The concentration of active ingredient in a pharmaceutical product is very impor

Question

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.

Explanation / Answer

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%

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