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. Determine the 90% confidence interval for the mean repair cost for the stereos. 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.

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

Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 90% confidence interval for the true mean repair cost, is ;

P(-2.015 < < 2.015) = 0.90  {As the critical value of t at 5 degree of

                                                 freedom are -2.015 & 2.015 with P = 5%}

P(-2.015 < < 2.015) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [ ,  ]

                                                 = [ , ]

                                                 = [45.688 , 85.552]

Therefore, 90% confidence interval for the true mean repair cost for the stereos is [45.688 , 85.552].

three-fourths of an hour

36÷48=0.75

0.75=

answer: it's d

step-by-step explanation: