Question

Exercise 11-21 (LO11-2) A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distributions follow the normal probability distribution and the population standard deviations are equal. The information is summarized below. Statistic Men Women Sample mean 24.26 22.45 Sample standard deviation 5.91 4.88 Sample size 35 39 At the 0.01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? State the decision rule for 0.01 significance level: H0: μMen= μWomen H1: μMen ≠ μWomen. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) Compute the value of the test statistic. (Round your answer to 3 decimal places.) What is your decision regarding the null hypothesis? What is the p-value? (Round your answer to 3 decimal places.)

Answer 1

Solution:-

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

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 1.2683
DF = 72
t = [ (x₁ - x₂) - d ] / SE

t = 1.427

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 72 degrees of freedom is more extreme than 1.427; that is, less than -1.427 or greater than 1.427.

Thus, the P-value = 0.158

Interpret results. Since the P-value (0.158) is greater than the significance level (0.01), hence we failed to reject null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that there a difference in the mean number of times men and women order take-out dinners in a month.