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?

1. 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.)

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

3. What is your decision regarding the null hypothesis?

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

Answer 1

Given or X = 23.82 S. = 5.91 X = 28.38 S2 = 4087 ni = 34 n2 = 36

Hypothesis. Ho. There is no difference in the mean. number of times mon and women order take-out dinners in monte je Holmen - Mwomen als hr Hi There is difference in the mean. number of times men and women order take-out dinners in a month j.e Ho : Mmen & women under Ho, Ntnit n-2 Sup (n-1) 5,2 + (12-1) 52 I nitha-2 Decision : At 0.01 significance if p-value 22=0.01 then we reject Ho. A150 If t 3 tcritical at 0.01 level then we reject Ho. 7338 (5.91) 27 (4.87) 2 x 35 - 347 36-2 Se = 5.3998

23. 82 - 21038 5.3998/34 36 t = 1.8995 Critical # using tools ; 68 = 2.6501 excel = T. INV. 25 (ool, 68) ato.o1 Here t=1.8805 level then we < tcritical = 206501 fail to reject Ho. P-value or using erecel, (1.88 , 68) = TO DIST. 27 = 0.064 392 P-value = 0.064392 Here fail P-value=0.0644 > 2=0.01 then to reject Ho. we Conclusion : - we conclude that there is no difference in the mean number of times men and women order take out dinners in a month. j.e. Momen = Mwomen.