A consumer affairs investigator records the repair cost for 44 randomly selected TVs. A sample mean of $54.41$ 54.41 and standard deviation of $28.89$ 28.89 are subsequently computed. Determine the 90%90% confidence interval for the mean repair cost for the TVs. 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.

($47.245; $61.575)

Step-by-step explanation:

Mean sample repair cost (X) = $54.41

Standard deviation (s) = $28.89

Sample size (n) =44

Z-score for a 90% confidence interval (z) = 1.645

The confidence interval, assuming a normal distribution, is given by:

Applying the given data, the lower (L) and upper (U) bounds of the interval are:

The confidence interval is I = ($47.245; $61.575)

90% confidence interval for the mean repair cost for the TV's is [47.086 , 61.734].

Step-by-step explanation:

We are given that a consumer affairs investigator records the repair cost for 44 randomly selected TVs. A sample mean of $54.41$ 54.41 and standard deviation of $28.89$ 28.89 are subsequently computed.

Firstly, the pivotal quantity for 90% confidence interval for the true mean repair cost for the TV's is given by;

        P.Q. = ~

where, = sample mean repair cost = $54.41

             s = sample standard deviation = $28.89

             n = sample of TV's = 44

              = 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(-1.6816 < < 1.6816) = 0.90  {As the critical value of t at 43 degree of

                                                freedom are -1.6816 & 1.6816 with P = 5%}

P(-1.6816 < < 1.6816) = 0.90

P( < < ) = 0.90

P( < < ) = 0.90

90% confidence interval for = [  ,  ]

                                             = [ , ]

                                             = [47.086 , 61.734]

Therefore, 90% confidence interval for the true mean repair cost for the TV's is [47.086 , 61.734].

the first one, all those points on the line just make a strait line.