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)

The lower bound = 35.443

The upper bound = 71.697

The error bound = 18.127

Step-by-step 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.  =    ~

where, = sample mean worth of coupons = = = 53.57 cents

            s = sample standard deviation = = 31.40 cents

            n = sample size = 14

             = 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, is ;

P(-2.16 < < 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 < < 2.16) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = [ , ]

 = [ , ]

 = [35.443, 71.697]

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

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