Question

1. In a survey of 500 U.S. adults, 45% of them said that lounging at the beach was their “dream vacation.’ Assuming this to be a simple random sample of U.S. adults, the lower confidence limit of the 95% confidence intervals for the proportion of U.S. adults who consider lounging at the beach to be their dream vacation would equal to;

Answer 1

Solution :

n = 500

= 45% = 0.45

1 - = 1 - 0.45 = 0.55

At 95% confidence level the z 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 * (((0.450 * 0.550) / 500)

= 0.044

At 95% confidence interval estimate of the population mean is,

- E < < + E

(0.450 - 0.044 < < 0.450 + 0.044 )

0.406 < < 499

lower confidence limit = 0.406

upper confidence limit = 0.499