Chapter 9.3, Problem 70E

(a)

To determine

To Explain: the sample result gives some evidence for the alternative hypothesis.

Answer to Problem 70E

The sample mean 4.7 is less than 5.

Explanation of Solution

Given:

  $\begin{array}{c}{H}_0:\mu =5\\ H_1:\mu <5\\ \alpha =0.05\\ n=20\\ \overline{x}=4.7\\ s=0.74\end{array}$

It is observed that the sample mean 4.7 is smaller than 5, which agrees with the alternative hypothesis that the mean is less than 5 and therefore the sample result gives some proof for the alternative hypothesis.

(b)

To determine

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

Answer to Problem 70E

t= -1.813

  $0.25<P<0.05$

Or P=0.04283

Explanation of Solution

Given:

  $\begin{array}{c}{H}_0:\mu =5\\ H_1:\mu <5\\ \alpha =0.05\\ n=20\\ \overline{x}=4.7\\ s=0.74\end{array}$

Formula used:

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

Calculation:

The test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{4.7-5}{0.74/\sqrt{20}}\\ =-1.813\end{array}$

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. $df=n-1=20-1=19.$

  $0.25<P<0.05$

Command for Ti83/84-calculator: tcdf (-1E99,-1.813, 19) which will return a P-value of 0.04283. It could replace -1E99 by any other very small negative number.