Chapter 10.4, Problem 5P

(a)

To determine

Check the variance of the first plot by using the calculator.

Check the variance of the second plot by using the calculator.

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 5P

The level of significance is 0.01.

Null hypothesis: $H_0:{\sigma }_1^2={\sigma }_2^2$.

Alternative hypothesis: $H_0:{\sigma }_1^2>{\sigma }_2^2$.

Explanation of Solution

Calculation:

Sample variance for first plot:

Use Ti 83 calculator to find the variance as follows:

• Select STAT > Edit > Enter the values of Wheat as L1.
• Click $\boxed{2^{\text{nd}}}$ button > STAT; take the arrow to the MATH menu, and then ‘8’ numbered key.
• Click $\boxed{2^{\text{nd}}}$ button; then ‘1’ numbered key to get L1, and close the ‘)’ bracket.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the variance of the first plot is approximately 0.332.

Sample variance for second plot:

Use Ti 83 calculator to find the variance as follows:

• Select STAT > Edit > Enter the values of Wheat as L2.
• Click $\boxed{2^{\text{nd}}}$ button > STAT; take the arrow to the MATH menu, and then ‘8’ numbered key.
• Click $\boxed{2^{\text{nd}}}$ button; then ‘2’ numbered key to get L2, and close the ‘)’ bracket.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the variance of the second plot is approximately 0.089.

The variance of the first plot is larger than second plot. The first plot must be considered as population 1 due to larger variance.

Let ${\sigma }_1^2$ denotes the variance in annual wheat production of the first plot, and ${\sigma }_2^2$ denotes the variance in annual wheat production of the second plot.

From the given information the value of $\alpha$ is 0.01, and to test the claim that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:{\sigma }_1^2={\sigma }_2^2$

That is, the population variance of annual wheat production for the first plot is equal to the second plot.

Alternative hypothesis:

$H_1:{\sigma }_1^2>{\sigma }_2^2$

That is, the population variance of annual wheat production for the first plot is larger than that for the second plot.

(b)

To determine

Find the value of the sample F statistic.

Find the degrees of freedom.

Mention the assumptions made about the original distribution.

Answer to Problem 5P

The value of the sample F statistic is 3.73.

The degrees of freedom for annual wheat production for the first plot are 15.

The degrees of freedom for annual wheat production for the second plot are 15.

Explanation of Solution

Calculation:

Test statistic:

The sample test statistic is,

$F=\frac{s_1^2}{s_2^2}$

In the formula $s_1^2$ is sample variance for population 1, $s_2^2$ is sample variance for population 2, $n_1,n_2$ denotes the sample sizes, with degrees of freedom for numerator is $d.f{.}_N=n_1-1$, and degrees of freedom for denominator is $d.f{.}_D=n_2-1$.

Substitute 0.332 for $s_1^2$, 0.089 for $s_2^2$ in the test statistic formula.

$\begin{array}{c}F=\frac{0.332}{0.089}\\ =3.73\end{array}$

Hence, the value of the sample F statistic is 3.73.

Substitute 16 for $n_1$ in the degrees of freedom for numerator formula.

$\begin{array}{c}d.f{.}_N=16-1\\ =15\end{array}$

Hence, the degrees of freedom for annual wheat production for the first plot are 15.

Substitute 16 for $n_2$ in the degrees of freedom for denominator formula.

$\begin{array}{c}d.f{.}_D=16-1\\ =15\end{array}$

Hence, the degrees of freedom for annual wheat production for the second plot are 15.

The two populations are assumed to be independent and normally distributed. The samples are selected randomly from each of the population.

(c)

To determine

Find the P-value of the sample test statistic.

Answer to Problem 5P

The P-value of the sample test statistic is 0.0076.

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 ‘F’ distribution.
• In Numerator df, enter the value as 15.
• In Denominator df, enter the value as 15.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 373.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0076.

Hence, the P-value of the sample test statistic is 0.0076.

(d)

To determine

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

Answer to Problem 5P

The null hypothesis is rejected.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.0076.

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 0.0076 and the level of significance is 0.01.

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

That is, $0.0076\left(=P\text{-value}\right)<0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the null hypothesis is rejected.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is rejected. This shows that, there is sufficient evidence that the population variance of annual wheat production for the first plot is larger than that for the second plot at level of significance 0.01.