Chapter 9.3, Problem 71E

(a)

To determine

To Calculate: the standardized test statistic.

Answer to Problem 71E

t= 2.409

Explanation of Solution

Given:

  $\begin{array}{c}n=45\\ \overline{x}=125.7\\ s=29.8\end{array}$

Formula used:

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

Calculation:

Find the hypotheses:

  $\begin{array}{c}{H}_0:\mu =115\\ H_1:\mu >115\end{array}$

The test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{125.7-115}{29.8/\sqrt{45}}\\ =2.409\end{array}$

(b)

To determine

To find: and interpret the P-value.

Answer to Problem 71E

  $0.01<P<0.02$

There is a 1.012% possibility of getting a sample mean SSHA score of 125.7 or higher in a sample of 45 students, when the population mean SSHA score is 115.

Explanation of Solution

Given:

  $\begin{array}{c}{H}_0:\mu =115\\ H_1:\mu >115\\ \alpha =0.05\\ n=45\\ \overline{x}=125.7\\ s=29.8\end{array}$

Formula used:

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

Calculation:

The test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{125.7-115}{29.8/\sqrt{45}}\\ =2.409\end{array}$

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. $df=n-1=45-1=44.$ Since the table is not having row with $df=44$ , so will use $df=40$ instead.

  $0.01<P<0.02$

Command Ti83/84-calculator: tcdf(2.409, 1E99, 44) which will return a P-value of 0.01012. it could replace 1E99 by any other very large positive number.

There is a 1.012% possibility of getting a sample mean SSHA score of 125.7 or higher in a sample of 45 students, when the population mean SSHA score is 115.

(c)

To determine

To find: the conclusion would make.

Answer to Problem 71E

There is enough convincing proof that the mean SSHA score in the population of students at her college who are at least 30 years old exceeds 115.

Explanation of Solution

Given:

  $\begin{array}{c}{H}_0:\mu =115\\ H_1:\mu >115\\ \alpha =0.05\\ n=45\\ \overline{x}=125.7\\ s=29.8\end{array}$

Formula used:

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

Calculation:

Determine the value of the test statistic:

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{125.7-115}{29.8/\sqrt{45}}\\ =2.409\end{array}$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. $df=n-1=45-1=44.$ Since the table is not having row with $df=44$ , so will use $df=40$ instead.

  $0.01<P<0.02$

Command Ti83/84-calculator: tcdf(2.409, 1E99, 44) which will return a P-value of 0.01012. Note: it could replace 1E99 by any other very large positive number.

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

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

There is enough convincing proof that the mean SSHA score in the population of students at her college who are at least 30 years old exceeds 115.