Question

To properly treat patients, drugs prescribed by physicians must have a potency that is accurately defined. Consequently, not only must the distribution of potency values for shipments of a drug have a mean value as specified on the drug's container, but also the variation in potency must be small. Otherwise, pharmacists would be distributing drug prescriptions that could be harmfully potent or have a low potency and be ineffective. A drug manufacturer claims that its drug is marketed with a potency of 5 ± 0.1 milligram per cubic centimetre (mg/cc). A random sample of four containers gave potency readings equal to 4.95, 5.09, 5.04, and 4.89 mg/cc. (a) Do the data present sufficient evidence to indicate that the mean potency differs from 5 mg/cc? (Use α-0.05. Round your answers to three decimal places.) 1-2. Nl and alternative hypotheses: O Ho: μ-5 versus Ha: μ < 5 O Ho: μ : 5 versus Ha: μ#5 O Ho: μ 5 versus Ha: μ-5 G, Ho: μ 5 versus Ha: μ > 5 O Ho: μ-5 versus Ha: μ > 5 3. Test statistic: 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. t>

Answer 1

The potency readings of a random sample of 4 containers are provided as: {4.85, 5.09, 5.04, 4.89) (a) The claim that is to be tested is that the mean potency differs significantly from 5 mg/cc at a level of significance of 5%. Let u represents the true mean potency of drug. Then the null hypothesis will assume that the mean potency does not differ significantly from 5 mg/cc and the alterative hypothesis will assume that the mean potency differs significantly from 5 mg/cc. Mathematically, the hypothesis is formulated as follows: Họ: u=5 H:45 As the sample size is small so a one-sample t-test statistic will be used to test the above hypothesis. Compute the sample mean and sample standard deviation by using the following formulas: Islal 4.95 +5.09 +5.04+ 4.89 19.97 4 = 4.9925 Ż (x,-F) V n-1

3(x, – 4.9925) = 0.0895 The test statistic is calculated by the following formula: -u t=1s 4.9925-5 0.0895 = -0.17 Hence, the calculated value of the test statistic is -0.17. The degrees of freedom for the above test statistic are one less the size of the sample, that is, Degrees of freedom = n-1 = 4-1 = 3 The two-tailed critical value of the t-test statistic at 3 degrees of freedom and 5% level of significance is obtained using MS-Excel function as shown below: =T.INV.2T(0.05,3)

The above function yields a critical value of 3.18. As the absolute value of test statistic is less than the critical value, so the null hypothesis will not be rejected at 5% level of significance. Therefore, it is concluded that mean potency does not differs significantly from 5 mg/cc. (b) It is specified by the manufacturer that the potency values will fall in the range 5.0 +0.1 mg/cc. Also, the error range of 0.2 denotes 60. So, the null and alternative hypothesis will be formulated as follows: H:= 0.0011 H,:02 <0.0011 The Chi-square test statistic is calculated by the following formula: 72 - (n-1) sa 3x0.08952 0.0011 = 21.85 The above test statistic will follow Chi-square distribution with (n - 1) = 3 degrees of freedom. The critical value for Chi-square test statistic at 3 degrees of freedom and 5% level of significance is obtained using MS-Excel function as shown in the screenshot below: =CHISQ.INV.RT(0.05,3)

The above function yields a critical value of 7.81. As the calculated value of the test statistic is greater than the critical value, so the null hypothesis will be rejected at 5% level of significance and it is concluded that the variance differs significantly from the specified limit.