Question

A consumer advocate group selects a random sample of 20 ink cartridges and finds

that the average number of printouts per ink cartridge is 460 with a standard
deviation of 52. Find the 98% confidence interval for the population mean of
printouts per ink cartridge. You may assume that the number of printouts per ink
cartridge is approximately normally distributed.
Enter the lower and upper bounds for the interval in the following boxes,
respectively. Round your answers to nearest whole number.

Answer 1

A 98% confidence interval for the population mean of printouts per ink cartridge is [430.5, 489.5].

Explanation:

We are given that a consumer advocate group selects a random sample of 20 ink cartridges and finds that the average number of printouts per ink cartridge is 460 with a standard deviation of 52.

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 average number of printouts per ink cartridge = 460

s = sample standard deviation = 52

n = sample of ink cartridges = 20

`\mu` = population mean printouts per ink cartridge

Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 98% confidence interval for the population mean,
`\mu` is ;

P(-2.54 <
`t_1_9` < 2.54) = 0.98 {As the critical value of t at 19 degrees of

freedom are -2.54 & 2.54 with P = 1%}

P(-2.54 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.54) = 0.98

P(
`-2.54 * {(s)/(√(n) ) }` <
`\bar X-\mu}` <
`2.54 * {(s)/(√(n) ) }` ) = 0.98

P(
`\bar X-2.54 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.54 * {(s)/(√(n) ) }` ) = 0.98

98% confidence interval for
`\mu` = [
`\bar X-2.54 * {(s)/(√(n) ) }` ,
`\bar X+2.54 * {(s)/(√(n) ) }` ]

= [
`460-2.54 * {(52)/(√(20) ) }` ,
`460+2.54 * {(52)/(√(20) ) }` ]

= [430.5, 489.5]

Therefore, a 98% confidence interval for the population mean of printouts per ink cartridge is [430.5, 489.5].