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 4.95, 5.10, 5.02, and 4.91 mg/cc.

. Test statistic:    t =

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

(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.)

3. Test statistic:    2 =

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

Explanation / Answer

a.
Given that,
population mean(u)=5.01
sample mean, x =4.995
standard deviation, s =0.0835
number (n)=4
null, Ho: =5.01
alternate, H1: >5.01
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.35
since our test is right-tailed
reject Ho, if to > 2.35
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.995-5.01/(0.0835/sqrt(4))
to =-0.359
| to | =0.359
critical value
the value of |t | with n-1 = 3 d.f is 2.35
we got |to| =0.359 & | t | =2.35
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > -0.3593 ) = 0.62841
hence value of p0.05 < 0.62841,here we do not reject Ho
ANSWERS
---------------
null, Ho: =5.01
alternate, H1: >5.01
test statistic: -0.359
critical value: 2.35
decision: do not reject Ho
p-value: 0.62841
we do not have enough evidence to support the claim

b.
Given that,
population variance (^2) =24.01
sample size (n) = 4
sample variance (s^2)=26.01
null, Ho: ^2 =24.01
alternate, H1 : ^2 >24.01
level of significance, = 0.05
from standard normal table,right tailed ^2 /2 =7.815
since our test is right-tailed
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(4 - 1 ) * 26.01 / 24.01 = 3*26.01/24.01 = 3.25
| ^2 cal | =3.25
critical value
the value of |^2 | at los 0.05 with d.f (n-1)=3 is 7.815
we got | ^2| =3.25 & | ^2 | =7.815
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.3547
ANSWERS
---------------
null, Ho: ^2 =24.01
alternate, H1 : ^2 >24.01
test statistic: 3.25
critical value: 7.815
p-value:0.3547
decision: do not reject Ho

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