Suppose Connie owns a pet store in a large community that has over 50,000 pet ow

Question

Suppose Connie owns a pet store in a large community that has over 50,000 pet owners. She believes that the proportion of men who own cats is different from the proportion of women who own cats. Connie decides to test this claim by performing a two-sample z-test for two proportions. Her hypotheses are given by

H0:pm=pwH1:pmpw

where H0 is the null hypothesis, and H1 is the alternative hypothesis. The variables pw and pm represents the population proportion of men and women cat owners, respectively.

She randomly samples 420 men and 580 women who are pet owners in the community and finds that 205 men own cats and 266 women own cats. Determine the pooled sample proportion, p . Provide your answer precise to three decimal places.

Compute the standard error of the difference of the sample proportions (SE). Provide your answer precise to three decimal places.

SE =

p =

Explanation / Answer

n1 = 420

p1 = 205/420 = 0.49

n2 = 580
p2 = 266/580 = 0.46

Since the null hypothesis states that P1=P2, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.
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 = (0.49 * 420 + 0.46 * 580) / (420 + 580) = 0.4726

Compute the standard error (SE) of the sampling distribution difference between two proportions.
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.

SE = sqrt( 0.4726 * ( 1 - 0.4726 ) * [ (1/420) + (1/580) ])
= 0.0319

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