Question

estion

The standard deviation of the daily demand for a product is an important factor for inventory control for the product. Suppose that a pharmacy wants to estimate the standard deviation of the daily demand for a certain antibiotic. It is known that the daily demand for this antibiotic follows an approximately normal distribution. A random sample of

23

days has a sample mean of

125

orders for this antibiotic with a standard deviation of

10.1

orders. Find a

95%

confidence interval for the population standard deviation of the daily demand for this antibiotic. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

What is the lower limit of the 95% confidence interval?
What is the upper limit of the 95% confidence interval?

Answer 1

From Chi-square table,

chi square critical values at 0.05 significance level with 22 df = 10.982 , 36.781

95% confidence interval for is

sqrt [( n - 1) S2 / U ] < < sqrt [( n - 1) S2 / L ]

sqrt [ 22 * 10.12 / 36.781] < < sqrt [ 22 * 10.12 / 10.982 ]  

7.811 < < 14.295

Lower limit = 7.811

Upper limit = 14.295

95% CI Is ( 7.811 , 14.295 )