Suppose you work in the quality assurance department of a chemical processing plant and it is your responsibility to ascertain whether the viscosity of a certain chemical remains within a prescribed limit. To that end you measure fluid samples using a viscometer. a. The viscosities of a sample of 10 fluids processed during the day shift were measured to have the following results (Pa.s): 71, 45, 54, 75, 50, 49, 63, 55, 48, 50. Determine the range within which the true mean of the fluid samples from the day shift will be with a 95% confidence.

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

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