The mean repair cost for the stereos is (63.04, 102.86) with an 80% confidence interval.
<h3>What is the critical factor?</h3>
The critical factor for a 90% confidence interval for the true mean repair cost for stereos is given by;
Critical factor = (x-μ)/(s/√n)
where, x = sample mean repair cost
s = standard deviation of a sample
n = sample of stereos
μ = critical value
We have been given that a student records the repair cost for 17 randomly selected washers. a sample mean of $82.95 and a standard deviation of $14.89 are subsequently computed.
Since we don't know the population standard deviation, we used t statistics to create the 80% confidence interval in this case.
So, the 80% confidence interval for the true mean repair cost, μ is ;
⇒ P(-1.337< t₅ < 1.337) = 0.80
{As the critical value of t at 5 degrees of freedom are -1.337& 1.337 with P = 5%}
⇒ P(-1.337< (x-μ)/(s/√n) < 1.337) = 0.80
⇒ P(-1.337×(s/√n) < (x-μ) < 1.337×(s/√n)) = 0.80
⇒ P(x - 1.337×(s/√n) < μ < x + 1.337×(s/√n)) = 0.80
80% confidence interval for
⇒ μ = (x - 1.337×(s/√n) , x + 1.337×(s/√n))
Here, x = $82.95 , s = $14.89 , and n = 17
⇒ μ = (82.95 - 1.337×(14.89 /√17) , 82.95 + 1.337×(14.89 /√17))
⇒ μ = (63.04, 102.86)
Hence, the mean repair cost for the stereos is (63.04, 102.86) with a 80% confidence interval.
Learn more about the critical value here:
brainly.com/question/15417413
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