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 4.89 mg/cc.

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

State the null and alternative hypotheses. (choose the correct letter)

a) H₀: μ = 5 versus H_a: μ < 5

b) H₀: μ = 5 versus H_a: μ ≠ 5    

c) H₀: μ = 5 versus H_a: μ > 5

d) H₀: μ < 5 versus H_a: μ > 5

e) H₀: μ ≠ 5 versus H_a: μ = 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. (choose the correct letter)

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

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

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

d) H₀ is rejected. There is sufficient 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 6σ, as suggested by the Empirical Rule. Use α = 0.05.)

State the null and alternative hypotheses. (choose the correct letter)

a) H₀: σ² = 0.2 versus H_a: σ² ≠ 0.2

b) H₀: σ² > 0.0011 versus H_a: σ² < 0.0011    

c) H₀: σ² = 0.2 versus H_a: σ² > 0.2

d) H₀: σ² = 0.0011 versus H_a: σ² > 0.0011

e) H₀: σ² = 0.0011 versus H_a: σ² < 0.0011

State the test statistic. (Round your answer to three decimal places.)
χ² =  

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

χ² >

χ² <

State the conclusion. (choose the correct letter)

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

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

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

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

Answer 1

The sample mean and standard devistion for given potency readings is,

sample mean = = 4.9875

sample standard deviation = s = 0.0922

sample size = n = 4

null and alternative hypotheses -

( claim )

two tailed test.

Test statistics -

Rejection Region -

Based on given significance level = 0.05 and df = n - 1 = 3 , the critical value for two tailed t test is tc = 3.1824.            { using Excel, =TINV( , DF ) , = TINV( 0.05,3 ) = 3.1824 }

So rejection region is,

R = {t : t < -3.1824 or t > 3.1824}

Decision about null hypothesis-

It is observed that   -3.1824 < -2.2712 < 3.1824

So fail to reject null hypothesis.

Conclusion :

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