Answer to Question #327834 in Statistics and Probability for Vithu

Question

A manufacturer of a brand of designer jeans has pitched her advertising to develop an

expensive and classy image, the suggested retail price is Rs 750. However, she is concerned

that retailers are undermining her image by offering jeans at discount prices. To better

understand what is happening, she randomly sampled 15 retailers who sell her product and

determines the price. The data collected is as follows:

670 720 700 570 780 770 670 720 730 770 720 810 730 790 750

Find the 95% confidence interval estimate for the prices.

Answer 1

Sample mean:

$\mu=\frac{1}{15}(670+720+700+570+780+770+\\ +670+720+730+770+720+810+\\ +730+790+750)=\\ =\frac{1}{15}(570+2\cdot670+700+3\cdot720+\\ +2\cdot730+750+2\cdot770+780+790+810)\approx\\ \approx726.67$

Variance:

$\sigma^2=\frac{1}{n-1}\sum_{i=0}^{n}(x_i-\mu)^2=\\ =\frac{1}{14}(156.67^2+2\cdot56.67^2+26.67^2+\\ +3\cdot6.67^2+2\cdot3.33^2+23.33^2+2\cdot43.33^2+\\ +53.33^2+63.33^2+83.33^2)\approx 3566.67$

Standard deviation:

$\sigma=\sqrt{\sigma^2}\approx59.72$

t-value for 14 degrees of freedom and 95% level of confidence is 2.131

Confidence interval:

$\mu\pm t\frac{\sigma}{\sqrt{n}}=726.67\pm 2.131\cdot\frac{59.72}{\sqrt{15}}=726.67\pm32.86$

The suggested retail price fits into this interval.