Question

In order for a new drug to be sold on the market, the variance of the active ingredient in each dose should be 0.01mg0.01⁢mg. A random sample of 2525 tablets with a dosage strength of 43.67mg43.67⁢mg is taken. The variance of the active ingredient from this sample is found to be 0.00030.0003. Does the data suggests at α=0.01α=0.01 that the variance of the drug in the tablets is less than the desired amount? Assume the population is normally distributed.

Step 1 of 5 :  

State the null and alternative hypotheses. Round to four decimal places when necessary.

Step 1 of 5: State the null and alternative hypotheses. Round to four decimal places when necessary. Step 2 of 5: Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places. Step 3 of 5: Determine the value of the test statistic. Round your answer to three decimal places. Step 4 of 5: Make the decision. Step 5 of 5: What is the conclusion?

Answer 1

Step 1 of 5 :

Ho:       σ2	=	0.0100
Ha:        σ2	<	0.0100

Step 2 of 5:

for 1 % level and given df critical values of X2 =

10.856

  Step 3 of 5:

test statistic X2 =(n-1)s2/ σ2=(25-1)*0.0003/0.01 =

0.720

Step 4 of 5:  

reject the null hypothesis since test statistic < critical value

Step 5 of 5: we can conclude that variance is less than 0.01