Chapter 8.1, Problem 23P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Identify the tail of test.

Answer to Problem 23P

The level of significance is 0.01.

The null hypothesis is $H_0:\mu =11$.

The alternative hypothesis is $H_1:\mu \ne 11$.

The tail of the test is two-tailed.

Explanation of Solution

Calculation:

Let $\mu$ denotes the average wheat hail damage.

From the given information the value of $\alpha$ is 0.01, and the percentage of wheat crop lost to hail in County W is different (either way) from the national mean of 11%.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:\mu =11$

Alternative hypothesis:

$H_1:\mu \ne 11$

In this situation the alternative hypothesis is not equal indicates that the test is two-tailed test.

Hence, the tail of the test is two-tailed test.

(b)

To determine

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the z value of the sample test statistic.

Answer to Problem 23P

The sampling distribution to be used is $\overline{x}$ distribution.

The $\overline{x}$ sampling distribution is chosen because the x distribution is normal and population standard deviation is known.

The z value of the sample test statistic is 1.20.

Explanation of Solution

Calculation:

Conditions:

• When the x distribution considered in the study has the normal distribution with the known population standard deviation $\sigma$ then the sampling distribution $\overline{x}$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
• When the x distribution considered in the study is not normally distributed and the population standard deviation $\sigma$ is known then the sampling distribution $\overline{x}$ has normal distribution if the sample size n is greater than or equal 30. That is, $n\ge 30$.

Test statistic for z:

The z statistic value for sample test statistic $\overline{x}$ is,

$z=\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}$

In the formula $\overline{x}$ is mean of a simple random sample, $\mu$ is value stated in $H_0$, $\sigma$ is known standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation $\sigma =5.0$. The x distribution considered in the study is normally distributed and population standard deviation is known. Hence, the sampling distribution to be used is $\overline{x}$ distribution.

Z-statistic:

Substitute $\overline{x}$ as 12.5, $\mu$ as 11, $\sigma$ as 5.0, and n as 16 in the test statistic formula

$\begin{array}{c}z=\frac{12.5-11.0}{\left(\frac{5.0}{\sqrt{16}}\right)}\\ =\frac{1.5}{1.25}\\ =1.20\end{array}$

Hence, the sample test statistic z is 1.20.

(c)

To determine

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

Answer to Problem 23P

The P-value is 0.2302.

Explanation of Solution

Calculation:

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Both Tails, for the region of the curve to shade.
• Enter the X value as 1.20.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.1151 which is one sided value.

The two-tailed P-value is,

$\begin{array}{c}P\text{-value}=2\times 0.1151\\ =0.2302\end{array}$

Hence, the P-value of the test statistic is 0.2302.

(d)

To determine

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

Answer to Problem 23P

The null hypothesis is failed to be rejected.

The data is not statistically significant at level 0.01.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.2302.

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.2302 and the level of significance is 0.01.

The P-value is greater than the level of significance.

That is, $0.2302\left(=P\text{-value}\right)>0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is failed to be rejected.

Hence, the data is not statistically significant at level 0.01.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

From part (d), the null hypothesis is failed to be rejected. This shows that, the average hail in County W is not different (either way) from the national mean of 11% at 0.01 level of significance.