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

Answer:

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.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean worth of coupons = $\frac{\sum X}{n}$ = $\frac{750}{14}$ = 53.57 cents

            s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 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 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.16) = 0.95

P( $-2.16 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}{$ < $2.16 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.16 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.16 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.16 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.16 \times {\frac{s}{\sqrt{n} } }$ ]

 = [ $53.57-2.16 \times {\frac{31.40}{\sqrt{14} } }$ , $53.57+2.16 \times {\frac{31.40}{\sqrt{14} } }$ ]

 = [35.443, 71.697]

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