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, 5.10, 5.02, and 4.90 mg/cc.

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

State the null and alternative hypotheses.

H₀: ? ? 5 versus H_a: ? = 5H₀: ? = 5 versus H_a: ? > 5    H₀: ? = 5 versus H_a: ? ? 5H₀: ? < 5 versus H_a: ? > 5H₀: ? = 5 versus H_a: ? < 5

State the test statistic. (Round your answer to three decimal places.)
t =  

State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three decimal places.)

t >

t <

State the conclusion.

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

State the null and alternative hypotheses.

H₀: ?² = 0.0011 versus H_a: ?² < 0.0011H₀: ?² = 0.2 versus H_a: ?² > 0.2    H₀: ?² > 0.0011 versus H_a: ?² < 0.0011H₀: ?² = 0.2 versus H_a: ?² ? 0.2H₀: ?² = 0.0011 versus H_a: ?² > 0.0011

State the test statistic. (Round your answer to three decimal places.)
?² =  

State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three decimal places.)

?² >

?² <

State the conclusion.

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

Answer 1

Q 1)

a) Null and Alternative Hypothesis :

H0: ? = 5 versus Ha: ? ? 5

The sample mean is

The sample standard deviation

Under H0, the test statistic is

Degrees of freedom = n-1= 3

Significance level

The critical value of t for 3 df at 5% significance level is 3.182

Decision Rule : Reject H0, if t > 3.182 or t< -3.182

Conclusion : Since calculated t fall with in the range of critical values. FAil to REject H0.

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