Answer:
90% confidence interval for the mean repair cost for the stereos is [45.688 , 85.552].
Step-by-step explanation:
We are given that a student records the repair cost for 6 randomly selected stereos. A sample mean of $65.62 and standard deviation of $24.23 are subsequently computed.
Firstly, the pivotal quantity for 90% confidence interval for the true mean repair cost for the stereos is given by;
P.Q. = ~
where, = sample mean repair cost = $65.62
s = sample standard deviation = $24.23
n = sample of stereos = 6
= true mean repair cost
<em>Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.</em>
So, 90% confidence interval for the true mean repair cost, is ;
P(-2.015 < < 2.015) = 0.90 {<u>As the critical value of t at 5 degree of </u>
<u>freedom are -2.015 & 2.015 with P = 5%</u>}
P(-2.015 < < 2.015) = 0.90
P( < < ) = 0.90
P( < < ) = 0.90
<u>90% confidence interval for</u> = [ , ]
= [ , ]
= [45.688 , 85.552]
Therefore, 90% confidence interval for the true mean repair cost for the stereos is [45.688 , 85.552].