Question

The weights (in pounds) of 20 adult US college students are listed below.

202 172 185 190 169 245 169 195 240 174 211 198 205 192 233 260 188 158 171 228

1. Construct an 90% confidence interval for the mean.

2. What's the sample size needed to reduce the error by 50%? (for 90% confidence level)

Answer 1

Solution :

Given data is 202 172 185 190 169 245 169 195 240 174 211 198 205 192 233 260 188 158 171 228

=> mean x = sum of terms/number of terms

= 3985/20

= 199.25

=> standard deviation s = 28.8241

1.
=> df = n - 1 = 19

=> for 90% confidence interval, t = 1.729

=> A 90% confidence interval of the mean is

=> x +/- t*s/sqrt(n)

=> 199.25 +/- 1.79*28.8241/sqrt(20)

=> (187.7130 , 210.7870)

2.
=> given that margin of error E = 0.5

=> for 90% confidence interval, Z = 1.645

=> Sample size n = (Z*s/E)^2

= (1.645*28.8241/0.5)^2

= 8992.9734

= 8993 (nearest integer)