Housework for women and men Do women tend to spend more time on housework than m

Question

Housework for women and men Do women tend to spend more time on housework than men? If so, how much more? Based on data from the National Survey of Families and Households, one study reported the results in the table for the number of hours spent in housework per week. (Source: Data from A. Lincoln, Journal of Marriage and Family, vol. 70, 2008, pp. 806–814.)

PLEASE!! Use the following code!!! Thank you!!

Housework Hours Gender Sample Size Mean Standard Deviation Women Men 476 496 33.0 19.9 21.9 14.6 a. Based on this study, calculate how many more hours, on the average, women spend on housework than men. b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two groups? c. Calculate the 95% confidence interval comparing the he population means for women and men. Interprett result including the relevance of 0 being within the interval or not. (Hint: For such large sample sizes, the t-score is practically identical to a z-score.) d. State the assumptions upon which the interval in part c is based.

Explanation / Answer

x1(bar) = 33.00

x2(bar) = 19.90

s1 = 21.90

s2 = 14.60

n1 = 476

n2 = 496

a)

x1bar - x2bar = 13.10

Hence, 13.1 hours women spend more on housework than men.

b)

SE = sqrt[ (s1^2/n1) + (s2^2/n2) ]

(s1^2/n1) = 1.0076

(s2^2/n2) = 0.4298

SE = 1.1989

Becuase of higher sample size, the SE is lower.

c)

x1bar - x2bar = 13.10

SE = 1.1989

CI = 95%

z-value = 1.96

ME = z*SE = 2.3498

Confidence Interval is (x1bar - x2bar) +/- ME

Lower bound = 10.75017

Upper bound = 15.44983

Confidence Interval (10.7502 , 15.4498)

0 is not within the interval, hence there are not sufficient evidence that the means are different.

d)

Assumption is random sample is drawn from normally distributed population.

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