Question

Dylan Jones kept careful records of the fuel efficiency of his new car. After the first seven times he filled up the tank, he found the mean was 20.9 miles per gallon (mpg) with a sample standard deviation of 0.7 mpg.

1. Compute the 95% confidence interval for his mpg. (Use t Distribution Table.) (Round your answers to 3 decimal places.)

confidence interval for is mpg ____ and _____.

2. How many times should he fill his gas tank to obtain a margin of error below 0.10 mpg? (Use z Distribution Table.) (Round your answer to the next whole number.)

sample size ___

Answer 1

Solution :

Given that,

= 20.9

s = 0.7

n = 7

Degrees of freedom = df = n - 1 = 7 - 1 = 6

a) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,6 = 2.447

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

= 2.447 * (0.7 / 7)

= 0.647

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

- E < < + E

20.9 - 0.647 < < 20.9 + 0.647

20.253 < < 21.547

Confidence Interval for mpg is 20.253 and 21.547

b) Margin of error (E) = 0.10

At 90% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Sample size = n = [(Z_/2 * ) / E]²

= [(1.96 * 0.7) /0.10 ]²

= 188.2384

Sample size = 189