In: Statistics and Probability
A survey was conducted that asked 1008 people how many books they had read in the past year. Results indicated that x overbarequals14.4 books and sequals16.6 books. Construct a 99% confidence interval for the mean number of books people read. Interpret the interval.
Solution :
Given that,
Point estimate = sample mean = = 14.4
sample standard deviation = s = 16.6
sample size = n = 1008
Degrees of freedom = df = n - 1 = 100 8- 1 = 1007
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2,df = t0.005,1007 = 2.581
Margin of error = E = t/2,df * (s /n)
= 2.581 * (16.6 / 1008)
= 1.35
The 99% confidence interval estimate of the population mean is,
- E < < + E
14.4 -1.35 < < 4.4 + 1.35
13.05 < < 15.75
(13.05 , 15.75)