A survey of the mean number of cents off that coupons give w

A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20

Solution

Since sample size is less than 30 and population standard deviation is unknown hence use t-distribution.

d.
sample size,n = 14
sample mean,xbar = 53.9286
standard deviation,s = 31.6336
alpha,a = 1-0.95 = 0.05

t(a/2,n-1) = t(0.025, 13) = 2.16

confidence interval:
xbar +/- [t(a/2,n-1) * (s/sqrt(n))]
53.93 +/- [2.16 * (31.6336/sqrt(14))]
53.9286 +/- 18.2616
= ( 35.6670 , 72.1902 )

We are 95% confident that population mean worth of coupons lies in the interval ( 35.6670 , 72.1902 ).

error bound = 18.2616

95% of the confidence intervals constructed should contain the population mean worth of coupons