Question

A survey of 158 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. The result of the survey is that

79 of the 158 students responded​ "yes." An approximate 95​% confidence interval is ​(0.422, 0.578​). Complete parts a through d below.

​a) How would the confidence interval change if the confidence level had been 90% instead of 95%?

b) How would the confidence interval change if the sample size had been 395 instead of 158?

​c) How would the confidence interval change if the confidence level had been 98% instead of 95%?

The new confidence interval would be NARROWER/WIDER

The new confidence interval would be (X,X)

​(Round to three decimal places as​ needed.)

​d) How large would the sample size have to be to make the margin of error half as big in the 95% confidence​ interval?

The new sample size would have to be _______.

Answer 1

a) The width of the confidence interval would decreases (or becomes narrower) if the confidence level is changed to 90% from 95%.

b) As the sample size increases the confidence interval width decreases. so the confidence interval would be narrower.

c)

The width of the confidence interval would increases (or becomes wider) if the confidence level is changed to 98% from 95%.

n = 158, x = 79, p̅ = x/n = 0.5

98% Confidence interval :

At α = 0.02, two tailed critical value, z_c = NORM.S.INV(0.02/2) = 2.326

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.5 - 2.326 *√(0.5*0.5/158) = 0.407

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.5 + 2.326 *√(0.5*0.5/158) = 0.593

d) Margin of error = 0.078/2

Confidence Level, CL = 0.95

Significance level, α = 1 - CL = 0.05

Critical value, z = NORM.S.INV(0.05/2) = 1.9600

Sample size, n = (z² * p * (1-p)) / E² = (1.96² * 0.5 * 0.5)/ 0.039²

= 631.4035 = 631