Question

7. In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. Calculate a 99 percent confidence interval for the proportion of persons who work less than 40 hours per week.

Answer 1

Solution :

n = 83

=30.1% = 0.301

1 - = 1 - 0.301 = 0.699

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.301 * 0.699) /83)

= 0.130

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.301 - 0.130 < p < 0.301 + 0.130

0.171 < p < 0.431

The 99% confidence interval for the population proportion p is : ( 0.171 , 0.431 )