Question

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¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal.

a. i. $\stackrel{-}{x}$= __________

ii. $s_x$= __________

iii. n = __________

iv. n-1 = __________

b. Define the random variables $X$and$\stackrel{-}{X}$ in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 95% confidence interval for the population mean worth of coupons.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

e. If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why.

Short answer

a. (i) $\stackrel{-}{x}=0.539$

(ii) $s_x=0.316$

(iii) $n=14$

(iv) $n-1=13$

b. $X$means one coupons discount amount and $\stackrel{-}{X}$means mean discount amount.

c. $t_{14-1}=13$

d. (i) $CI=(0.357,0.722)$

(ii) the graph is shown

(iii) Error bound$=0.183$

e.$95\mathrm{\%}$

Step-by-step solution

part (a) explanation

(i) Mean time $\stackrel{-}{x}=0.539$

(ii) Standard deviation $s_x=0.316$

(iii) Sample size $n=14$

(iv)$n-1=14-1=13$

part (b) explanation

$X$means one coupons discount amount and$\stackrel{-}{X}$ means mean discount amount.

part (c) explanation

Since we don't know the standard deviation population we will use $t_{14-1}=13$.

part (d) explanation

(i) We have mean time $\stackrel{-}{x}=0.539$and standard deviation$s_x=0.316$

$n=14$

$\alpha =1-CL$$=1-0.95=0.05$

nowlocalid="1650877272733" $\Rightarrow 0.025t_{\frac{a}{2}}=2.16$

Error bound= $=t\frac{a}{2}\left(\frac{s}{\sqrt{n}}\right)$

$=2.16\left(\frac{0.316}{\sqrt{14}}\right)$

Confidence interval =$\stackrel{-}{x}$$\pm$error bound

$\Rightarrow 0.539\pm 2.16\left(\begin{array}{c}0.316\\ \sqrt{14}\end{array}\right)$

Confidence interval=$(0.357,0.722)$

(ii) The graph is shown

(iii) Error bound is calculated as

$t_{n-1}\left(\frac{\alpha }{2}\right)\frac{s}{\sqrt{n}}$

$t_{14-1}\left(\frac{0.05}{2}\right)\frac{0.316}{\sqrt{14}}$

$=0.183$

part (e) explanation

If many random samples were taken of size$14$, $95$per cent of the confidence intervals constructed should contain the population mean worth of coupons.

As the level of confidence is defined $95\%$samples contain the population mean worth of coupons.