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 4.94, 5.09, 5.02, and 4.91 mg/cc.

(a) Do the data present sufficient evidence to indicate that the mean potency differs from 5 mg/cc? (Use = 0.05. Round your answers to three decimal places.)

1-2. Null and alternative hypotheses:

H0: = 5 versus Ha: 5

H0: < 5 versus Ha: > 5

H0: 5 versus Ha: = 5

H0: = 5 versus Ha: < 5

H0: = 5 versus Ha: > 5

3. Test statistic: t =

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

t > =

t < =

5. Conclusion:

H0 is rejected. There is sufficient evidence to indicate that the mean potency differs from 5 mg/cc.

H0 is not rejected. There is sufficient evidence to indicate that the mean potency differs from 5 mg/cc.

H0 is not rejected. There is insufficient evidence to indicate that the mean potency differs from 5 mg/cc.

H0 is rejected. There is insufficient evidence to indicate that the mean potency differs from 5 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 = 0.2 versus Ha: 2 > 0.2

H0: 2 = 0.2 versus Ha: 2 0.2

H0: 2 > 0.0011 versus Ha: 2 < 0.0011

3. Test statistic: 2 =

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

2 > =

2 < =

5. Conclusion:

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

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

Explanation / Answer

a)

It is a case of right tailed/one tailed test
z value corresponding to alpha = 0.05 applicable to one tailed test is + 2.33

The area under the standard normal curve left to z = 2.33 represents acceptance region and right to z = 2.33 indicates the rejection region.
Acceptance region = 0.9900
Rejection region is NONE in the left tail
Rejection region = 0.0100 in the right tail.
Rejection region lies in the extreme right tail of the normal curve.

B) It is a case of two tailed test.
z value corresponding to alpha = 5% or 0.05 applicable to two tailed test is 1.645
Area under the standard normal curve between z = - 1.645 and z = + 1.645 represents the acceptance region.
Area under the standard normal curve left to z = - 1.645 represents rejection region. It lies in the extreme left tail of the normal curve.
AND
Area under the standard normal curve right to z = + 1.645 represents rejection region. It lies in the extreme right tail of the normal curve.

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