Question

The retail price for 10 randomly selected agriculture books are shown in the table. Assume that the population is normally distributed. Construct a 98​% confidence interval for the standard deviation.​ (Note: s=5.644.)

63.79

66.39

64.99

78.73

67.98

71.78

63.92

65.57

70.77

78.65

The 98% confidence interval is from _ to _.

Answer 1

solution:-

given that n = 10 , s = 5.644

and given that 98% confidence interval

degree of freedom df = n - 1 = 10 - 1 = 9

from chi-square distribution table with 98% confidence and degree of freedom

x^2 L = 21.666

X^2 R = 2.088

confidence interval formula for standard deviation

=> sqrt((n-1)s^2/X^2 L) < σ < sqrt((n-1)s^2/X^2 R)

= sqrt((10-1)*5.644^2/21.666) < σ < sqrt((10-1)*5.644^2/2.088)

= 3.638 < σ < 11.718 (rounded to three decimals)