Question

1. Do women tend to spend more time on housework then men? A recent study reports the following descriptive statistics for the number of hours spent in housework per week for men and women? Determine whether women tend to spend more time on housework than men? You are expected to perform a significance test (a= .05), obtain a 99% confidence interval and interpret your results. (10 points)
   1. identify the claim and state H0 and Ha
   2. find the critical value(s)/p-values and identify the rejection region(s),
   3. find the standardized test statistic z, and sketch a graph
   4. decide whether to reject or fail to reject the null hypothesis, and interpret the decision in the context of the original claim.
   5. Obtain a 99% confidence interval and interpret your results.

               Cooking and Whashing Up Minutes

Sex          Sample Size       Mean        Standard Deviation

Men               1219                   23                       32

Women           733                   37                       16

Answer 1

Solution

Part (a)

Let X =time spent on house work by women.

      Y = time spent on house work by men.

Let (µ₁, σ₁) and (µ₂, σ₂) be the mean and SD of X and Y respectively.

Claim:

Women tend to spend more time on house work than men. Answer 1

Hypotheses:

Null: H₀: µ₁ = µ₂ Vs Alternative: H_A: µ₁ > µ₂  Answer 2

Part (b)

Critical Value, p-value and rejection region

Under H₀, t ~ t_{n1 + n2 - 2}. Hence, for level of significance α%,

Critical Value = upper α% point of t_{n1 + n2 - 2} = upper 2.5% point of t₁₉₅₀ = 1.6456 Answer 3

p-value = P(t_{n1 + n2 - 2} > tcal) = P(t₁₉₅₀ > 11.1431) < 0.05 [actual value = 1.52E-27] Answer 4

[The above are found Using Excel Function: Statistical TINV and TDIST]

Rejection region:

Reject H₀ if tcal > tcrit, or equivalently since p-value < α Answer 5

Part (c)

Test statistic = (Xbar - Ybar)/[s√{(1/n₁) + (1/n₂)}] = 11.0431 Answer 6

where

s² = {(n₁ – 1)s₁² + (n₂ – 1)s₂²}/(n₁ + n₂ – 2);

Xbar and Ybar are sample averages and s₁,s₂ are sample standard deviations based on n₁ observations on X and n₂ observations on Y respectively.

Part (d)

Decision: reject H₀ Answer 7

Interpretation: There is sufficient evidence to support the claim and hence we conclude that

Women tend to spend more time on house work than men. Answer 8

Part (e)

100(1 - α) % Confidence Interval for (μ₁ - μ₂) is:

(Xbar – Ybar) ± MoE,

where

MoE{(t_{2n – 2, α/2})(s)√{(1/n₁) + (1/n₂)}

Thus,

99% confidence interval for the difference in the mean time spent on housework by women and men = [11.18, 16.81] Answer 9

Details of calculations

Given	n1	733
	n2	1219
	Xbar	37.0000
	Ybar	23
	s1	32.0000
	s2	16.0000
	s^2	544.2954
	s	23.3301
	α	0.01
	tα/2	2.5784
	MoE	2.8115
Lower Bound		11.1885
Upper Bound		16.8115

Done