Question

quality control specialist for a restaurant chain takes a random sample of size 13 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.54. Assume the underlying population is normally distributed. Find the 95% confidence interval for the true population mean for the amount of soda served. (Round your answers to two decimal places.) ,

Answer 1

95% confidence interval for the true population mean for the amount of soda served is [12.37 , 14.23].

Explanation:

We are given that quality control specialist for a restaurant chain takes a random sample of size 13 to check the amount of soda served in the 16 oz. serving size.

The sample mean is 13.30 with a sample standard deviation of 1.54.

Firstly, the pivotal quantity for 95% confidence interval for the true population mean is given by;

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

where,
`\bar X` = sample mean = 13.30

s = sample standard deviation = 1.54

n = sample size = 13

`\mu` = true population mean

Here for constructing 95% confidence interval we have used 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(-2.179 <
`t_1_2` < 2.179) = 0.95 {As the critical value of t at 12 degree of

freedom are -2.179 & 2.179 with P = 2.5%}

P(-2.179 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.179) = 0.95

P(
`-2.179 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.179 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.179 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.179 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-2.179 * {(s)/(√(n) ) }` ,
`\bar X+2.179 * {(s)/(√(n) ) }` ]

= [
`13.30-2.179 * {(1.54)/(√(13) ) }` ,
`13.30+2.179 * {(1.54)/(√(13) ) }` ]

= [12.37 , 14.23]

Therefore, 95% confidence interval for the true population mean for the amount of soda served is [12.37 , 14.23].