the greater denver area chamber of commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. the chamber does not know what the population standard deviation is. using a 98% confidence interval, what are the lower and upper values of the confidence interval of the population mean for the minutes spent getting to work?



Answer :

The lower and upper values of the confidence interval of the population mean for the minutes spent getting to work is (31.74,38.4)

What is Standard Deviation?

The square root of the variance is used to calculate the standard deviation, a statistic that expresses how widely distributed a dataset is in relation to its mean. By calculating the departure of each data point from the mean, the standard deviation may be determined as the square root of variance. The bigger the deviation within the data collection, the more the data points deviate from the mean; Hence, the higher the standard deviation, the more dispersed the data.

We know that:

[tex]\text { Standard Deviation }=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n-1}}[/tex]

where [tex]x_{i}[/tex]  is the mean and n is the number of observations.

[tex]\text { Standard Deviation }=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n-1}}[/tex]

98% Confidence interval:  

[tex]\bar{x} \pm t_{\text {critical }} \frac{s}{\sqrt{n}}[/tex]

Putting the values, we get,  

[tex]t_{\text {critical }} \alpha_{0.02}\\=\pm 2.14535.07 \pm 2.145\left(\frac{6.02}{\sqrt{15}}\right)\\=35.07 \pm 3.33\\=(31.74,38.4)[/tex]

Hence, The lower and upper values of the confidence interval of the population mean for the minutes spent getting to work is (31.74,38.4)

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