Question

In a random sample of 8 people, the mean commute time to work was 34.5minutes and the standard deviation was 7.3 minutes. A 90% confidence intervalusing the t-distribution was calculated to be (29.6,39.4). After researching commutetimes to work, it was found that the population standard deviation is 8.9 minutes.Find the margin of error and construct a 90% confidence interval using thestandard normal distribution with the appropriate calculations for a standarddeviation that is known. Compare the results.

Answer 1

Given;

sample, n = 8 people

sample mean, x' = 34.5 min

sample standard deviation, s = 7.3 min

90% confidence interval, CI = (29.6,39.4)

population standard deviation is 8.9 min

Find;

Margin of error, M

90% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known.

Margin of error formula:

`M=t*(s)/(√(n))`

t is the critical value for a 2 tailed test at a 10% level of significance

degree of freedom = sample size - 1, df = 8 - 1 = 7

and the level of significance α = 10%

The value of t from a t distribution table is; t = 1.8946

`M=1.8946*(7.3)/(√(8))=4.9`

Another way to calculate M, is by substracting CI - x' = 39.4 - 34.5 = 4.9

Now, let's construct the 90% CI using s = 8.9

The new Margin of error, M* is:

`M*=1.8946*(8.9)/(√(8))=5.9616\cong5.96`

Therefore, the new Confidence interval, CI* is

`\begin{gathered} CI*=x^(\prime)\pm M=34.5\pm5.96 \\ CI=\left(28.54,40.46\right) \end{gathered}`

Answer: New margin of error, M* = 5.96, New confidence interval, CI* = (28.54,40.46)

comparison:

Initial margin of error, M = 4.9

New margin of error, M* = 5.96

Initial confidence interval, CI = (29.6,39.4)

New confidence interval, CI* = (28.54,40.46)