Chapter 23, Problem 23.25E

(a)

To determine

To find: The proportions of failures for businesses headed by women and businesses headed by men.

To test: Whether the same proportion of women’s and men’s businesses fails or not.

To find: The hypotheses, the test statistic, P-value and conclusion.

Answer to Problem 23.25E

The proportion of failures for businesses headed by men is 0.1415.

The proportion of failures for businesses headed by women is 0.1666.

The null hypothesis is $p_{\text{1}}=p_{\text{2}}$.

The alternative hypothesis is $p_{\text{1}}\ne p_{\text{2}}$.

The value of the z-statistic is –0.39.

The P-value is 0.698.

The conclusion is that, the same proportion of women’s and men’s businesses fails.

Explanation of Solution

Given info:

In the survey, 15 out of 106 were headed by men failed and 7 out of 42 were headed by women failed.

Calculation:

PLAN:

Check whether the same proportion of women’s and men’s businesses fails or not.

State the test hypotheses.

Null hypothesis:

$H_0:p_{\text{1}}=p_{\text{2}}$

Alternative hypothesis:

$H_a:p_{\text{1}}\ne p_{\text{2}}$

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain proportion, test statistic and P-value using the MINITAB software:

• Choose Stat > Basic Statistics > 2 Proportions.
• Choose Summarized data.
• In First sample, enter Trials as 106 and Events as 15.
• In Second sample, enter Trials as 42 and Events as 7.
• Choose Options.
• Choose Pooled proportions.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output,

The proportion of failures for businesses headed by men is 0.1415.

The proportion of failures for businesses headed by women is 0.1666.

The value of the z-statistic is –0.39.

The P-value is 0.698.

CONCLUDE:

The P-value is 0.698 and the significance level is 0.05.

Here, the P-value is greater than the significance level.

That is, $0.698\left(=P\text{-value}\right)>0.05\left(=\alpha \right)$.

Therefore, using the rejection rule, it can be concluded that there is no evidence to reject $H_0$ at $\alpha =0.05$.

Thus, the same proportion of women’s and men’s businesses fails.

(b)

To determine

To find: The proportions of failures for businesses headed by women and businesses headed by men.

To test: Whether the same proportion of women’s and men’s businesses fails or not.

To find: The hypotheses, the test statistic, P-value and conclusion.

Answer to Problem 23.25E

The proportion of failures for businesses headed by men is 0.1415.

The proportion of failures for businesses headed by women is 0.1666.

The null hypothesis is $p_{\text{1}}=p_{\text{2}}$.

The alternative hypothesis is $p_{\text{1}}\ne p_{\text{2}}$.

The value of the z-statistic is –2.12.

The P-value is 0.034.

The conclusion is that, there is no evidence that the same proportion of women’s and men’s businesses fails.

Explanation of Solution

Given info:

In the survey, 210 out of 1,260 were headed by women failed and 450 out of 3,180 were headed by men failed.

Calculation:

PLAN:

Check whether the same proportion of women’s and men’s businesses fails or not.

State the test hypotheses.

Null hypothesis:

$H_0:p_{\text{1}}=p_{\text{2}}$

Alternative hypothesis:

$H_a:p_{\text{1}}\ne p_{\text{2}}$

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain proportion, test statistic and P-value using the MINITAB software:

• Choose Stat > Basic Statistics > 2 Proportions.
• Choose Summarized data.
• In First sample, enter Trials as 3,180 and Events as 450.
• In Second sample, enter Trials as 1,260 and Events as 210.
• Choose Options.
• Choose Pooled proportions.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output,

The proportion of failures for businesses headed by men is 0.1415.

The proportion of failures for businesses headed by women is 0.1666.

The value of the z-statistic is –2.12.

The P-value is 0.034.

CONCLUDE:

The P-value is 0.034 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, $0.034\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

Therefore, using the rejection rule, it can be concluded that there is evidence to reject $H_0$ at $\alpha =0.05$.

Thus, there is no evidence that the same proportion of women’s and men’s businesses fails.

(c)

To determine

To find: The 95% confidence interval for the difference between the proportions of women’s and men’s businesses that fail for the settings of part (a).

To find: The 95% confidence interval for the difference between the proportions of women’s and men’s businesses that fail for the settings of part (b).

To identify: The effect of larger samples on the confidence interval.

Answer to Problem 23.25E

The 95% confidence interval for the difference between the proportions of women’s and men’s businesses that fail for the settings of part (a) is $\left(-0.1559,0.1056\right)$.

The 95% confidence interval for the difference between the proportions of women’s and men’s businesses that fail for the settings of part (b) is $\left(-0.0490,-0.0013\right)$.

The effect of larger samples on the confidence interval is that the confidence interval is narrow.

Explanation of Solution

Given info:

In the survey, 210 out of 1,260 were headed by women failed and 450 out of 3,180 were headed by men failed.

Calculation:

From part (a), the 95% confidence interval for the difference between the proportions of women’s and men’s businesses that fail for the settings of part (a) is $\left(-0.1559,0.1056\right)$.

From part (b), the 95% confidence interval for the difference between the proportions of women’s and men’s businesses that fail for the settings of part (b) is $\left(-0.0490,-0.0013\right)$.

Justification:

From the results, it can be observed that the confidence interval is narrow if the sample is 30 times large.