Question

Never forget that even small effects can be statistically significant if the samples are large. To illustrate this fact, consider a sample of 76 small businesses. During a three-year period, 8 of the 59 headed by men and 3 of the 17 headed by women failed.

(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 test of the hypothesis that the same proportion of women's and men's businesses fail. (Use the two-sided alternative). What can we conclude (Use α=0.05α=0.05)?
The P-value was ____ so we conclude that
Choose a conclusion. The test showed strong evidence of a significant difference. The test showed no significant difference.

(b) Now suppose that the same sample proportion came from a sample 30 times as large. That is, 90 out of 510 businesses headed by women and 240 out of 1770 businesses headed by men fail. Verify that the proportions of failures are exactly the same as in part (a). Repeat the test for the new data. What can we conclude?
The P-value was___ so we conclude that
Choose a conclusion. The test showed strong evidence of a significant difference. The test showed no significant difference.

(c) It is wise to use a confidence interval to estimate the size of an effect rather than just giving a P-value. Give 95% confidence intervals for the difference between proportions of men's and women's businesses (men minus women) that fail for the settings of both (a) and (b). (Be sure to check that the conditions are met. If the conditions aren't met for one of the intervals, use the same type of interval for both)
Interval for smaller samples: ___ to ___
Interval for larger samples: ___ to ___

What is the effect of larger samples on the confidence interval?
Choose an effect. The confidence interval is unchanged. The confidence interval's margin of error is reduced. The confidence interval's margin of error is increased.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ P₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = 0.14474

SE = 0.09685
z = (p₁ - p₂) / SE

z = - 0.42

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.42 or greater than 0.42.

P-value = P(z < - 0.42) + P(z > 0.42)

Use z-calculator to find the p-values.

P-value = 0.337 + 0.337

Thus, the P-value = 0.674

Interpret results. Since the P-value (0.674) is greater than the significance level (0.05), we failed to reject the null hypothesis.

The P-value was is greater than 0.05, so we conclude that the test showed no significant difference.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P₁ = P₂
Alternative hypothesis: P₁ P₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = 0.145175

SE = 0.017704
z = (p₁ - p₂) / SE

z = -2.31

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 2.31 or greater than 2.31.

P-value = P(z < - 2.31) + P(z > 2.31)

Use z-calculator to find the p-values.

P-value = 0.01 + 0.01

Thus, the P-value = 0.02

Interpret results. Since the P-value (0.02) is less than the significance level (0.05), we have to reject the null hypothesis.

The P-value was is less than 0.05, so we conclude that the test showed strong evidence of a significant difference.

c)

Confidence interval for the smaller samples is C.I = (0.1489, 0.2307).

C.I = (0.1356 - 0.17647) + 1.96*0.09685

C.I = - 0.04087 + 0.18983

C.I = (- 0.2307, 0.1489)

Confidence interval for the larger samples is C.I = (- 0.0756, - 0.0062)

C.I = (0.1356 - 0.17647) + 1.96*0.017704

C.I = - 0.04087 + 0.03469

C.I = (- 0.0756, - 0.0062)

The confidence interval's margin of error is reduced.