Question

A drug, which is used for treating cancer, has potentially dangerous side effects if it is taken in doses which are larger than 46.33mg, the required dosage for the treatment. It is important that the variance of the amount of the active ingredient is 0.03. 10 tablets are randomly selected and the amount of the drug in each tablet is measured. It is determined that the variance of the amount of active ingredient is 0.0061mg. Does the data suggests at α=0.1 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 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.

Reject Null Hypothesis

Fail to Reject Null Hypothesis

Step 5 of 5: What is the conclusion?

There is sufficient evidence to show that the amount of active ingredient has a variance that is less than the desired amount.

There is not sufficient evidence to show that the amount of active ingredient has a variance that is less than the desired amount.

Answer 1

Step 1 of 5

$\small \sigma ^{2}$ : variance of the amount of the active ingredient

Hypothesized population variance :  $\small \sigma ^{2}_{o}$ = 0.03

Null hypothesis : H_o :  the variance of the drug in the tablets is same as the desired amount ; $\small \sigma ^{2}$ = $\small \sigma ^{2}_{o}$ (=0.03)

Alternate hypothesis : H₁ :  the variance of the drug in the tablets is less than the desired amount $\small \sigma ^{2}$ < $\small \sigma ^{2}_{o}$ (=0.03) : Left tailed test;

Step 2 of 5:

Sample size : n= 10

Degrees of freedom = n-1 =10-1=9

For Left Tailed Test : Reject null hypothesis if Calculated value of$\small \chi ^{2}$ is less than Critical Value

EX1-a)

for 9 degrees of freedom ; (From chi-square tables)

xi-a = x -0.1= xốp = 4.168

Step 3 of 5:

Test Statistic: 2 (n-1)s

2 Xstat = (n-1)s? 0.0549 (10 - 1) < 0.0061 0.03 0.02 =1.83

Step 4 of 5:

As Calculated Value of \chi ^{2} is less than Critical Value i.e. ( 1.83<4.1682 ); Reject Null Hypothesis

Reject null hypothesis;

Step 5 of 5: What is the conclusion?

There is sufficient evidence to show that the amount of active ingredient has a variance that is less than the desired amount.