Chapter 8.2, Problem 63E

a.

To determine

State the null and alternate hypotheses.

Answer to Problem 63E

Null hypothesis: $H_0:\mu =225$.

Alternate hypothesis: $H_1:\mu \ne 225$.

Explanation of Solution

Calculation:

The output obtained using MINITAB for the hypothesis test of a population mean is given.

From the output, the given information is sample size $n=50$, sample mean $\overline{x}=235.32$, standard error of mean $SE\left(\overline{x}\right)=4.9497$, confidence interval (CI)$95\%\text{ CI}=\left(225.619,245.021\right)$, test statistic value $z=2.085$, P-value $P=0.037$.

Denote $\mu$ as the true population mean.

The given test hypotheses are:

Null hypothesis:

 $H_0:\mu =225$.

That is, the true population mean is 225.

Alternate hypothesis:

$H_1:\mu \ne 225$.

That is, the true population mean is different from 225.

b.

To determine

Find the value of test statistic.

Answer to Problem 63E

The value of test statistic z is 2.085.

Explanation of Solution

Calculation:

Test statistic:

The z-test statistic is:

$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$,

Where, $\overline{x}$ be the sample mean, $\mu$ be the hypothesized mean and $\sigma$ be the standard deviation and n be the sample size.

From the given output, the value of test statistic z is 2.085.

c.

To determine

Find the P-value.

Answer to Problem 63E

The P-value is 0.037.

Explanation of Solution

Calculation:

P-value:

P-value is defined as the probability of obtaining a result at least as extreme as the observed value, given that the null hypothesis, $H_0$ is true. The test is said to be statistically significant, when $P\text{-value}<\alpha$, for level of significance $\alpha$.

From the given output, the P-value is 0.037.

d.

To determine

Decide whether the null hypothesis $H_0$ is rejected at $\alpha =0.05$ level.

Answer to Problem 63E

The null hypothesis $H_0$ is rejected at $\alpha =0.05$ level.

Explanation of Solution

It is given that the significance level $\alpha =0.05$.

From previous part (a), the P-value is 0.037.

Decision based on the P-value method level of significance $\alpha$:

• If $P-\text{value}\le \alpha$, reject $H_0$.
• If $P-\text{value}>\alpha$, fail to reject $H_0$.

Conclusion:

The significance level is, $\alpha =0.05$ and the P-value is 0.037.

Here, the P-value of 0.037 is less than the significance level 0.05.

That is, $P\text{-value}\left(=0.037\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

e.

To determine

Decide whether the null hypothesis $H_0$ is rejected at $\alpha =0.01$ level.

Answer to Problem 63E

The null hypothesis $H_0$ is not rejected at $\alpha =0.01$ level.

Explanation of Solution

It is given that the significance level $\alpha =0.01$.

From previous part (a), the P-value is 0.037.

Decision based on the P-value method level of significance $\alpha$:

• If $P-\text{value}\le \alpha$, reject $H_0$.
• If $P-\text{value}>\alpha$, fail to reject $H_0$.

Conclusion:

The significance level is, $\alpha =0.01$ and the P-value is 0.037.

Here, the P-value of 0.037 is greater than the significance level 0.01.

That is, $P\text{-value}\left(=0.037\right)>\alpha \left(=0.01\right)$.

Therefore, the null hypothesis is not rejected.

f.

To determine

Find the value of test statistic.

Answer to Problem 63E

The value of test statistic is 1.07.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis:

 $H_0:\mu =230$.

That is, the true population mean is 230.

Alternate hypothesis:

$H_1:\mu >230$.

That is, the true population mean is greater than 230.

Test statistic:

The z-test statistic is:

$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$,

Where, $\overline{x}$ be the sample mean, $\mu$ be the hypothesized mean and $\sigma$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

• Choose Stat > Basic Statistics > 1-Sample Z.
• In Summarized data select the Sample size as 50, Sample mean as 235.32.
• Select Known standard deviation as 35.
• Enter Hypothesized mean as 230.
• Select Options.
• Choose $\text{Mean}>\text{hypothesized mean}$ in Alternative hypothesis.
• Click OK in all dialogue boxes.

The output using Minitab is given below:

From the MINITAB output, the test statistic, that is, the z-value is 1.07.

Thus, the value of test statistic z is 1.07.

g.

To determine

Identify the P-value.

Answer to Problem 63E

The P-value is 0.1412.

Explanation of Solution

From the previous part (f), it is observed from the MINITAB output, the P-value is 0.1412.

Thus, the P-value is 0.1412.

h.

To determine

Decide whether the null hypothesis $H_0$ in the part (f) is rejected at $\alpha =0.05$ level.

Answer to Problem 63E

The null hypothesis $H_0$ is not rejected at $\alpha =0.05$ level.

Explanation of Solution

It is given that the significance level $\alpha =0.05$.

From previous part (a), the P-value is 0.1412.

Decision based on the P-value method level of significance $\alpha$:

• If $P-\text{value}\le \alpha$, reject $H_0$.
• If $P-\text{value}>\alpha$, fail to reject $H_0$.

Conclusion:

The significance level is, $\alpha =0.05$ and the P-value is 0.037.

Here, the P-value of 0.1412 is greater than the significance level 0.05.

That is, $P\text{-value}\left(=0.1412\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.