A simple random sample of 60 households in City 1 is taken.In the sample,there are 45 households that decorate their houses with lights for the holidays.A simple random sample of 50 households is also taken from the neighboring City 2.In the sample,there are 40 households that decorate their houses.We wish to estimate the difference in proportions of households that decorate their houses with lights for the holidays,with a 95% confidence interval.What is the standard error of the difference in sample proportions?

A) 0.040
B) 0.0795
C) 0.0802
D) 0.1125

Question

Punam Verma · Accepted Answer

Final Answer : B, Explanation :   The standard error (SE) of the difference in sample proportions is calculated using the formula:   S E = p 1 ( 1 − p 1 ) n 1 + p 2 ( 1 − p 2 ) n 2 SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} SE = n 1 ​ p 1 ​ ( 1 − p 1 ​ ) ​ + n 2 ​ p 2 ​ ( 1 − p 2 ​ ) ​ ​   where   p 1 p_1 p 1 ​   and   p 2 p_2 p 2 ​   are the sample proportions, and   n 1 n_1 n 1 ​   and   n 2 n_2 n 2 ​   are the sample sizes for the two groups. For City 1,   p 1 = 45 60 = 0.75 p_1 = \frac{45}{60} = 0.75 p 1 ​ = 60 45 ​ = 0.75  , and for City 2,   p 2 = 40 50 = 0.8 p_2 = \frac{40}{50} = 0.8 p 2 ​ = 50 40 ​ = 0.8  . Plugging these values into the formula gives:   S E = 0.75 ( 1 − 0.75 ) 60 + 0.8 ( 1 − 0.8 ) 50 = 0.1875 60 + 0.16 50 = 0.003125 + 0.0032 = 0.006325 ≈ 0.0795. SE = \sqrt{\frac{0.75(1-0.75)}{60} + \frac{0.8(1-0.8)}{50}} = \sqrt{\frac{0.1875}{60} + \frac{0.16}{50}} = \sqrt{0.003125 + 0.0032} = \sqrt{0.006325} \approx 0.0795. SE = 60 0.75 ( 1 − 0.75 ) ​ + 50 0.8 ( 1 − 0.8 ) ​ ​ = 60 0.1875 ​ + 50 0.16 ​ ​ = 0.003125 + 0.0032 ​ = 0.006325 ​ ≈ 0.0795.

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