Chapter 9.5, Problem 1AC

1.

To determine

To write: The null hypothesis.

Answer to Problem 1AC

The null hypothesis is,

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

Explanation of Solution

The given data shows that the data values from the automobile magazine of 1996 June. The article compared various parameters of U.S. and Japanese-made sports cars.

The greater variability in the price of automatic transmissions for the countries is tested between the Japanese cars and the U.S. cars.

Thus, the null hypothesis can be written as,

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

2.

To determine

To identify: The test is used to test for any significant difference in the variances.

Answer to Problem 1AC

The test is used to test for any significant difference in the variances is F-test.

Explanation of Solution

Justification:

F-test:

The F-test is used to test for the significant difference in the variances.

Here, it is observed that F-test is used to test for any significant difference in the variances.

3.

To determine

To test: Whether there is any significant difference in the variability in the prices between the Japanese cars and the U.S. cars.

Answer to Problem 1AC

There is no sufficient evidence to support the claim that there is a significant difference in the variability in the prices between the German cars and the U.S. cars.

Explanation of Solution

Justification:

Step by step procedure for finding the sample variances using Minitab procedure is,

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns Japanese and U.S. Cars.
• Choose option statistics, and select Variance.
• Click OK.

Output using Minitab procedure is,

Here, the sample variances of $\mathrm{U}.S.\left(265,021\right)>\text{Japanese}\left(6,043.4\right)$. Thus, $s_1^2=6,043.4$ and $s_2^2=265,021$.

Where, $s_1^2$ and ${\sigma }_1^2$ represent the sample variance and population variance for the prices of cars in Japanese, $s_2^2$ and ${\sigma }_2^2$ represents the sample variance and population variance for the prices of cars in U.S. The samples sizes $n_1=6$ and $n_2=6$.

Here, the variability in the price of automatic transmissions for the countries is tested. Hence, the claim is that, ${\sigma }_1^2\ne {\sigma }_2^2$.

Test statistic:

The formula to find the F-statistic is,

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

Substitute $s_1^2=6,043.4$ and $s_2^2=265,021$

$\begin{array}{c}F=\frac{6,043.4}{265,021}\\ =0.0228\end{array}$

Thus, the F-statistic value is 0.0228.

Degrees of freedom:

The degrees of freedom are $\text{d}\text{.f}{\text{.}}_{\text{N}}=n_1-1\text{ and d}\text{.f}{\text{.}}_{\text{D}}=n_2-1$.

That is,

$\begin{array}{c}\text{d}\text{.f}{\text{.}}_{\text{N}}=n_1-1\\ =6-1\\ =5\end{array}$

$\begin{array}{c}\text{d}\text{.f}{\text{.}}_{\text{D}}=n_2-1\\ =6-1\\ =5\end{array}$

Critical value:

The level of significance is $\alpha =0.05$.

Divide the level of significance by 2.

That is,

$\begin{array}{c}\frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

The critical F-value for a two-tailed test is obtained using the Table H: The F-Distribution with the level of significance 0.025.

Procedure:

• Locate 5 in the degrees of freedom, denominator row of the Table H.
• Obtain the value in the corresponding degrees of freedom, numerator column below 5.

That is, $F_0=14.94$

Rejection region:

The null hypothesis would be rejected if $F>14.94$.

Conclusion:

Here, the F-value is lesser than the critical value.

That is, $0.0228<14.94$.

Thus, it can be concluding that the null hypothesis is not rejected.

Hence, there is no sufficient evidence to support the claim that there is a significant difference in the variability in the prices between the Japanese cars and the U.S. cars.

4.

To determine

The effect of the small sample sizes have on the standard deviations.

Answer to Problem 1AC

The effect of the small sample sizes have on the standard deviations is the small sample sizes are impacted by outliers.

Explanation of Solution

Justification:

Here, it is observed that taking the small sample sizes gives the do not give appropriate results because the small sample sizes are impacted by outliers. Thus, it can be conclude that effect of the small sample sizes have on the standard deviations is the small sample sizes are impacted by outliers.

5.

To determine

To find: The degrees of freedom are used for the statistical test.

Answer to Problem 1AC

The degrees of freedom are used for the numerator and denominator is 5.

Explanation of Solution

Justification:

Form the part (3), it can be observed that the degrees of freedom are used for the numerator and denominator is 5.

6.

To determine

To check: Whether the two sets of data have significantly different variances without having significantly different means.

Answer to Problem 1AC

Yes, the two sets of data have significantly different variances without having significantly different means.

Explanation of Solution

From the given information, it can be observed that the mean of the two sets of data is same in the center but the standard deviations of the two sets are different.

Thus, it can be conclude that the two sets of data have significantly different variances without having significantly different means.