Question

2. A survey organization takes a simple random sample of 200 households from a city of 50,000 households. On the average there are 2.3 persons per sample household, and the SD is 1.8.

a) Find a 90%-confidence interval for the average household size in the city

b) Find a 80%-confidence interval for the average household size in the city.

c) If the researcher decreased the sample size to only 50 people, the length of the 80% confidence interval would be:

(i) multiplied by 2 (ii)multiplied by 4 (iii) divided by 2 (iv) divided by 4

Answer 1

n = 200

sample average = 2.3

sample sd = 1.8

CI = mean + / - E

a) 90% CI

t value for 90% CI = TINV(0.1, 199) = 1.653

CI = 2.3 +/- 0.21

CI = (2.09 ,    2.51)

b) 80% CI

t value for 80% CI = TINV(0.2, 199) = 1.286

CI = 2.3 +/- 0.164

CI = (2.14 ,    2.46)

length of CI = 0.33

(c) If sample size is reduced to 50:

E = 0.331

CI = (1.97 , 2.63)

length of CI = 0.66

If the researcher decreased the sample size to only 50 people, the length of the 80% confidence interval would be : multiplied by 2