Question

To properly treat patients, drugs prescribed by physicians must have a potency that is accurately defined. Consequently, not only must the distribution of potency values for shipments of a drug have a mean value as specified on the drug's container, but also the variation in potency 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 its drug is marketed with a potency of 5 ± 0.1 milligram per cubic centimetre (mg/cc). A random sample of four containers gave potency readings equal to 4.93, 5.08, 5.03, 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. Round your answers to three decimal places.)

1-2. 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

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 not 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 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 it implies that the potency values will fall into the interval 5.0 ± 0.1 mg/cc with very high probability—the implication is always—let us assume that the range 0.2; or (4.9 to 5.1), represents 6σ, as suggested by the Empirical Rule. Note that letting the range equal 6σ rather than 4σ places a stringent interpretation on the manufacturer's claim. We want the potency to fall into the interval

5.0 ± 0.1

with very high probability.) (Use α = 0.05. Round your answers to three decimal places.)

1-2. Null and alternative hypotheses:

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

3. Test statistic:    χ² =  

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

χ2 >

χ2 <

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.

Answer 1

a.
Given that,
population mean(u)=5
sample mean, x =4.9825
standard deviation, s =0.0877
number (n)=4
null, Ho: μ=5
alternate, H1: μ!=5
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =3.182
since our test is two-tailed
reject Ho, if to < -3.182 OR if to > 3.182
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.9825-5/(0.0877/sqrt(4))
to =-0.399
| to | =0.399
critical value
the value of |t α| with n-1 = 3 d.f is 3.182
we got |to| =0.399 & | t α | =3.182
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.3991 ) = 0.7165
hence value of p0.05 < 0.7165,here we do not reject Ho
ANSWERS
---------------
1.
null, Ho: μ=5
2.
alternate, H1: μ!=5
3.
test statistic: -0.399
critical value: -3.182 , 3.182
4.
test two tailed
decision: do not reject Ho
p-value: 0.7165
5.
we do not have enough evidence to indicate that the mean potency differs from 5 mg/cc

b.
Given that,
population variance (σ^2) =0.0011
sample size (n) = 4
sample variance (s^2)=0.00769
null, Ho: σ^2 =0.0011
alternate, H1 : σ^2 !=0.0011
level of significance, α = 0.05
from standard normal table, two tailed ᴪ^2 α/2 =7.815
since our test is two-tailed
reject Ho, if ᴪ^2 o < - OR if ᴪ^2 o > 7.815
we use test statistic chisquare ᴪ^2 =(n-1)*s^2/o^2
ᴪ^2 cal=(4 - 1 ) * 0.00769 / 0.0011 = 3*0.00769/0.0011 = 20.973
| ᴪ^2 cal | =20.973
critical value
the value of |ᴪ^2 α| at los 0.05 with d.f (n-1)=3 is 7.815
we got | ᴪ^2| =20.973 & | ᴪ^2 α | =7.815
make decision
hence value of | ᴪ^2 cal | > | ᴪ^2 α| and here we reject Ho
ᴪ^2 p_value =0.0001
ANSWERS
---------------
1.
null, Ho: σ^2 =0.0011
2.
alternate, H1 : σ^2 !=0.0011
3.
test statistic: 20.973
critical value: -7.815 , 7.815
p-value:0.0001
4.
test two tailed
decision: reject Ho
5.
we have enough evidence to indicate that the variation in potency differs from the specified error limits