# Normal distribution of the dosage

I am having an issue with the following task: For the dose of the MemPro medicine it is known that has normal distribution. Sample of 30 parameters has been taken into consideration and based on the obtained data, we know that mean of a dose of this medicine is 5grams. With possibility that alpha error is 5%, test the hypothesis that mathematical expectation is 6grams.

Could you please elaborate this, I have no idea how to solve this matter. Thanks in advance.

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## 1 Answer

This is a test of a null hypothesis that the mean is 6 grams. Since you have 30 observations from a normal idstribution the appropriate test is a t test. The test considers the pivotal (sample mean - 6)/(sample std dev/√30) in your case since n=30 and the hypothesized mean = 6. Since sample mean=5 this pivotal quantity is -√30/sample std dev. This value is to be compared to a t distribution with 29 degrees of freedom. You need to calculate the sample standard deviation from your sample of 30 observation to arrive at this number. You did not specify whether this is a one-sided or two-sided test. For the moment I will assume it is one-sided at level alpha=0.05. Then you look up the 5th percentile for a t distribution with 29 degrees of freedom and reject the null hypothesis that the true mean parameter is 6 or more if the test statistic is below the 5th percentile. From the table I see that the critical value (5th percentile) is -1.699. So compare -30/sample std dev to -1.699 to determine whether or not to reject.

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(sample mean - 6)/(sample std dev/√30)= (5-6)/(sample std dev/√30)= -1/(sample std dev/√30)= -30/sample std dev. in your case since sample mean =5. –  Michael Chernick Jul 3 '12 at 15:27
Thank you @Michael! Very well explained. :) –  Takarakaka Jul 3 '12 at 15:28
Yeah, I have just figured it out. :( Many thanks for everything well explained. –  Takarakaka Jul 3 '12 at 15:29