Question

1. (0.8) two areas of a large city are being considered as the seat of
day care centres. Of 200 families interviewed in one section.
the proportion of mothers working full - time was 0.52. In
the other section, 40 % of the 150 families interviewed had mothers who
they worked full-time jobs. With a level of significance
of o =0.04. is there is a significant difference in the proportion of mothers
who work full - time in the two areas of the city?.

Answer 1

Solution:-

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 0.04. 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 = 0.4686

SE = 0.05389
z = (p1 - p2) / SE

z = 2.23

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 -2.23 or greater than 2.23.

P-value = P(z < - 2.23) + P(z > 2.23)

Use z-calculator to find the p-values.

P-value = 0.013 + 0.013

Thus, the P-value = 0.026

Interpret results. Since the P-value (0.026) is less than the significance level (0.04), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is a significant difference in the proportion of mothers who work full - time in the two areas of the city.