Use this information for questions 8-10. Bayer produces pharmaceuticals. They ar

Question

Use this information for questions 8-10. Bayer produces pharmaceuticals. They are doing quality control on their new antidepressant. The intended mean active drug content is 25 mg per pill. They produce a sample batch of 16 pills with a sample variance of 4mg. 8. What is the probability that another sample batch of 4 pills will have a mean weight of active drug of 21.72 mg or less? (You may round to the nearest t value and use the sample variance acquired from the other sample) 9. Create a 99% confidence interval for the variance of the active drug content of the 16 pill batch. 10. A prior scientist estimated a confidence interval for the variance of the active content in these 16 pills as (2.4004,8.2634.) What level of confidence (confidence coefficient) was he using? Use this information for questions 8-10. Bayer produces pharmaceuticals. They are doing quality control on their new antidepressant. The intended mean active drug content is 25 mg per pill. They produce a sample batch of 16 pills with a sample variance of 4mg. 8. What is the probability that another sample batch of 4 pills will have a mean weight of active drug of 21.72 mg or less? (You may round to the nearest t value and use the sample variance acquired from the other sample) 9. Create a 99% confidence interval for the variance of the active drug content of the 16 pill batch. 10. A prior scientist estimated a confidence interval for the variance of the active content in these 16 pills as (2.4004,8.2634.) What level of confidence (confidence coefficient) was he using?

Explanation / Answer

Mean active drug content = 25 mg per pill

sample size n = 16 pills

sample variance = 4 mg

(8) Here we have to find the probability that another sample batch of 4 pills will have a mean weight of active drug of 21.72 mg or less

Pr(x < 21.72 mg) = ?

where it can be assumed normally distributed where population mean = 25 mg

estimated variance of population s = 4 mg

Standard error of sample mean = s/ sqrt(n) = 4/ sqrt(4) = 1

t = (21.72 - 25)/ 1 = -3.28

so p - value = Pr(t < -3.28, dF = 3) = 0.0232

(9) 99% confidence interval for the variance of the active drug content

99% confidence interval Lower limit = (n-1)s2/X21-a/2 = (16 -1) * 4 / 32.8013 = 1.83 [ here a = 0.01]

99% confidence interval Upper limit = (n-1)s2/X2a/2 = (16 -1) * 4 / 4.600 = 13.04

so 99% confidence interval = (1.83, 13.04)

(10) Here lower limit = 2.4004 =  (n-1)s2/X21-a/2

X21-a/2 = (16 -1) * 4/ 2.4004 = 24.9958

so for dF = 15

CHIDIST (24.9958, 15) = 0.05

so here the confidence level is 90% that he was using.

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