Question

How often do you go out dancing? This question was asked by a professional survey group on behalf of the National Arts Survey. A random sample of

n₁ = 97

single men showed that

r₁ = 24

went out dancing occasionally. Another random sample of

n₂ = 91

single women showed that

r₂ = 19

went out dancing occasionally. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value a small amount and thereby produce a slightly more "conservative" answer.

(a) Do these data indicate that the proportion of single men who go out dancing occasionally is higher than the proportion of single women who do so? Use a 5% level of significance. List the assumptions you made in solving this problem. Do you think these assumptions are realistic?

(i) What is the level of significance?

What is the value of the sample test statistic? (Round your answer to three decimal places.)
(b) Compute a 90% confidence interval for the population difference of proportions

p₁ − p₂

of single men and single women who occasionally go out dancing. (Round your answers to three decimal places.)

lower limit
upper limit

Answer 1

Solution:-

a)

The level of significance is 0.05

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

Null hypothesis: P_Men> P_Women
Alternative hypothesis: P_Men < P_Women

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. 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 = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

p = 0.3707

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.0705
z = (p₁ - p₂) / SE

z = 0.55

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 one-tailed test, the P-value is the probability that the z-score is more than 0.55. We use the Normal Distribution Calculator to find P(z > 0.55) = 0.2912

Thus, the P-value = 0.2912

Interpret results. Since the P-value (0.2912) is greater than the significance level (0.05), we cannot reject the null hypothesis.

b) 90% confidence interval for the population difference of proportions p₁ − p₂ is C.I = (- 0.0773, 0.1546).

C.I = 0.03863 + 1.645 × 0.0705

C.I = 0.03863 + 0.1159725

C.I = (- 0.0773, 0.1546)