Question

Estimate the mean mpg of all cars with 86% confidence. After collecting a random sample of 26 cars you find a sample mean of 31 miles. Assume the distribution of mpg for all cars is normal with a standard deviation of 1.2 miles. ( Round your answers to two decimal places.)

1. Z table value =

2. Margin of Error =

3: You estimate with 86% confidence that the population mean falls between the lower value of and the upper value of

*please show the work

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 31

Population standard deviation = = 1.2

Sample size = n = 26

1)

At 86% confidence level the z is ,

= 1 - 86% = 1 - 0.86 = 0.14

/ 2 = 0.14 / 2 = 0.07

Z_/2 = Z_0.07 = 1.48

2)

Margin of error = E = Z_/2* ( /n)

= 1.48* (1.2 / 26)

= 0.348

3)

At 86% confidence interval estimate of the population mean is,

- E < < + E

31 - 0.348 < < 31 + 0.348

30.65 < < 31.35

(30.65 , 31.35 )