Question

In a sample of 70 randomly selected students, 29 favored the amount being budgeted for next year's intramural and interscholastic sports. Construct a 98% confidence interval for the proportion of all students who support the proposed budget amount. (Give your answers correct to three decimal places.)

Lower Limit
Upper Limit

Answer 1

Solution :

Given that,

n = 70

x = 29

Point estimate = sample proportion = = x / n = 29/70=0.414

1 - = 1-0.414=0.586

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 (((0.414*0.586) / 70)

= 0.137

A 98% confidence interval is ,

- E < p < + E

0.414-0.137 < p < 0.414+0.137

0.277 < p < 0.551

Lower Limit 0.277
Upper Limit 0.551