Question

suppose that cellulose content in the normally distributed population has standard deciation of 8 mg/g. Asimple random sample of 30 cuttings has mean cellulose content of 148 mg/g. Construct and interpret an 84.5% confidence interval for the mean cellulose content in the population

Answer 1

Solution:

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 148

σ = 8

n = 30

Confidence level = 85%

Critical Z value = 1.4221

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 148 ± 1.4221*8/sqrt(30)

Confidence interval = 148 ± 2.0771

Lower limit = 148 - 2.0771 = 145.92

Upper limit = 148 + 2.0771 =150.08

Confidence interval = (145.92, 150.08)

We are 84.5% confident that the population mean cellulose content will lies between 145.92 mg/g and 150.08 mg/g.