Question

A random sample of size 18 taken from a normally distributed population revealed a sample mean of 100 and a sample variance of 49. The 95% confidence interval for the population mean would equal:

a)93.56 - 103.98

b)97.65 – 107.35

c)95.52 -104.65

d)96.52 – 103.48

Answer 1

Solution:
Given in the question
Number of sample (n) = 18
Sample mean(Xbar) = 100
Sample variance(S^2) = 49
Sample standard deviation(S) = 7
95% confidence interval can be calculateda s
Mean +/- talpha/2 * S/sqrt(n)
Here we will use a t-test as sample size is less than 30 and population standard deviation is not known so
confidence level = 0.95
alpha = 0.05, alpha/2 = 0.025 and df = 18-1 = 17
from t table we found talpha/2 = 2.11
100 +/- 2.11*7/sqrt(18)
100 +/- 3.48
So 95% confidence interval is 96.52 - 103.48, so its correct answer is D. i.e. 96.52 - 103.48