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 : $E=z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

$={2.33}\dfrac{8.6}{\sqrt{88}}\\\\=2.13605797717\approx2.14$

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) .