Question

In a random sample of 88 ​people, the mean commute time to work was 36.536.5 minutes and the standard deviation was 7.47.4 minutes. A 9898​% confidence interval using the​ t-distribution was calculated to be (28.7 comma 44.3 )(28.7,44.3). After researching commute times to​ work, it was found that the population standard deviation is 8.68.6 minutes. Find the margin of error and construct a 9898​% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results.

Answer 1

Answer with explanation:

As per given , we have

n= 88

`\overline{x}=36.5\text{ minutes}`

Population standard deviation :
`\sigma=8.6\text{ minutes}`

Critical value for 98% confidence interval :
`z_(\alpha/2)=2.33`n

Margin of error :
`E=z_(\alpha/2)(\sigma)/(√(n))`

`={2.33}(8.6)/(√(88))\\\\=2.13605797717\approx2.14`

98% confidence interval :
`36.5\pm 2.14`

`(36.5-2.14,\ 36.5+2.14)=(34.36,\ 38.64)`

The 98% confidence interval (34.36, 38.64) found by normal distribution with the appropriate calculations for a standard deviation that is known is narrower than the interval (28.7,44.3) found by the t-distribution (population standard deviation is unknown) .