Respuesta :
Answer:
[tex]L = 3.975[/tex]
Step-by-step explanation:
Given
See comment for data
Required
The lower limit 94% confidence interval
First, calculate the mean
[tex]\bar x =\frac{\sum x}{n}[/tex]
[tex]\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}[/tex]
[tex]\bar x =\frac{74.1}{12}[/tex]
[tex]\bar x =6.175[/tex]
Next, calculate the standard deviation using:
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}[/tex]
So, we have:
[tex]\sigma = \sqrt{\frac{(0.9 - 6.175)^2 +(10- 6.175)^2 +.................+(8.9- 6.175)^2 +(4.6- 6.175)^2}{12}}[/tex]
[tex]\sigma = \sqrt{\frac{197.2625}{12}}[/tex]
[tex]\sigma = \sqrt{16.4385416667}[/tex]
[tex]\sigma = 4.054[/tex]
For 94% interval, the z score is
[tex]z = 1.88[/tex]
So, the margin of error (E) is:
[tex]E = z * \frac{\sigma}{\sqrt n}[/tex]
[tex]E = 1.88 * \frac{4.054}{\sqrt {12}}[/tex]
[tex]E = 1.88 * \frac{4.054}{3.464}[/tex]
[tex]E = 2.200[/tex]
The confidence interval of the true mean is:
[tex]CI = (\bar x \± E)[/tex]
The lower limit (L) is:
[tex]L = \bar x -E[/tex]
[tex]L = 6.175 -2.200[/tex]
[tex]L = 3.975[/tex]
and the upper limit (U) is
[tex]U = \bar x + E[/tex]
[tex]U = 6.175 + 2.200[/tex]
[tex]U = 8.375[/tex]
The lower limit is 3.975 and the upper limit is 8.375. Then the confidence interval for the true mean is (3.975, 8.375).
What is the margin of error?
The probability or the chances of error while choosing or calculating a sample in a survey is called the margin of error.
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 website.
z-value for 94% of confidence level is 1.88.
z = 1.88
The mean will be
[tex]\mu = \dfrac{\Sigma x}{n}\\\\\mu = \dfrac{0.9+10+0.9+8.7+8.4+2.6+1+13.3+9.5+8.9+8.9+4.6}{12}\\\\\mu = \dfrac{74.1}{12}\\\\\mu = 6.175[/tex]
Then the stabndard devistion will be
[tex]\sigma = \sqrt{ \dfrac{\Sigma (x - \mu )^2}{n}}\\\\\sigma = \sqrt{ \dfrac{\Sigma (0.9-6.175 )^2+(10-6.175 )^2+...+(4.6-6.175 )^2}{12}}\\\\\sigma = \sqrt{ \dfrac{197.2625}{12}}\\\\\sigma = \sqrt{16.438}\\\\\sigma = 4.054[/tex]
The margin of error (E) will be
[tex]E = z*\dfrac{\sigma}{\sqrt{n}}\\\\E = 1.88*\dfrac{4.054}{\sqrt{12}}\\\\E = 2.200[/tex]
Then the confidence interval of the true mean will be
[tex]CI = (\mu \pm E)[/tex]
The lower limit (L) will be
[tex]L = \mu - E \\\\L = 6.175 - 2.200\\\\L = 3.975[/tex]
The upper limit (U) will be
[tex]U = \mu + E \\\\U = 6.175 + 2.200\\\\U = 8.375[/tex]
More about the margin of error link is given below.
https://brainly.com/question/6979326
