Question

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.

Answer 1

Answer:

($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:

$X \pm z\frac{s}{\sqrt{n}}$

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

$L=54.41-1.645*\frac{28.89}{\sqrt{44}} \\L=\ 47.245\\U=54.41+1.645*\frac{28.89}{\sqrt{44}} \\U=\ 61.575$

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

Answer 2

Answer:

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. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean repair cost = $54.41

             s = sample standard deviation = $28.89

             n = sample of TV's = 44

             $\mu$ = 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, $\mu$ is ;

P(-1.6816 < $t_4_3$ < 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 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.6816) = 0.90

P( $-1.6816 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.6816 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.6816 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.6816 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.6816 \times {\frac{s}{\sqrt{n} } }$  , $\bar X+1.6816 \times {\frac{s}{\sqrt{n} } }$ ]

                                             = [ $54.41-1.6816 \times {\frac{28.89}{\sqrt{44} } }$ , $54.41+1.6816 \times {\frac{28.89}{\sqrt{44} } }$ ]

                                             = [47.086 , 61.734]

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