Question

A manager records the repair cost for 14 randomly selected dryers. A sample mean of $88.34 and standard deviation of $19.22 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the dryers. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

Hence the 90% confidence interval estimate of the population mean is $(79.24 , 97.44)$

Step-by-step explanation:

Given that,  

Point estimate = sample mean = $\bar x$ = 88.34  

sample standard deviation = s = 19.22  

sample size = n = 14  

Degrees of freedom = df = n - 1 = 13  

Critical value =$t\alpha /2,$ df = 1.771

 

Margin of error

$E = t\alpha/2,df \times (\frac{s}{\sqrt{n} } )\\= 1.771 \times (19.22 / \sqrt 14)$  

Margin of error = E = 9.10  

The 90% confidence interval estimate of the population mean is,  

$\bar x - E < \mu < \bar x + E\\\\88.34 - 9.10 < \mu < 88.34 + 9.10\\\\79.24 < \mu < 97.44\\(79.24 , 97.44)$