Question

1. A university planner is interested in determining the percentage of spring semester students who will attend summer school. She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school.
   
   1. Construct a 90% confidence interval estimate for the percentage of spring semester students who will return to summer school.
   2. Construct a 95% confidence interval estimate for the percentage of spring semester students who will return to summer school.

Answer 1

Solution :

n = 160

x = 56

= x / n = 56 / 160 = 0.350

1 - = 1 - 0.350 = 0.650

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.350 * 0.650) / 160)

= 0.062

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.350 - 0.062 < p < 0.350 + 0.062

0.288 < p < 0.412

The percentage of spring semester students minimum  28.8% and maximum 41.2% will return to summer school.

b ) At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (((0.350 * 0.650) / 160)

= 0.074

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.350 - 0.074 < p < 0.350 + 0.074

0.277 < p < 0.424

The percentage of spring semester students minimum  27.7% and maximum 42.4% will return to summer school