MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Question
8.83 Drug content assessment. Refer to Exercise 8.16 (p. 443)
and the Analytical Chemistry (Dec. 15, 2009) study in
which scientists used high-performance liquid chromatog-
HPLC
raphy to determine the amount of drug in a tablet. Recall
that 25 tablets were produced at each of two different,
independent sites. The researchers want to determine if the
two sites produced drug concentrations with different vari-
ances. A Minitab printout of the analysis follows. Locate
the test statistic and p-value on the printout. Use these
values and a = .05 to conduct the appropriate test for the
researchers.
Test and CI for Two Variances: Content vs Site
Method
o (1) / o(2) = 1
Null hypothesis
Alternative hypothesis o(1) / o(2) # 1
Significance level
a = 0.05
F method was used. This method is accurate for normal data only.
Statistics
95% CI for
site
N
StDev
Variance
StDevs
9.406
11.147
(2.395, 4.267)
(2.607, 4.645)
25
3.067
2
25 3.339
Ratio of standard deviations = 0.919
Ratio of variances = 0.844
95% Confidence Intervals
CI for StDev
Ratio
(0.610, 1.384)
CI for
Variance
Ratio
Method
F
(0.372, 1.915)
Tests
Test
DF1
DF2
Statistic
0.84
Method
P-Value
F
24
24
0.681
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Transcribed Image Text:8.83 Drug content assessment. Refer to Exercise 8.16 (p. 443) and the Analytical Chemistry (Dec. 15, 2009) study in which scientists used high-performance liquid chromatog- HPLC raphy to determine the amount of drug in a tablet. Recall that 25 tablets were produced at each of two different, independent sites. The researchers want to determine if the two sites produced drug concentrations with different vari- ances. A Minitab printout of the analysis follows. Locate the test statistic and p-value on the printout. Use these values and a = .05 to conduct the appropriate test for the researchers. Test and CI for Two Variances: Content vs Site Method o (1) / o(2) = 1 Null hypothesis Alternative hypothesis o(1) / o(2) # 1 Significance level a = 0.05 F method was used. This method is accurate for normal data only. Statistics 95% CI for site N StDev Variance StDevs 9.406 11.147 (2.395, 4.267) (2.607, 4.645) 25 3.067 2 25 3.339 Ratio of standard deviations = 0.919 Ratio of variances = 0.844 95% Confidence Intervals CI for StDev Ratio (0.610, 1.384) CI for Variance Ratio Method F (0.372, 1.915) Tests Test DF1 DF2 Statistic 0.84 Method P-Value F 24 24 0.681
Expert Solution
Check Mark
Step 1

The aim is to test whether the two sites produced drug concentrations with different variance.

The test hypotheses are given below:

H0:σ1=σ2H1:σ1σ2

With the help of MINITAB output, the F-statistic is 0.84 and the p-value is 0.681.

Here, the level of significance is 0.05.

 

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