Chapter 10.2, Problem 10.37E

(a)

To determine

To Explain: the sampling distribution of the mean score and sketch the density curve of this distribution.

Answer to Problem 10.37E

Therefore the sampling distribution of the sample mean is about normal with mean 115 and standard deviation 1.2

Explanation of Solution

Given:

  $\begin{array}{c}{H}_0:\mu =115\\ H_0:\mu >115\\ \sigma =6\\ n=25\end{array}$

Calculation:

The sampling distribution of the sample mean $\overline{x}$ is about normal when the sample is large. The mean of the sampling distribution of the sample mean is equal to the population mean

  ${\mu }_{\overline{x}}=\mu =115$

The standard deviation of the sampling distribution is

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{6}{\sqrt{25}}\\ =\frac{6}{5}\\ =1.2\end{array}$

Therefore the sampling distribution of the sample mean is about normal with mean 115 and standard deviation 1.2

Graph:

  

(b)

To determine

To Explain: in simple language that one outcome is good evidence that the mean score of all older students is greater than 115 and the other outcome is not.

Explanation of Solution

Given:

  $\begin{array}{c}\overline{x}=118.6\\ \overline{x}=125.7\end{array}$

The sample mean $\overline{x}=125.7$ giving good evidence that the population mean is greater than 115, because the sample mean is removed from the peak in the drawn distribution that is unlikely that the population mean is certain 115.

The sample mean $\overline{x}=118.6$ does not giving good evidence that the population mean is greater than 115, because the sample mean is 3 standard deviations from the mean and even think that sample mean is unlikely to done by chance when the population mean is 115, it can happen.

Graph:

  

(c)

To determine

To Explain: the area under the curve that is the p-value for the sample result.

Answer to Problem 10.37E

Area right of 118.6

Explanation of Solution

Graph:

  

  $\overline{x}=118.6$

The P-value is the probability of getting a value more extreme or equal to the sample mean $\overline{x}$ , suppose that the null hypothesis is true.

Since the alternative hypothesis claims that the mean is large than 115, the P-value is then the area containing all sample means large than the sample mean of $\overline{x}=118.6$ .