Question

Sodoku puzzles have become very popular in recent years: 31.1% of members of households an there differences between the genders? The proportion of women and men from these households who worked Sodoku puzzles in 2012 can be estimated with annusl income of at least $100,000 worked Sodoku puzzles in 2012 (Stetistica. com, March 10, 2013) Are from the following sample data Gender Men Sample size 1200 1600 Worked Sadoka Puzzles 312 512 If required, enter negative values as negative numbers a. State the hypotheses that can be used to test for a difference between the preportion for the population of men and the proportion for the population of women who worked Sodoku puzzles Ha: pl-pl not equal to 0 b. What is the sample proportion of men who worked Soosku puzzies (to 2 decimalis) 32 what is the sample proportion et women (to 2 decimals)? 26 63 c. Conduct the hypothesis test and compute the p-value (to 4 decimals) p-vaiue Hypothesis test At a.05 leve of significance, what is your cenclusion The propert-is higher d. what is the margin of error and 95% cotidenee interval estimate e, ehe drirence betanente nglate.gooriast,eenei Margin of error Confidence interval population proportions (to 4 decimals)?

Answer 1

Solution:-

11.33)

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

Null hypothesis: P₁ - P₂ < 0
Alternative hypothesis: P₁ - P₂ > 0

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 = 0.30

SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.05684
z = (p₁ - p₂) / SE

z = 2.815

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 greater than 2.815.

Thus, the P-value = 0.0024.

Interpret results. Since the P-value (0.024) is greater than the significance level (0.01), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the proportion of managers expecting the higher rates diminished in July.