Question

3. Federal law requires that 250 mg doses of Divalproex differ by at most 1 part in 100, as measured by the standard deviation. For any reasonable manufacturing process, the strength of a dose will have an approximately normal distribution. a. Find a 95% confidence interval for that standard deviation, if a sample of 50 pills has a mean strength of 250.22 mg, with a standard deviation of 2.07 mg. b. Find a 95% confidence interval for the mean strength. C. On the basis of these confidence intervals, is it reasonable to assume that the 1-part-in-100 variability bound is met? (Hint: Using both parts (a) and (b), how do you get the highest ratio of standard deviation to mean?]

Answer 1

3.

a.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.05
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.025 = 0.975
the two critical values ᴪ^2 left, ᴪ^2 right at 49 df are 70.2224 , 31.555
s.d( s )=2.07
sample size(n)=50
confidence interval for σ^2= [ 49 * 4.2849/70.2224 < σ^2 < 49 * 4.2849/31.555 ]
= [ 209.9601/70.2224 < σ^2 < 209.9601/31.5549 ]
[ 2.9899 < σ^2 < 6.6538 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (2.9899) < σ < sqrt(6.6538), ]
= [ 1.7291 < σ < 2.5795 ]

b.
TRADITIONAL METHOD
given that,
sample mean, x =250.22
standard deviation, s =2.07
sample size, n =50
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.07/ sqrt ( 50) )
= 0.293
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 49 d.f is 2.01
margin of error = 2.01 * 0.293
= 0.588
III.
CI = x ± margin of error
confidence interval = [ 250.22 ± 0.588 ]
= [ 249.632 , 250.808 ]
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DIRECT METHOD
given that,
sample mean, x =250.22
standard deviation, s =2.07
sample size, n =50
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 49 d.f is 2.01
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 250.22 ± t a/2 ( 2.07/ Sqrt ( 50) ]
= [ 250.22-(2.01 * 0.293) , 250.22+(2.01 * 0.293) ]
= [ 249.632 , 250.808 ]
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interpretations:
1) we are 95% sure that the interval [ 249.632 , 250.808 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

c.
95%confidence interval for standard deviation [ 1.7291 < σ < 2.5795 ]
95% sure that the interval [ 249.632 , 250.808 ] mean strength.
highest ratio between standard deviation and mean = (2.07/250.22)*100 = 0.00827*100
= 0.827%