A researcher records the repair cost for 8 randomly selected refrigerators. A sample mean of $57.89 and standard deviation of $23.69 are subsequently computed. Determine the 95% confidence interval for the mean repair cost for the refrigerators. 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.

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Question:

A researcher records the repair cost for 8 randomly selected refrigerators. A sample mean of $57.89 and standard deviation of $23.69 are subsequently computed. Determine the 95% confidence interval for the mean repair cost for the refrigerators. 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.

Step 2 of 2 : Construct the 95% confidence interval. Round your answer to two decimal places.

Given Information:  

Sample mean repair cost  = $57.89

Sample standard deviation = σ = $23.69

Sample size = 8

Confidence level = 95%

Required Information:  

step 1: critical value = ?

step 2: 95% confidence interval = ?

step 1: critical value = 2.365

step 2: 95% confidence interval = ($38.08, $77.70)

Step-by-step explanation:

Since the sample size is less than 30 and the standard deviation of the population is also unknown therefore, we can use the t-distribution to find the required confidence interval.

The confidence interval is given by

Where is the mean repair cost and MoE is the margin of error that is given by

Where n is the sample size, s is the sample standard deviation, and  is the t-score corresponding to 95% confidence level.

The t-score corresponding to 95% confidence level is

Significance level = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom (DoF) = n - 1 = 8 - 1 = 7

From the t-table at α = 0.025 and DoF = 7

t-score = 2.365

Therefore, the critical value that should be used in constructing the confidence interval is 2.365

So the required 95% confidence interval is

Therefore, we are 95% confident that the mean repair cost for the refrigerators is within the range of ($38.08, $77.70)

a. the are of the rectangle is 18.25

b. 18.25 times two is 36.50

rae me : d

i will pick c put check it over to be sure

step-by-step explanation: