Question

Company A is trying to sell its website to Company B. As part of the sale, Company A claims that the average user of their site stays on the site for 10 minutes. To test this claim Company B collects the times (in minutes) below for a sample of 12 users. Assume normality. Assignment 6q3 data Construct a 94% confidence interval for the true mean time spent on the web site. a) What is the lower limit of the 94% interval

Answer 1

Answer:

$L = 3.975$

Step-by-step explanation:

Given

See comment for data

Required

The lower limit 94% confidence interval

First, calculate the mean

$\bar x =\frac{\sum x}{n}$

$\bar x =\frac{0.9+10+0.9+8.7+4.8+2.6+1+13.3+9.5+8.9+8.9+4.6}{12}$

$\bar x =\frac{74.1}{12}$

$\bar x =6.175$

Next, calculate the standard deviation using:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

So, we have:

$\sigma = \sqrt{\frac{(0.9 - 6.175)^2 +(10- 6.175)^2 +.................+(8.9- 6.175)^2 +(4.6- 6.175)^2}{12}}$

$\sigma = \sqrt{\frac{197.2625}{12}}$

$\sigma = \sqrt{16.4385416667}$

$\sigma = 4.054$

For 94% interval, the z score is

$z = 1.88$

So, the margin of error (E) is:

$E = z * \frac{\sigma}{\sqrt n}$

$E = 1.88 * \frac{4.054}{\sqrt {12}}$

$E = 1.88 * \frac{4.054}{3.464}$

$E = 2.200$

The confidence interval of the true mean is:

$CI = (\bar x \± E)$

The lower limit (L) is:

$L = \bar x -E$

$L = 6.175 -2.200$

$L = 3.975$

and the upper limit (U) is

$U = \bar x + E$

$U = 6.175 + 2.200$

$U = 8.375$