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 4.93, 5.10, 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 α = 0.05.)

State the null and alternative hypotheses.

H₀: μ ≠ 5 versus H_a: μ = 5H₀: μ = 5 versus H_a: μ < 5    H₀: μ = 5 versus H_a: μ > 5H₀: μ < 5 versus H_a: μ > 5H₀: 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.

H₀ is rejected. There is insufficient 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 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.

H₀: σ² > 0.0011 versus H_a: σ² < 0.0011H₀: σ² = 0.2 versus H_a: σ² > 0.2    H₀: σ² = 0.2 versus H_a: σ² ≠ 0.2H₀: σ² = 0.0011 versus H_a: σ² > 0.0011H₀: 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.

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

You may need to use the appropriate appendix table or technology to answer this question.

Answer 1

From the given data we can easily calculate

sample mean =m=4.99

Sample variance =SV =0.007667

Sample standard deviation =SD=0.0875

n=4

a)

Here we have to test that  the mean potency differs from 5 mg/cc

Hence

H0: μ = 5 versus Ha: μ ≠ 5

now

Since sample mean < population mean hence

we reject the null hypothesis when

or

t_0.025=3.182 <t

but t value is > -3.187

Hence we failed to reject the null hypothesis

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

b)

given

interval is (4.9,5.1)

so

so

we reject the null hypothesis

when

or

Since

Hence we reject the null hypothesis

hence

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