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 2.50 gallons. From his records, he selects a random sample of 70 sales and finds the mean number of gallons sold is 7.80.

1. What is the point estimate of the population mean? (Round your answer to 2 decimal places.)
2. Develop a 98% confidence interval for the population mean. (Use z Distribution Table.) (Round your answers to 2 decimal places.)

Answer 1

Solution :

Given that,

Point estimate = sample mean =   = =7.80

Population standard deviation =    = 2.50

Sample size = n =70

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z/2    * ( /n)

= 2.326 * ( 2.50/  70 )

=0.70
At 98% confidence interval estimate of the population mean
is,

- E < < + E

7.80- 0.70   <   < 7.80+ 0.70

7.10 <   < 8.50

( 7.10 , 8.50 )
Point estimate = =7.80