Question

A random sample of 10 employees of a company was selected to estimate the mean one-way commute time for all employees at the company. The mean and standard deviation of the sample were 38 minutes and 6 minutes, respectively. Assuming all conditions for inference are met, which of the following is the margin of error, in minutes, for a 95 percent confidence interval for the population mean one-way commute time? 1.812(610√) 1.812 times the fraction with numerator 6, and denominator the square root of 10 A 1.833(610√) 1.833 times the fraction with numerator 6, and denominator the square root of 10 B 1.96(610√) 1.96 times the fraction with numerator 6, and denominator the square root of 10 C 2.228(610√) 2.228 times the fraction with numerator 6, and denominator the square root of 10 D 2.262(610√)

Answer 1

Answer:

The margin of error is of $2.262\frac{6}{\sqrt{10}} = 4.29$

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 $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.262

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.262\frac{6}{\sqrt{10}} = 4.29$

In which s is the standard deviation of the sample and n is the size of the sample.