Question

- publisher wants to estimate the mean length of time in minutes) all adults spend reading newspapers. To determine this estimate the publisher takes a random sample of 15 people and obtain the results below. From past studies, the bisher assumes als 1.9 minutes and that the population of times is normally distributed 10 12 10 7 10 11 6 7 6 Construct the 90% and 99% confidence intervals for the population mean. Which interval is wider convenient, use technology to construct the confidence interval The 90% confidence interval in CD Round to ore decinto place is rended)

Answer 1

Here
X=

ΣΧι n
=
129 15 = 8.6
,$\sigma =1.9$ , n = 15 ( n < 30, here we use t test confidence interval)

Confidence interval for the population mean =

$\bar{X} \pm t * \frac{\sigma }{\sqrt{n}}$

Degrees of freedom = n-1 = 14

Confidence interval = 90%

Therefore, t = 1.761 ... Using t distribution table

$8.6 \pm 1.761 * \frac{1.9}{\sqrt{15}}$

$8.6 \pm 1.761 * 0.4906$

$8.6 \pm 0.8639$

90% confidence interval for population mean

Similarly,

99% confidence interval for population mean

Confidence interval = 99%

Therefore, t = 2.977

$8.6 \pm 2.977 * \frac{1.9}{\sqrt{15}}$

99% confidence interval (7.1 , 10.1) is wider than 90% confidence interval (7.7, 9.5).