Answer:
The margin of error is of [tex]2.262\frac{6}{\sqrt{10}} = 4.29[/tex]
Step-by-step explanation:
We have the standard deviation for the sample. So the T-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 10 - 1 = 9
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.262
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.262\frac{6}{\sqrt{10}} = 4.29[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The equation that represents the margin of error is: [tex]E = 2.262 \times \frac{6}{\sqrt {10}}[/tex]
The given parameters are:
The margin of error is calculated using:
[tex]E = z_y \times \frac{\sigma}{\sqrt n}[/tex]
So, we have:
[tex]E = z_y \times \frac{6}{\sqrt {10}}[/tex]
The quantile at 95% confidence interval is 2.262
So, the equation becomes
[tex]E = 2.262 \times \frac{6}{\sqrt {10}}[/tex]
Hence, the equation that represents the margin of error is: [tex]E = 2.262 \times \frac{6}{\sqrt {10}}[/tex]
Read more about margin of error at:
https://brainly.com/question/14396648
