Question

Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 1.80 gallons. From his records, he selects a random sample of 80 sales and finds the mean number of gallons sold is 6.90.

1. What is the point estimate of the population mean? (Round your answer to 2 decimal places.)

2. Develop a 95% confidence interval for the population mean. (Use NORM.S.INV() for z.) (Round your answers to 2 decimal places.)

Answer 1

Answer:

a)

Given,

To determine the point estimate of the population mean

sigma = 1.80 gallons

n = 80

Sample mean = 6.90

Here the point estimate for population mean is the Sample mean x bar

i.e.,

point estimate of the population mean p^ = 6.90

b)

To determine the 95% confidence interval for the population mean

Now for 95% confidence interval,

z alpha/2 = 1.96

Standard error = sigma/sqrt (n)

= 1.80 / sqrt(80)

= 0.20125

Margin of error = +/- Z*standard error

= +/- 1.96*0.20125

Margin of error = +/- 0.3945

Now let us consider,

Confidence interval = point estimate +/- 0.3945

= 6.90 +/- 0.3945

= (6.5055 , 7.2945)

Confidence interval = (6.51 , 7.30)