Question

The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of the 15 workers reveals the following number of minutes spent traveling. 29 38 38 33 38 21 45 34 40 37 37 42 30 29 35Develop a 98 percent confidence interval for the population mean. Interpret the results.

Answer 1

Answer:

98% Confidence interval:  (31.74,38.4)

Step-by-step explanation:

We are given the following data set:

29, 38, 38, 33, 38, 21, 45, 34, 40, 37, 37, 42, 30, 29, 35

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{526}{15} = 35.07$

Sum of squares of differences = 506.93

$S.D = \sqrt{\displaystyle\frac{506.93}{14}} = 6.02$

98% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.02} = \pm 2.145$  

$35.07 \pm 2.145(\frac{6.02}{\sqrt{15}} ) = 35.07 \pm 3.33 = (31.74,38.4)$  

Thus, there is 98% confidence that the population mean number of minutes spent traveling by workers is between 31.74 mins and 38.40