Question

# simple random sample of size n=400 individuals who are currently employed is asked if they work...

simple random sample of size n=400 individuals who are currently employed is asked if they work at home at least once per week. Of the 400 employed individuals​ surveyed,44 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.

Solution :

Given that,

n = 400

x = 44

Point estimate = sample proportion = = x / n = 0.11

1 - = 0.89

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 * (((0.11 * 0.89) / 400)

= 0.040

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.11 - 0.040 < p < 0.11 +0.040

0.070 < p < 0.150

The 99% confidence interval is : (0.070 , 0.150)

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