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.
( _______, ___________ )
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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