-10 points MendStat14 10.E.053 My Notes Ask Your Teache To properly treat patien

Question

-10 points MendStat14 10.E.053 My Notes Ask Your Teache 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 patency af 5 0.1 milligram per cubic centimeter (mg/cc). A random sample of four containers gave potency readings equal to 4.95, 5.10, 5.04, 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 Round your answers to three decimal places. 1-2. Null and alternatlve hypotheses: Ha: -5 versus Ha: #5 O Ho: = 5 versus H.: g13 5 Ha: # 5 versus Ha: -5 3. Test statistic: - 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. t2 5. Conclusion: O Ha is not rejected. There is insufficient evidence to indicate that the mean potency differs from 5 mg/cc. Ho is rejected. There is sufficient evidence to indicate that the mean potency differs from 5 mg/cc. Ho is rejected. There is insufficient evidence to indicate that the mean potency differs from 5 mg/cc. Ifo is not rejected. There is sufficient evidence to indicate that the mean potency differs from 5 mg/cc.

Explanation / Answer

(a) 1. Null and alternative hypothesis

H0 : = 5

Ha : 5

3. Here sample mean x = 4.9975 mg/cc

sample standard deviation s = 0.0896 mg/cc

standard error of sample mean se0 = s/sqrt(n) = 0.0896/sqrt (4) = 0.0448

Test statistic

t = (x - H)/ se0 = (4.9975 - 5)/ 0.0448 = -0.056

4. Here dF = 4-1 =3 ; alpha = 0.05

t3,0.05  = 3.1824

t < - 3.1824

5. here l t l < tcritical so we shallnot reject the null hypothesis and there is insufficient evidence that the mean potency differs from 5mg/cc

(b) Here the 6 = 0.2

2 = 0.01111

1. H0 : 2 = 0.0011

Ha : 2 0.0011

02  = 0.0011

2. here s =  0.0896 mg/cc

sample variance s2  = 0.008025

X2 = (n-1)(s/0)2 = (4 -1)* (0.008025/0.0011) = 21.8864

here,

dF = 3, X20.05,3 = 7.8147

so here rejection region is X2 > 7.8147

so here as we see that X2 > X2critical so we shall reject the null hypothesis and conclude that there is sufficient evidence that the variation i potency differes from the specificaton limits.

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