Question

A survey of the average amount of cents off that coupons give was done by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20cents; 70cents; 50cents; 65cents; 30cents; 55cents; 40cents; 40cents; 30cents; 55cents; $1.50; 40cents; 65cents; 40cents. Assume the underlying distribution is approximately normal.

Construct a 95% confidence interval for the population mean worth of coupons .
What is the lower bound? ( Round to 3 decimal places )
What is the upper bound? ( Round to 3 decimal places )
What is the error bound? (Round to 3 decimal places)

Answer 1

The lower bound = 35.443

The upper bound = 71.697

The error bound = 18.127

Explanation:

We are given that a survey of the average amount of cents off that coupons gives was done by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News.

The following data were collected (X): 20cents; 70cents; 50cents; 65cents; 30cents; 55cents; 40cents; 40cents; 30cents; 55cents; 150 cents; 40cents; 65cents; 40cents.

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

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

where,
`\bar X` = sample mean worth of coupons =
`(\sum X)/(n)` =
`(750)/(14)` = 53.57 cents

s = sample standard deviation =
`\sqrt{(\sum (X-\bar X)^(2) )/(n-1) }` = 31.40 cents

n = sample size = 14

`\mu` = population mean worth of coupons

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(-2.16 <
`t_1_3` < 2.16) = 0.95 {As the critical value of t at 13 degrees of

freedom are -2.16 & 2.16 with P = 2.5%}

P(-2.16 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.16) = 0.95

P(
`-2.16 * {(s)/(√(n) ) }` <
`{\bar X-\mu}{` <
`2.16 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.16 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.16 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-2.16 * {(s)/(√(n) ) }` ,
`\bar X+2.16 * {(s)/(√(n) ) }` ]

= [
`53.57-2.16 * {(31.40)/(√(14) ) }` ,
`53.57+2.16 * {(31.40)/(√(14) ) }` ]

= [35.443, 71.697]

Therefore, a 95% confidence interval for the population mean worth of coupons is [35.443, 71.697].