Question

A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sample of 33 bins, the sample mean amount was 50.11 pounds and the sample standard deviation was 3.641 pounds. Construct a 94.5% confidence interval for the expected amount of garbage per bin for all bins in the city.
Answer to 3 decimals
(a) What is the lower limit of the 94.5% interval? Give your answer to three decimal places.

(b) What is the upper limit of the 94.5% interval? Give your answer to three decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 50.11

sample standard deviation = s = 3.641

sample size = n = 33

Degrees of freedom = df = n - 1 = 33 - 1 = 32

At 94.5% confidence level

= 1 - 94.5%

= 1 - 0.945 = 0.055

/2 = 0.0275

t/2,df = t0.0275,32  = 1.992

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

= 1.992 * ( 3.641 / 33)

Margin of error = E = 1.263

(a) The lower limit of the 94.5% confidence interval estimate of the population mean is,

   - E

= 50.11 - 1.263

= 48.847

(b) The upper limit of the 94.5% confidence interval estimate of the population mean is,

   + E

= 50.11 + 1.263

= 51.373