Question

for the data values below construct a 95 confidence interval if the sample mean is known to be 12898 and the standard deviation is 7719

Answer 1

A 95% confidence interval for the population mean is [3315.13, 22480.87] .

Explanation:

We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample proportion of defective items = 12,898

s = sample standard deviation = 7,719

n = sample size = 5

`\mu` = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean,
`\mu` is ;

P(-2.776 <
`t_4` < 2.776) = 0.95 {As the critical value of t at 4 degrees of

freedom are -2.776 & 2.776 with P = 2.5%}

P(-2.776 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.776) = 0.95

P(
`-2.776 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.776 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.776 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.776 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-2.776 * {(s)/(√(n) ) }` ,
`\bar X+2.776 * {(s)/(√(n) ) }` ]

= [
`12,898-2.776 * {(7,719)/(√(5) ) }` ,
`12,898+2.776 * {(7,719)/(√(5) ) }` ]

= [3315.13, 22480.87]

Therefore, a 95% confidence interval for the population mean is [3315.13, 22480.87] .