Question

Reserve Problems Chapter 8 Section 2 Problem 1 Consider results of 20 randomly chosen people who have run a marathon. Their times, in minutes, are as follows: 141, 144, 150, 161, 169, 179, 186, 194, 199, 209, 219, 220, 226, 237, 254, 261, 275, 278, 286, 295. Calculate a 99% upper confidence bound on the mean time of the race. Assume distribution to be normal. Round your answer to the nearest integer (e.g. 9876).

Answer 1

Answer:

99% Confidence interval:  (183,246)

Step-by-step explanation:

We are given the following data set:

141, 144, 150, 161, 169, 179, 186, 194, 199, 209, 219, 220, 226, 237, 254, 261, 275, 278, 286, 295

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.  

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

$Mean =\displaystyle\frac{4283}{20} = 214.15$

Sum of squares of differences = 45922.55

$S.D = \sqrt{\frac{45922.55}{19}} = 49.16$

99% Confidence interval:  

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

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.01} = \pm 2.86$  

$214.15 \pm 2.86(\frac{49.16}{\sqrt{20}} ) = 214.15 \pm 31.44 =(182.71,245.59) \approx (183,246)$