64, which agrees with the alternative hypothesis that the mean is not 64 and therefore the sample result gives some proof for the alternative hypothesis.

(b)

To determine

To Calculate: the standardized test static and P-value.

Answer to Problem 69E

t= -1.119

0.20 < P < 0.30 or

P=0.2742

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:\mu =64\\ H_1:\mu \ne 64\\ \alpha =0.05\\ n=25\\ \overline{x}=62.8\\ s=5.36\end{array}$

Formula used:

  $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

Calculation:

The t-test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{62.8-64}{5.36/\sqrt{25}}\\ =-1.119\end{array}$

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. $df=n-1=25-1=24.$

  $0.20=2\left(0.10\right)<P<2\left(0.15\right)=0.30$

Command for the Ti83/84-calculator: 2*tcdf (-1E99,-1.119, 24) which will return a P-value of 0.2742.