Question

On this problem, the answer has been worked out, but you must fill in the blanks in the solution.A recent study of 28 randomly selected employees of a company showed that the mean of the distance they traveled to work was 14.3 miles. The standard deviation of this sample was 2.0 miles. Find the 95% confidence interval for µ (the true mean time for all employees of the company). Round your answer to one place after the decimal point.

Answer 1

`\begin{gathered} \bar{X}\text{ }\pm\text{ }t_{(\alpha)/(2)}(\frac{s}{\sqrt[]{n}})\text{ = }14.3\text{ }\pm\text{ 2.052(}\frac{2}{\sqrt[]{28}})\text{ = 1-}\alpha \\ 13.5\text{ < }\mu\text{ < }15.1 \end{gathered}`

The 1st box: 14.3

second box: 2.052

Third box: 2

fourth box: 28

fifth box: 13.5

sixth box: 15.1

Step-by-step explanation:
`\begin{gathered} nu\text{mber in survey = 28} \\ n\text{ = 28} \\ \text{degr}ee\text{ of fr}eedom\text{ = n - 1 = 28 - 1} \\ \text{degr}ee\text{ of freedom = }27 \end{gathered}`

`\begin{gathered} To\text{ }find_{}t_{(\alpha)/(2)},\text{ }wewould\text{ use the degr}ee\text{ of fr}eedom,\text{ }\frac{\alpha}{2\text{ }}\text{ and t -table} \\ \text{from the table }t_{(\alpha)/(2)}\text{ = 2.052} \end{gathered}`
`\begin{gathered} \bar{X}\text{ }\pm\text{ }t_{(\alpha)/(2)}(\frac{s}{\sqrt[]{n}})\text{ = }14.3\text{ }\pm\text{ 2.052(}\frac{2}{\sqrt[]{28}}) \\ \end{gathered}`
`\begin{gathered} =14.3\text{ - 2.052(}\frac{2}{\sqrt[]{28}})<\text{ }\mu<\text{ }14.3\text{ +2.052(}\frac{2}{\sqrt[]{28}}) \\ =\text{ }14.3\text{ - 0.775 }<\text{ }\mu<\text{ 14.3 +0}.775 \\ =\text{ }14.3\text{ }\pm\text{ 0.775} \\ =\text{ }14.3\text{ }\pm\text{ 0.8} \end{gathered}`
`\begin{gathered} So,\text{ the confidence interval is:} \\ \text{ 14.3 - 0.8 < }\mu\text{ <}14.3\text{ + 0.8} \\ 13.5\text{ < }\mu\text{ < }15.1 \end{gathered}`

The 1st box: 14.3

second box: 2.052

Third box: 2

fourth box: 28

fifth box: 13.5

sixth box: 15.1