Question

Drug that is used for treating cancer has potentially dangerous side effects if it is taken in doses that are larger than the required dosage for the treatment. The pharmaceutical company that manufactures the drug must be certain that the standard deviation of the drug content in the tablet is not more than 0.1 mg. Twenty-five tablets are randomly selected and the amount of drug in each tablet is measured. The sample has a mean of 20 mg and a variance of 0.02 mg. The hypotheses for the test are H0: ?2 ? 0.01 vs Ha: ?2 > 0.01.

Step 1 of 2:

Calculate the test statistic. Round your answer to two decimal places.

Answer 1

`t=(25-1) [(0.141)/(0.1)]^2 =47.71`

Explanation:

Data given

`\bar X=20` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`s^2=0.02` represent the sample variance

`s=0.141` represent the sample deviation

n=25 represent the sample size

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 0.01, so the system of hypothesis are:

H0:
`\sigma \leq 0.1`

H1:
`\sigma >0.1`

In order to check the hypothesis we need to calculate the statistic given by the following formula:

`t=(n-1) [(s)/(\sigma_o)]^2`

This statistic have a Chi Square distribution distribution with n-1=25-1=24 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

`t=(25-1) [(0.141)/(0.1)]^2 =47.71`

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 24 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,24)". And our critical value would be
`\chi^2 =36.415`

Since our calculated value is higher than the critical value we reject the null hypothesis at 5% of significance.