Question

and the standard deviation, s, of sample data summarized in the frequency distribution table given below by using the formula below, where represents the class midpoint, frepresents the class frequency, and n represents the total number of sample values. Also, compare the omputed standard deviation to the standard deviation obtained from the original list of data values, 9.0. [(1•x?)] - [>«*-x)] n(n-1) Interval 30-36 37-43 44-50 51-57 58-64 65-71 Frequency 5 19 45 25 5 Sa 1 tandard deviation = 6.8 (Round to one decimal place as needed.) consider a difference of 20% between two values of a standard deviation to be significant. How does this computed value compare with the ven standard deviation, 9.0? A. The computed value is not significantly different from the given value. B. The computed value is significantly greater than the given value. C. The computed value is significantly less than the given value. Ick to select your answer(s).

Answer 1

Interval Mid x f 30 - 36 33 5 37 43 40 19 44 50 47 45 51 57 54 25 58 64 61 5 71 68 1 Total 100 N = f = 100 fx 165 760 2115 1350 305 68 4763 f(x-x)2 1070.18 1106.12 17.8605 1014.42 893.785 414.937 4517.31 Сл Σfx_ mean = x = = 47.63 N Sample Variance = Ef(x - 2)2 = 45.6294 N-1 Sample S.D= f(x - 2)2 N - 1 = 6.8

20% of 9 = 1.8

Difference = 9-6.8 = 2.2

Significant, because 2.2>1.8.

Ans:

B. The computed value is significantly greater than the given value.