Question

In the 2017 Community Health Status Survey, area residents were asked how much sitting they do in a day. Of 2036 men who responded, 560 (27.5%) reported sitting at least 8 hours a day. 2183 respondents were female; 502 (23%) reported sitting at least 8 hours. Is there a difference in the proportion of males (population #1) and females (population #2) who spend at least 8 hours a day seated? Test H0: p1 = p2 versus Ha: p1not equal to p2 using αλπηα = 0.01.

z-test statistic =

p-value =

conclusion =

Give a 99% confidence interval for the difference between the proportions (p1 − p2) of men and women who sit more than 8 hours a day.

lower limit = u

pper limit =

Answer 1

The statistical software output for this problem is:

Two sample proportion summary hypothesis test:

p₁ : proportion of successes for population 1
p₂ : proportion of successes for population 2
p₁ - p₂ : Difference in proportions
H₀ : p₁ - p₂ = 0
H_A : p₁ - p₂ ≠ 0

Hypothesis test results:

Difference	Count1	Total1	Count2	Total2	Sample Diff.	Std. Err.	Z-Stat	P-value
p1 - p2	560	2036	502	2183	0.045090344	0.013371466	3.3721316	0.0007

99% confidence interval results:

Difference	Count1	Total1	Count2	Total2	Sample Diff.	Std. Err.	L. Limit	U. Limit
p1 - p2	560	2036	502	2183	0.045090344	0.013381046	0.010623053	0.079557634

Hence,

z test statistic = 3.3721

p-value = 0.0007

Conclusion: Reject Ho: Sufficient evidence for difference in proportion

99% confidence interval:

Lower limit = 0.0106

Upper limit = 0.0796