Question

In a survey sample of 100 people, about 28% work less than 40 hours per week. Calculate a 90% confidence interval for the proportion of persons who work less than 40 hours per week.

Report the lower limit and the upper limit. Round each to the second decimal place.

( _______, ___________ )

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = 0.28

1 - = 1 - 0.28 = 72

Z/2 = 1.645

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

= 1.645 * (((0.28 * 0.72) / 100)

Margin of error = E = 0.07

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.28 - 0.07 < p < 0.28 + 0.07

0.21 < p < 0.35

lower limit and upper limit **(0.21 , 0.35)**

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