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, 5.10, 5.04, and 4.90 mg/cc.

(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.)

1-2. Null and alternative hypotheses:

H0: 2 = 0.0011 versus Ha: 2 > 0.0011

H0: 2 > 0.0011 versus Ha: 2 < 0.0011     

H0: 2 = 0.0011 versus Ha: 2 < 0.0011

H0: 2 = 0.2 versus Ha: 2 0.2

H0: 2 = 0.2 versus Ha: 2 > 0.2

3. Test statistic:    2 =  

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

5. Conclusion:

H0 is rejected. There is insufficient evidence to indicate that the variation in potency differs from the specified error limits.

H0 is rejected. There is sufficient evidence to indicate that the variation in potency differs from the specified error limits.     

H0 is not rejected. There is insufficient evidence to indicate that the variation in potency differs from the specified error limits.

H0 is not rejected. There is sufficient evidence to indicate that the variation in potency differs from the specified error limits.

Explanation / Answer

here 6=0.2 i.e. =0.0333 and 2=0.0011,

here it is  variation in potency values must be small, so it left one tailed test

(first part 1-2) H0: 2 = 0.0011 versus Ha: 2 < 0.0011

(second part 3) answer is 21.82

here we use chi-square test and chi-square=(n-1)s2/=(4-1)*0.008/0.0011=21.82 with n-1=4-1=3 df

(third part) 2 <7.81

critical chi-square(0.05,3)=7.81

(fourth part)

H0 is not rejected. There is insufficient evidence to indicate that the variation in potency differs from the specified error limits.

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