To properly treat patients, drugs prescribed by physicians must not only have a

Question

To properly treat patients, drugs prescribed by physicians must not only have a mean potency value as specified on the drug's container, but also the variation in potency values 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 his drug has a potency of 5 ± 0.1 milligram per cubic centimeter (mg/cc). A random sample of four containers gave potency readings equal to 4.95, 5.09, 5.03, and 4.90 mg/cc.

3. Test statistic:    t =  

4. Rejection region: If the test is one-tailed, enter NONE for the unused region.

t <

(b) Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer? (HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as specified by a manufacturer. Since he implies that the potency values will fall into the interval 5 ± 0.1 mg/cc with very high probability—the implication is almost always—let us assume that the range 0.2; or 4.9 to 5.1, represents 6, as suggested by the Empirical Rule). (Use = 0.05. Round your answers to three decimal places.)3. Test statistic:    2 =  

4. Rejection region: If the test is one-tailed, enter NONE for the unused region.

t <

(b) Do the data present sufficient evidence to indicate that the variation in potency differs from the error limits specified by the manufacturer? (HINT: It is sometimes difficult to determine exactly what is meant by limits on potency as specified by a manufacturer. Since he implies that the potency values will fall into the interval 5 ± 0.1 mg/cc with very high probability—the implication is almost always—let us assume that the range 0.2; or 4.9 to 5.1, represents 6, as suggested by the Empirical Rule). (Use = 0.05. Round your answers to three decimal places.)3. Test statistic:    2 =  

4. Rejection region: If the test is one-tailed, enter NONE for the unused region.

Explanation / Answer

(a) 1. Null and alternative hypothesis

H0 : = 5

Ha : 5

3. Here sample mean x = 4.9975 mg/cc

sample standard deviation s = 0.0896 mg/cc

standard error of sample mean se0 = s/sqrt(n) = 0.0896/sqrt (4) = 0.0448

Test statistic

t = (x - H)/ se0 = (4.9975 - 5)/ 0.0448 = -0.056

4. Here dF = 4-1 =3 ; alpha = 0.05

t3,0.05  = 3.1824

t < - 3.1824

5. here l t l < tcritical so we shallnot reject the null hypothesis and there is insufficient evidence that the mean potency differs from 5mg/cc

(b) Here the 6 = 0.2

2 = 0.01111

1. H0 : 2 = 0.0011

Ha : 2 0.0011

02  = 0.0011

2. here s =  0.0896 mg/cc

sample variance s2  = 0.008025

X2 = (n-1)(s/0)2 = (4 -1)* (0.008025/0.0011) = 21.8864

here,

dF = 3, X20.05,3 = 7.8147

so here rejection region is X2 > 7.8147

so here as we see that X2 > X2critical so we shall reject the null hypothesis and conclude that there is sufficient evidence that the variation i potency differes from the specificaton limits.

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