Question

A drug which is used for treating high cholesterol has potentially dangerous side effects if it is taken at an improper dosage. The drug is manufactured at two different plants, plant A and plant B. A medical researcher believes that tablet A has a smaller variance in dosage than the variance in dosage from tablet B. The sample variance of a sample of 6 dosages of tablet A is 0.016. The sample variance of a sample of 20 dosages of tablet B is 0.017. Test the claim using a 0.05 level of significance. Let σ12 represent the population variance for tablet A.

Step 1 of 5 :  State the null and alternative hypotheses for the test.

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 to 4 decimal places

step 3 of 5: compute the value of the test statistic. round to 4 decimal place

step 4 of 5: reject null hypothesis or fail to reject null hypothesis

step 5 of 5: state the test conclusion. does the evidence support the claim? yes or no

Answer 1

The provided sample variances are sî = 0.016 and sî = 0.017 and the sample sizes are given by ni = 6 and n2 = 20. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:= ož Ha : 07= ož This corresponds to a two-tailed test, for which a F-test for two population variances needs to be used. (2) Rejection Region - Based on the information provided, the significance level is a - 0.05, and the the rejection region for this two-tailed test is R = {F:F< 0.158 or F > 3.333}. (3) Test Statistics The F-statistic is computed as follows: 0.016 F 0.941 1993 0.017 (4) Decision about the null hypothesis Since from the sample information we get that Fl = 0.158 < F = 0.941 < Fu = 3.333, it is then concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population variance oſ is different than the population variance oż, at the a = 0.05 significance level.

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