Question

In a random sample of

8

​people, the mean commute time to work was

35.5

minutes and the standard deviation was

7.3

minutes. A

98%

confidence interval using the​ t-distribution was calculated to be

left parenthesis 27.8 comma 43.2 right parenthesis(27.8,43.2).

After researching commute times to​ work, it was found that the population standard deviation is

8.7

minutes. Find the margin of error and construct a

98%

confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 35.5

Population standard deviation = = 7.3

Sample size = n = 8

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

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

= 2.326 * (7.3 / 8)

= 6

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

- E < < + E

35.5 - 6.0 < < 35.5 + 6.0

29.5 < < 41.5

(29.5 , 41.5)

Using the t distribution the margin of error is high .