Question

The weights of a random sample of cereal boxes that are supposed to weigh 1 pound are given below. Estimate the standard deviation of the entire population with 91.5% confidence. Assume the population is normally distributed.

1.03, 0.99, 1.03, 1.03, 1.04, 1.03, 0.95, 1.01

LCL =

UCL =

Answer 1

SOLUTION:

From given data,

The weights of a random sample of cereal boxes that are supposed to weigh 1 pound are given below. Estimate the standard deviation of the entire population with 91.5% confidence. Assume the population is normally distributed.

1.03

0.99

1.03

1.03

1.04

1.03

0.95

1.01

= x / n = (1.03+0.99+1.03+1.03+1.04+1.03+0.95+1.01 ) / 8 = 8.11/8 = 1.01375

s = ( x- )² / (n-1)

= (1.03-1.01375)^2+(0.99-1.01375)^2+(1.03-1.01375)^2+(1.03-1.01375)^2+(1.04-1.01375)^2+(1.03-1.01375)^2+( 0.95-1.01375)^2+(1.01-1.01375)^2 / (8-1)

= 0.0009125

91.5 % confidence interval

Confidence interval is 91.5%

91.5% = 91.5/100 = 0.915

= 1 - Confidence interval = 1-0.915 = 0.085

/2 = 0.085/ 2

= 0.0425

1 - /2 = 1-0.0425 = 0.9575

df = n-1 = 8-1 = 7

91.5 % confidence interval for population variance

= ( ,     )

= ( ,     )

= ( ,   ​​​​​​  )

= (0.0016896 , 0.0006333)

91.5 % confidence interval for population standard deviation

= [ (sqrt(0.0016896) , sqrt(0.0006333)) ]

= [ 0.0411047 , 0.0251655 ]

LCL = 0.0251655

UCL = 0.0411047