Chapter 10, Problem 7CRP

(a)

To determine

(i)

Identify the problem based on the scenario.

Find the level of significance.

State the null and alternative hypothesis.

(ii)

Find the sample test statistic.

(iii)

Find the P-value of the sample test statistic.

(iv)

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

(v)

Interpret the conclusion in the context of the application.

Answer to Problem 7CRP

(i)

The problem is classified as chi-square test of variance.

The level of significance is 0.01.

Null hypothesis: $H_0:{\sigma }^2=1,040,400$.

Alternative hypothesis: $H_1:{\sigma }^2>1,040,400$.

(ii)

The sample test statistic is 51.03.

(iii)

The P-value of the sample test statistic is 0.007.

(iv)

The null hypothesis is rejected.

Explanation of Solution

Calculation:

(i)

A study is conducted by consumer agency for determining the blowout pressures of Soap Stone tires. The claim of the test is that the variance of blowout pressures is more than Soap Stone. The chi-square test of variance is appropriate because the variance of the population has to be tested.

Hence, the problem is classified as chi-square test of variance.

Let ${\sigma }^2$ denotes the variance of blowout pressures.

From the given information the value of $\alpha$ is 0.01, and to test the claim that the variance of blowout pressures is more than Soap Stone claims. The standard deviation of 1,020 foot-pounds than the variance is 1,040,400$\left(=1,{020}^2\right)$.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:{\sigma }^2=1,040,400$

That is, the variance of blowout pressures is 1,040,400.

Alternative hypothesis:

$H_1:{\sigma }^2>1,040,400$

That is, the variance of blowout pressures is more than 1,040,400.

(ii)

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 30 Soap Stone tires were inflated to the recommended pressure with a sample standard deviation of blowout forces was 1353 foot-pounds.

Step by step procedure to obtain test statistic using MINITAB software is given below:

• Choose Stat > Basic Statistics > 1 Variance.
• Under Data, choose Sample standard deviation.
• Enter Sample size as 30.
• Enter Sample standard deviation as 1,353.
• Check Perform hypothesis test.
• Choose Hypothesized standard deviation.
• Enter the Value as 1,020.
• Select Options and enter the Confidence level as 99.0 and Alternative as greater than. Click OK.
• Click OK.

Output using MINITAB software is given below:

From the MINITAB output, the sample test statistic value is 51.03.

Hence, the sample test statistic is 51.03.

(iii)

From part (ii) Minitab output, the P-value is 0.007.

Hence, the P-value of the sample test statistic is 0.007.

(iv)

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 0.007 and the level of significance is 0.01.

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

That is, $0.007\left(=P\text{-value}\right)<0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the null hypothesis of independence is rejected.

(v)

There is sufficient evidence that the variance of blowout pressures is more than Soap Stone claims at level of significance 0.01.

(b)

To determine

Find the 95% confidence interval for the variance of blowout pressures.

Answer to Problem 7CRP

The 95% confidence interval for the variance of blowout pressures is $1,161,147.4<{\sigma }^2<3,307,642.4$ square foot-pounds.

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,

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

From part (a) MINITAB output, the degrees of freedom are 29.

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

• In d.f. column locate the value 29.
• In Right-tail Area row of locate the value 0.025.
• The intersecting value of row and columns is 45.72.

The value of ${\chi }_U^2$ is 45.72.

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 29.
• In Right-tail Area row of locate the value 0.975.
• The intersecting value of row and columns is 16.05.

The value of ${\chi }_L^2$ is 16.05.

Substitute 30 for n, 1,353 for s, 45.72 for ${\chi }_U^2$, and 16.05 for ${\chi }_L^2$ in the confidence interval formula.

$\begin{array}{c}\frac{\left(30-1\right){\left(1,353\right)}^2}{45.72}<{\sigma }^2<\frac{\left(30-1\right){\left(1,353\right)}^2}{16.05}\\ \frac{53,087,661}{45.72}<{\sigma }^2<\frac{53,087,661}{16.05}\\ 1,161,147.4<{\sigma }^2<3,307,642.4\end{array}$

Hence, the 95% confidence interval for the variance of blowout pressures is $1,161,147.4<{\sigma }^2<3,307,642.4$ square foot-pounds.