Chapter 10.3, Problem 3P

(a)

To determine

Find the level of significance.

State the null and alternative hypothesis.

Answer to Problem 3P

The level of significance is 0.05.

Null hypothesis: $H_0:{\sigma }^2=42.3$.

Alternative hypothesis: $H_1:{\sigma }^2>42.3$.

Explanation of Solution

Calculation:

Let ${\sigma }^2$ denotes the variance in the new section, the variable x denotes the number of sites per transect.

From the given information the value of $\alpha$ is 0.05, and to test the claim that the variance in the new section is greater than 42.3.

Hence, the level of significance is 0.05.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:{\sigma }^2=42.3$

That is, the variance in the new section is equal to 42.3.

Alternative hypothesis:

$H_1:{\sigma }^2>42.3$

That is, the variance in the new section is greater than 42.3.

(b)

To determine

Find the value of the chi-square statistic for the sample.

Find the degrees of freedom.

Mention the assumptions made about the original distribution.

Answer to Problem 3P

The value of the chi-square statistic for the sample is 23.98.

The degrees of freedom are 22.

Explanation of Solution

Calculation:

Test statistic:

The sample chi-square test statistic is,

${\chi }^2=\frac{\left(n-1\right)s^2}{{\sigma }^2}$

In the formula ${\sigma }^2$ is given in the null hypothesis $H_0$, $s^2$ is the sample variance, n denotes the sample size, with degrees of freedom $d.f.=n-1$.

From the given information, a random sample of 23 transects gave a sample variance $s^2=46.1$ for the number of sites per transect, and the population variance is ${\sigma }^2=42.3$.

Substitute 23 for n, 46.1 for $s^2$, 42.3 for ${\sigma }^2$ in the test statistic formula.

$\begin{array}{c}{\chi }^2=\frac{\left(23-1\right)46.1}{42.3}\\ =\frac{1,014.2}{42.3}\\ =23.98\end{array}$

Hence, the value of the chi-square statistic for the sample is 23.98.

Substitute 23 for n in the degrees of freedom formula.

$\begin{array}{c}d.f.=23-1\\ =22\end{array}$

Hence, the degrees of freedom are 22.

The assumption made about the original distribution is assuming a normal population distribution.

(c)

To determine

Find the P-value of the sample test statistic.

Answer to Problem 3P

The P-value is 0.3483.

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 ‘Chi-Square’ distribution.
• In Degrees of freedom, enter the value as 22.
• Click the Shaded Area tab.
• Choose X Value and Right Tail, for the region of the curve to shade.
• Enter the X value as 23.98.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.3483.

Hence, the P-value is 0.3483.

(d)

To determine

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

Answer to Problem 3P

The null hypothesis of independence is not rejected.

Explanation of Solution

Calculation:

From part (c), the P-value is 0.3483.

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 0.3483 and the level of significance is 0.05.

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

That is, $0.3483\left(=P\text{-value}\right)>0.05\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is not rejected.

Hence, the null hypothesis of independence is not rejected.

(e)

To determine

Interpret the conclusion in the context of the application.

Explanation of Solution

Calculation:

From part (d), the null hypothesis is not rejected. This shows that, there is no sufficient evidence that the variance in the new section is greater than 42.3 at level of significance 0.05.

(f)

To determine

Find the 95% confidence interval for the population variance.

Interpret the confidence interval.

Answer to Problem 3P

The 95% confidence interval for the population variance is $27.57<{\sigma }^2<92.37$.

Explanation of Solution

Calculation:

Confidence interval:

The confidence interval for the population variance ${\sigma }^2$ is,

$\frac{\left(n-1\right)s^2}{{\chi }_U^2}<{\sigma }^2<\frac{\left(n-1\right)s^2}{{\chi }_L^2}$

In the formula c is confidence level, $s^2$ is the sample variance, n denotes the sample size, ${\chi }_U^2$ is chi-square value using $d.f.=n-1$ and right-tail area $\frac{\left(1-c\right)}{2}$, and ${\chi }_L^2$ is chi-square value using $d.f.=n-1$ and right-tail area $\frac{\left(1+c\right)}{2}$.

Critical value for ${\chi }_U^2$:

The right-tail area is,

$\begin{array}{c}\frac{\left(1-c\right)}{2}=\frac{\left(1-0.95\right)}{2}\\ =\frac{0.05}{2}\\ =0.025\end{array}$

Use the Appendix II: Tables, Table 7: The ${\chi }^2$ distribution:

• In d.f. column locate the value 22.
• In Right-tail Area row of locate the value 0.025.
• The intersecting value of row and columns is 36.78.

The value of ${\chi }_U^2$ is 36.78.

Critical value for ${\chi }_L^2$:

The right-tail area is,

$\begin{array}{c}\frac{\left(1+c\right)}{2}=\frac{\left(1+0.95\right)}{2}\\ =\frac{1.95}{2}\\ =0.975\end{array}$

Use the Appendix II: Tables, Table 7: The ${\chi }^2$ distribution:

• In d.f. column locate the value 22.
• In Right-tail Area row of locate the value 0.975.
• The intersecting value of row and columns is 10.98.

The value of ${\chi }_L^2$ is 10.98.

Substitute 23 for n, 46.1 for $s^2$, 36.78 for ${\chi }_U^2$, and 10.98 for ${\chi }_L^2$ in the confidence interval formula.

$\begin{array}{c}\frac{\left(23-1\right)36.78}{36.78}<{\sigma }^2<\frac{\left(23-1\right)36.78}{10.98}\\ \frac{1,014.2}{36.78}<{\sigma }^2<\frac{1,014.2}{10.98}\\ 27.57<{\sigma }^2<92.37\end{array}$

Hence, the 95% confidence interval for the population variance is $27.57<{\sigma }^2<92.37$.

The confidence interval can be interpreted as; there is 95% confidence that the population variance in the new section lies within the interval 27.57 to 92.37.