Question

Dylan Jones kept careful records of the fuel efficiency of his new car. After the first twelve times he filled up the tank, he found the mean was 22.9 miles per gallon (mpg) with a sample standard deviation of 1.2 mpg.

1. Compute the 98% confidence interval for his mpg. (Use t Distribution Table.) (Round your answers to 3 decimal places.)

2. How many times should he fill his gas tank to obtain a margin of error below 0.15 mpg? (Use z Distribution Table.) (Round your answer to the next whole number.)

Answer 1

Solution :

Given that,

= 22.9

s =1.2

n = 12

Degrees of freedom = df = n - 1 = 12- 1 = 11

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t _/2,df = t_0.01,11 = 2.718

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

=2.718 * ( 1.2/ 12) =0.942

The 98% confidence interval estimate of the population mean is,

- E < < + E

22.9 - 0.942 < <22.9 + 0.942

21.958 < < 23.842

( 21.958 , 23.842)

(B)

Solution :

Given that,

standard deviation =s =   =1.2

Margin of error = E = 0.15

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z_/2 = Z_0.01 = 2.326

sample size = n = [Z_/2* / E] ²

n = ( 2.326* 1.2 / 0.15 )²

n =346

Sample size = n =346