Answer:
(0.9464,1.034)
Step-by-step explanation:
First find the mean μ of all values as;
(0.95+1.02+ 1.01+ 0.98 ) /4 =0.99
μ=0.99
Find the standard deviation as;
-For each value subtract the mean and square the results
0.95-0.99 = -0.04 , -0.04² = 0.0016
1.02-0.99= 0.03, 0.03²=0.0009
1.01-0.99= 0.02, 0.02²= 0.0004
0.98-0.99 = -0.01, -0.01²=0.0001
Find mean of squared deviations as;
(0.0016+0.0009+0.0004+0.0001)/4 =0.00075 ----variance
Standard deviation = √0.00075 = 0.0274
δ = 0.0274
n=4
Thus n< 30 , confidence interval for μ = x ± t*δ/√n ---------(i)
Degree of freedom (df) = n-1 =4-1 =3
The t value for 95% confidence with df= 3 is t=3.182
Substituting the values in equation (i)
0.99 ± 3.182 * 0.0274/√4
0.99 ± 3.182 * 0.0274/2
0.99 ± 3.182 * 0.0137
0.99 ± 0.0436
(0.9464,1.034)
