Question

Significant does not mean important. Never forget that even small effects can
be statistically significant if the samples are large. To illustrate this fact, consider
a sample of 148 small businesses. During a three-year period, 15 of the 106
headed by men and 7 of the 42 headed by women failed.
22

(a) Find the proportions of failures for businesses headed by women and
businesses headed by men. These sample proportions are quite close to
each other. Give the P-value for the z test of the hypothesis that the same
proportion of women’s and men’s businesses fail. (Use the two-sided alternative.) The test is very far from being significant.
(b) Now suppose that the same sample proportions came from a sample 30
times as large. That is, 210 out of 1260 businesses headed by women and
450 out of 3180 businesses headed by men fail. Verify that the
proportions of failures are exactly the same as in part (a). Repeat the z test
for the new data, and show that it is now significant at the α = 0.05 level.
(c) It is wise to use a confidence interval to estimate the size of an effect
rather than just giving a P-value. Give the large sample 95% confidence
intervals for the difference between the proportions of women’s and men’s
businesses that fail for the settings of both parts (a) and (b). What is the
effect of larger samples on the confidence interval?

Answer 1

(a)

Define the hypothesis as:

where and represents the proportion of failed business headed by men and women respectively.

It has been given that out of the businesses held by men failed, thus, the sample proportion, , is:

Also,   out of the businesses held by women failed, thus, the sample proportion, , is:

The test statistic, , is computed using the formula:

where, represents the pooled proportion and is calculated using the formula:

Substitute the appropriate values in the formula to compute the test statistic:

The p-value is computed using the formula:

The value of is computed using the formula "=NORMSDIST()" in MS-excel as shown here:

Therefore,

This implies:

Since, this is a two tailed test, therefore the p-value is multiplied by 2, that is:

It is known that .

Therefore, there is an insufficient indication to discard the null hypothesis at 5% level of significance, because , that is the p-value is greater than the level of significance.

CONCLUSION:

The null hypothesis that is is true implying that the result is not significant at 5% level of significance, that is there is no difference between the proportion .

(b)

Define the hypothesis as:

where and represents the proportion of failed business headed by men and women respectively.

It has been given that out of the businesses held by men failed, thus, the sample proportion, , is:

Thus,

Also,   out of the businesses held by women failed, thus, the sample proportion, , is:

Thus,

Hence, it has been verified that the proportion of failures in part(a) and part(b) are same.

The test statistic, , is computed using the formula:

where, represents the pooled proportion and is calculated using the formula:

Substitute the appropriate values in the formula to compute the test statistic:

It is known that . The critical value for the left tail of the normal distribution is fornd using the command "=NORMSINV()" in MS-EXCEL as shown below:

This implies,

Therefore, there is a sufficient indication to discard the null hypothesis at 5% level of significance, because , that is the value of the test statistic lies outside the acceptance region.

CONCLUSION:

The null hypothesis that is is not accepted in part (b) implying that there is a significant difference between the two proportions at 5% level of significance.

(c)

The % cofidence interval for the difference of proportions, say , is calculated using the formula:

where, is the right tail value of the normal distribution at significance level

The critical value for the left tail of the normal distribution is found using the command "=NORMSINV()" in MS-EXCEL as shown below:

This implies,

Since the normal distribution is symmetric, therefore, .

The 95% confidence interval for part (a) is calculated below:

The 95% confidence interval for difference in proportion, say , for part (b) is calculated below:

Therefore, the required 95% confidence intervals for part (a) and part (b) is and respectively.

It can be concluded that as a result of increasing sample size in part (b) as compared to part (a), the width of the confidence interval has been reduced due to overall fall in the value of the standard deviation of the difference in the proportions.