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

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

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. =
`(\bar X - \mu)/((s)/(√(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 <
`(\bar X - \mu)/((s)/(√(n) ) )` < 1.6816) = 0.90

P(
`-1.6816 * {(s)/(√(n) ) }` <
`{\bar X - \mu}` <
`1.6816 * {(s)/(√(n) ) }` ) = 0.90

P(
`\bar X-1.6816 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+1.6816 * {(s)/(√(n) ) }` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X-1.6816 * {(s)/(√(n) ) }` ,
`\bar X+1.6816 * {(s)/(√(n) ) }` ]

= [
`54.41-1.6816 * {(28.89)/(√(44) ) }` ,
`54.41+1.6816 * {(28.89)/(√(44) ) }` ]

= [47.086 , 61.734]

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

Answer 2

($47.245; $61.575)

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(s)/(√(n))`

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

`L=54.41-1.645*(28.89)/(√(44)) \\L=\$47.245\\U=54.41+1.645*(28.89)/(√(44)) \\U=\$61.575`

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