Question

You intend to estimate a population mean mu with the following sample. 45.3, 42.3, 53, 49, 15.2, 52.3, 45.6, 39.6, 39.4, 16.1, 54.4.You believe the population is normally distributed. Find the 99.9% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).

Answer 1

A 99.9% confidence interval for the population mean is [22.31, 59.91] .

Explanation:

We are given the following sample values;

X = 45.3, 42.3, 53, 49, 15.2, 52.3, 45.6, 39.6, 39.4, 16.1, 54.4.

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 mean =
`(\sum X)/(n)` = 41.11

s = sample standard deviation =
`\sqrt{(\sum (X - \bar X)^(2) )/(n-1) }` = 13.59

n = sample size = 11

Here for constructing a 99.9% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 99.9% confidence interval for the population mean,
`\mu` is;

P(-4.587 <
`t_1_0` < 4.587) = 0.999 {As the critical value of t at 10 degrees of

freedom are -4.587 & 4.587 with P = 0.05%}

P(-4.587 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 4.587) = 0.999

P(
`-4.587 * {(s)/(√(n) ) }` <
`{\bar X-\mu}{` <
`4.587 * {(s)/(√(n) ) }` ) = 0.999

P(
`\bar X-4.587 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+4.587 * {(s)/(√(n) ) }` ) = 0.999

99.9% confidence interval for
`\mu` = [
`\bar X-4.587 * {(s)/(√(n) ) }` ,
`\bar X+4.587 * {(s)/(√(n) ) }` ]

= [
`41.11-4.587 * {(13.59)/(√(11) ) }` ,
`41.11+4.587 * {(13.59)/(√(11) ) }` ]

= [22.31, 59.91]

Therefore, a 99.9% confidence interval for the population mean is [22.31, 59.91] .