Question

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	23.82	21.38
Sample standard deviation	5.91	4.87
Sample size	34	36

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?

a. State the decision rule for 0.01 significance level: H₀: μ_Men= μ_Women   H₁: μ_Men ≠ μ_Women. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.)

b. Compute the value of the test statistic. (Round your answer to 3 decimal places.)

c. What is your decision regarding the null hypothesis?

d. What is the p-value? (Round your answer to 3 decimal places.)

Answer 1

Solution:

Here, we have to use two sample t test for the difference between two population means assuming equal population variances. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: There is no any significant difference in the mean number of times men and women order take-out dinners in a month.

Alternative hypothesis: H₁: There is a significant difference in the mean number of times men and women order take-out dinners in a month.

H₀: μ_Men= μ_Women  

H₁: μ_Men ≠ μ_Women

This is a two tailed test.

Part a

We are given

Level of significance = α = 0.01

n1 = 34

n2 = 36

df = n1 + n2 – 2 = 34 + 36 -2 = 68

Critical values = -2.6501 and 2.6501

(by using t-table)

Decision rule: Reject the null hypothesis H₀ when test statistic t is not lies within the critical values -2.650 and 2.650.

Reject H₀ if t < -2.650 or t > 2.650

Part b

The test statistic formula for this test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 23.82

X2bar = 21.38

S1 = 5.91

S2 = 4.87

n1 = 34

n2 = 36

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(34 – 1)*5.91^2 + (36 – 1)*4.87^2]/(34 + 36 – 2)

S_p² = 29.1576

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (23.82 – 21.38) / sqrt[29.1576*((1/34)+(1/36))]

t = 2.44/1.2913

t = 1.8895

Test statistic = t = 1.890

Part c

We have

Test statistic = 1.890

Test statistic is lies within the critical values -2.650 and 2.650.

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that there is a significant difference in the mean number of times men and women order take-out dinners in a month.

Part d

We have

t = 1.8895

df = 68

So, the P-value by using t-table (or calculator or excel) is given as below:

P-value = 0.063