Homework SourseThe concentration of active ingredient in a pharmaceutical p
The concentration of active ingredient in a pharmaceutical product is very important, since either too much or little can affect the drug\'s preformance. A particular product is supposed to contain 5.0% of the active ingredient. During production, 4 specimens from each batchc are taken for assay. The mean of the 4 samples from the last batch is x bar = 5.3%. You can assume based on past data that the concentration for this production process varies according to a normal distribution with standar deviation of 0.2%
A) do a test to see if the assay data give evidence that the mean concentration of the batch is not 5%
Use a value of a=1% ( all the 6 steps)
B) calculate the power of the test against the alternative mean 5.1.
The concentration of active ingredient in a pharmaceutical product is very important, since either too much or little can affect the drug\'s preformance. A particular product is supposed to contain 5.0% of the active ingredient. During production, 4 specimens from each batchc are taken for assay. The mean of the 4 samples from the last batch is x bar = 5.3%. You can assume based on past data that the concentration for this production process varies according to a normal distribution with standar deviation of 0.2%
A) do a test to see if the assay data give evidence that the mean concentration of the batch is not 5%
Use a value of a=1% ( all the 6 steps)
B) calculate the power of the test against the alternative mean 5.1.
A) do a test to see if the assay data give evidence that the mean concentration of the batch is not 5%
Use a value of a=1% ( all the 6 steps)
B) calculate the power of the test against the alternative mean 5.1.
Solution
Set up hypotheses as
H0: x bar = 5%=0.05
Ha: x bar not equals 0.05
Two tailed test for mean
n =4
As n is very small, t test can be done
Mean diff = 0.03
Std error =0.2/2 =0.1
t = 0.3
df = 3
The P-Value is .783763.
The result is not significant at p < .01.
There is no evidence to show that sample mean means from 0.05
B) POwer of test=Prob (Reject H0/H1 is true)
=1%
