Q 8.129. - Elementary Statistics

Question

Northeast Commutes. According to Scarborough Research, more than $85\%$ of working adults commute by car. Of all U.S. cities. Washington. D.C. and New York City have the longest commute times. A sample of $30$ commuters in Washington. D.C., area yielded the following commute times, in minutes. 

Find and interpret a $90\%$ confidence interval for the mean commute time of all commuters in Washington. D.C. (Note: $\overline{x}=27.97$ minutes and $s=10.04$ minutes.)

Step-by-Step Solution

Answer

The $90\%$confidence interval for all commuters in Washington's mean commute time is$(24.856,31.084)$

Step 1: Given information

The table is: 

24	28	31	29	54	28
27	38	24	14	46	38
31	16	21	11	21	15
30	29	17	23	27	18
29	44	19	35	34	38

Step 2: Concept

The formula used: The confidence interval $\overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)$

Step 3: Calculation

Calculate the $90\%$ confidence interval for the average travel time of all Washington commuters.

Consider the variables $\overline{x}=27.97, n=30,$and $s=10.04$.

The needed value of $t_{\frac{\alpha }{2}}$ for $90\%$ confidence with$29 (=30-1)$ degrees of freedom is $1.699$, according to "Table IV Values of $t_{\alpha }$

Thus, the confidence interval is,

$$\begin{gathered} \overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=27.97\pm 1.699\left(\frac{10.04}{\sqrt{30}}\right) \\ =27.97\pm 1.699(1.833) \\ =27.97\pm 3.114 \\ =(24.856,31.084) \end{gathered}$$

As a result, the $90\%$ confidence interval for all commuters in Washington's mean commute time is $(24.856,31.084)$