Question

A simple random sample of size n=200 individuals who are currently employed is asked if they work at home at least once per week. Of the 200 employed individuals surveyed, 26 responded that they did work at home at least once per week. Construct a 99% confidence interval for the population proportion of employed individuals who work at home at least once per week. The lower bound is 0.069 (Round to three decimal places as needed) The upper bound is 0.191 (Round to three decimal places as needed.)

Answer 1

sample size : n=200

sample proportion : p = 26/200 = 0.13

99% confidence interval for the population proportion is

$(p-\tau _{\alpha /2}*\sqrt{\frac{p(1-p)}{n}},p+\tau _{\alpha /2}*\sqrt{\frac{p(1-p)}{n}})$$=(p-\tau _{0.005}*\sqrt{\frac{p(1-p)}{n}},p+\tau _{0.005}*\sqrt{\frac{p(1-p)}{n}})$

$=(0.13-2.576*\sqrt{\frac{0.13(1-0.13)}{200}},0.13+2.576*\sqrt{\frac{0.13(1-0.13)}{200}})$

$=(0.13-0.06126,0.13+0.06126)$

$=(0.0687,0.1912)$

$\approx (0.069,0.191)$

The lower bound is 0.069

The upper bound is 0.191

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