Answer to Question #231734 in Statistics and Probability for Fairy

Question

In a study conducted by the Department of Me- chanical Engineering at Virginia Tech, the steel rods supplied by two different companies were compared. Ten sample springs were made out of the steel rods supplied by each company, and a measure of flexibility was recorded for each The data are as follows:
Company A: 9.3, 8.8, 6.8, 8.7, 8.5, 6.7, 8.0, 6.5, 9.2, 7.0
Company B: 11.0, 9.8, 9.9, 10.2, 10.1, 9.7, 11.0, 11.1, 10.2, 9.6
a) (b)
Calculate the sample mean and median for the data for the two companies.
Plot the data for the two companies on the same line and give your impression regarding any apparent differences between the two companies.

Answer 1

(a) To calculate mean use the formula

$\bar{x} = \frac{\sum x_i}{n}$

where n = total number of observations

For company A , mean $= \frac{77.5 }{ 10} = 7.7$

For company B, mean $= \frac{103}{10} = 10.3$

To calculate median arrange the observations in ascending order . Find the $\frac{(n+1)}{2}th$ observation , which is the median

For company A : n=10 which is even. Therefore there are two middle observations 5th and 6th . The average of these two is the median. The 5th and 6th observations are 7 and 8 . Average of these $= \frac{(7+8)}{2} = \frac{15}{2} = 7.5$

Therefore, median for company A = 7.5

For company B : The 5th and 6th observations are 10.2 and 10.2

Therefore median for company B = 10.2

b) Plot of measure of flexibility of two companies is shown below:

The plot shows that the steel rods supplied by company B are more flexible than that of company A.