Question

c=0.95,x=14.3,s=4.0,n=5

Answer 1

it is given that

c=0.95 confidence level, so significance level =1 - c = 1-0.95 = 0.05

x=14.3 sample mean

s=4.0 sample standard deviation

n=5 sample size

degree of freedom = n-1 = 5- 1 = 4

using excel function T.INV.2T(probability, df)

setting probability = 0.05 and df= 4

we get, t critical = 2.776

Confidence interval is

setting the values, we get

So, required 95% confidence interval is (9.33,19.27)(rounded to 2 decimals) or (9.3,19.3)(rounded to 1 decimals)