A sample of 45 small generators produce fumes that was observed to be 700 ppm with a sample standard deviation of 165.

 

a)      Calculate and interpret 95% two-sided confidence interval for true average amount of fumes for a population of all similar generators.

 

b)     Assume the initial observations were based on an initial estimate of 170 for the value of the sample standard deviation. What is the necessary sample size that would result in an interval width of 40 ppm for a confidence level of 95%?

a) The 95% confidence interval estimate of the population mean is (650.44, 749.56)

b) The necessary sample size, n = 294

Step-by-step explanation

Given that,

a) Point estimate = sample mean, x? = 700

sample standard deviation, s = 165

sample size, n = 45

Degrees of freedom = df = n - 1 = 45 - 1 = 44

At 95% confidence level

 α = 1 - 95%

 α = 1 - 0.95 = 0.05

α/2 = 0.025

t_α/2,_df = t_0.025,44 = 2.015

Margin of error, E = t_α/2,df * (s /√n)

= 2.015 * ( 165/√45) = 2.015*24.5967

= 49.5624

Margin of error, E = 49.56

The 95% confidence interval estimate of the population mean is,

 x? ± E

= 700 ± 49.56

=(700 - 49.56, 700 + 49.56)

= (650.44, 749.56)

b) Margin of error, E = 40/2 = 20

sample standard deviation, s = 170

sample size, n = [ t_α/2,df * s/E]^2

n = [ 2.015 * 170 / 20 ]^2

= 17.1275^2

n = 293.35 ≈ 294

Thus, Sample size, n = 294