Answer:
48.668 ≤ μ ≤ 63.332
Step-by-step explanation:
Given the data:
Sample (X) : 71, 45, 54, 75, 50, 49, 63, 55, 48, 50
Number of samples (n) = 10
α = 95% = 0.05
Determine the range within which the true mean of the fluid samples from the day shift will be with a 95% confidence
True mean = μ
Confidence interval :
m ± t(α/2 ; df) * s/√n
m - t(α/2 ; df) * s/√n ≤ μ ≤ m + t(α/2 ; df) * s/√n
m = sample mean ; s = sample standard deviation ; df = degree of freedom =(n - 1) = (10 - 1) = 9
Calculating the sample mean and standard deviation using calculator to save computation time :
71, 45, 54, 75, 50, 49, 63, 55, 48, 50
m = 56 ; s = 10.25
m - t(0.05/2 ; 9) * s/√n ≤ μ ≤ m + t(0.05/2 ; df) * s/√n
From t table : t(0.05/2 ; 9) = t(0.025, 9) = 2.262
56-2.262 * (10.25/√10) ≤ μ ≤ 56+2.262 * (10.25/√10)
48.668 ≤ μ ≤ 63.332
