# A statistics practitioner took a random sample of 43 observations from a population whose standard deviation is 26 and computed the sample mean to be

A statistics practitioner took a random sample of 43 observations from a population whose standard deviation is 26 and computed the sample mean to be 108. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 95% confidence, changing the population standard deviation to 58, Confidence Interval = C. Estimate the population mean with 95% confidence, changing the population standard deviation to 8, Confidence Interval =
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Step 1

a. It is given that, Sample mean = 108. Population standard deviation, $\sigma =26$. The sample size, n is 43. The z-critical value for 95% confidence interval is 1.96. The 95% confidence interval can be calculated as follows: $CI=\stackrel{―}{x}±z\left(\frac{\sigma }{\sqrt{n}}\right)$
$=108±1.96\left(\frac{26}{\sqrt{43}}\right)$
$=108±7.7713$
$=\left(100.2287,115.7713\right)$ The 95% confidence interval is (100.2287, 115.7713).

Step 2

b. Population standard deviation, $\sigma$ is 58. The 95% confidence interval can be calculated as follows: $CI=\stackrel{―}{x}±z\left(\frac{\sigma }{\sqrt{n}}\right)$
$=108±1.96\left(\frac{58}{\sqrt{43}}\right)$
$=108±17.3360$
$=\left(90.664,125.336\right)$ The 95% confidence interval is (90.664,125.336).

Step 3

c. Population standard deviation, $\sigma =8$. The 95% confidence interval can be calculated as follows: $CI=\stackrel{―}{x}±z\left(\frac{\sigma }{\sqrt{n}}\right)$
$=108±1.96\left(\frac{8}{\sqrt{43}}\right)$
$=108±2.3911$
$=\left(105.6089,110.3911\right)$ The 95% confidence interval is (105.6089,110.3911).