somebody please help me! 4. A drug manufacturer states that the mean amount of N
ID: 3132541 • Letter: S
Question
somebody please help me!
4. A drug manufacturer states that the mean amount of Naproxen Sodium, the active ingredient in reducing pain, in a tablet 220mg. A pharmacist wonders if the mean amount of Naproen Sodium is different from 220mg? She randomly selects 17 tablets, measures the amount of Naproxen Sodium in each, and obtains the following information. Use significance 0.01.
Box plot= All points inside fences.
Normal probability plot= All points inside bounds.
xbar(sample average)= 221.24 mg
s = 2.42mg
4A. What is the appropriate test procedure?
z-test of the mean
t-test of the mean
z-test of the proportion
none of these
4B. what is the appropriate set of hypotheses (H0, H1)?
mu=220, mu does not equal 220
mu does not equal 220, mu= 220
mu= 220, mu is less than 220
mu=220, mu is greater than 220
4C. what is the correct magnitude of the critical value?
2.583
2.921
2.235
2.120
4D. what is the correct value of the test statistic?
-2.11
+2.11
+0.49
+8.43
4E. what is the correct p-value for this test statistic?
0.025
0.050
0.020
0.100
4F. what is the correct confidence interval?
220.0, 222.4
219.2, 223.2
220.2, 222.2
219.5, 222.9
4G. what is the correct conclusion of this hypothesis test?
do not reject the null
reject the null
sometimes reject the null
do not reject the alternative
4H. is the mean amount of Naproxen Sodium significantly different from 220 mg?
yes
no
sometimes
depends on the distribution
Explanation / Answer
A.
t-test of the mean [ANSWER]
As we do not know the population standard deviation.
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b.
Formulating the null and alternative hypotheses,
Ho: u = 220
Ha: u =/ 220 [ANSWER, A]
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c.
As we can see, this is a two tailed test.
Thus, getting the critical t,
df = n - 1 = 16
tcrit = +/- 2.920781622 [ANSWER, B]
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d.
Getting the test statistic, as
X = sample mean = 221.24
uo = hypothesized mean = 220
n = sample size = 17
s = standard deviation = 2.42
Thus, t = (X - uo) * sqrt(n) / s = 2.112665692 [ANSWER, B]
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e.
Also, the p value is
p = 0.050697963 [ANSWER, B]
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f.
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 221.24
t(alpha/2) = critical t for the confidence interval = 2.920781622
s = sample standard deviation = 2.42
n = sample size = 17
df = n - 1 = 16
Thus,
Margin of Error E = 1.714312503
Lower bound = 219.5256875
Upper bound = 222.9543125
Thus, the confidence interval is
( 219.5256875 , 222.9543125 ) [ANSWER, D]
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g.
As P > 0.01, we DO NOT REJECT THE NULL HYPOTHESIS. [ANSWER, A]
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h.
NO. [ANSWER]
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