5. Difference in proportions, problem 4 Aa Aa The following sample statistics ar

Question

5. Difference in proportions, problem 4 Aa Aa The following sample statistics are given to you: The first sample proportion is 0.251, from a sample size of 1,000. The second sample proportion is 0.199 from a sample size of 1,000. Perform a two-sample z-test for proportions to test wh ther the first population gre on proportion is greater than the second population proportion by more than 5%. an For a 95% CI for the difference in population proportions, the lower and upper bounds are and , respectively.

Explanation / Answer

H0: P1 = P2
Ha: P1 > P2

pooled sample proportion,

p = (p1 * n1 + p2 * n2) / (n1 + n2)

where p1 is the sample proportion from population 1, p2 is the sample proportion from population 2, n1 is the size of sample 1, and n2 is the size of sample 2.

p = (0.251*1000 + 0.199*1000)/2000 = 0.225

Standard error, SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } = sqrt{0.225 * 0.775*[(1/1000)+(1/1000)]} = 0.0186748

z = (p1 - p2)/SE = 2.78449

This is a right sided z test, with significance value 0.05

p-value = 0.002681 < 0.05, There we reject H0, as there is sufficient evidence to support the claim that proportion of sample 1 is greater than proportion of sample 2.

Confidence, CI

SEp1-p2 = sqrt{ [p1 * (1 - p1) / n1] + [p2 * (1 - p2) / n2] } = 0.0186386

For 95%, zcrit = +/-1.96

95% CI = (p1 - p2) +/- zcrit * SE
= (0.251 - 0.199) +/- 1.96*0.0186386 = (0.01547, 0.08853)

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