13)A sample of 20 cigarettes is tested to determine nicotine content and the ave
ID: 3152079 • Letter: 1
Question
13)A sample of 20 cigarettes is tested to determine nicotine content and the average value observed was 1.2mg. Compute a 99 percent two-sided confidence interval for the mean nicotine content of a cigarette if it is known that the standard deviation of a cigarette's nicotine content is .2mg.
14)In problem 13, suppose that the population variance is not known in advance of the experiment. If the sample variance is .04, compute a 99 percent two-sided confidence interval for the mean nicotine content.
15) In Problem 14, compute a value c for which we can assert "with 99 percent confidence" that c is larger than the mean nicotine content of a cigarette.
Explanation / Answer
13.
We know sigma, so we can use z distirbution.
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 1.2
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 0.2
n = sample size = 20
Thus,
Margin of Error E = 0.115194588
Lower bound = 1.084805412
Upper bound = 1.315194588
Thus, the confidence interval is
( 1.084805412 , 1.315194588 ) [ANSWER]
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14.
As this is just a sample standard deviation, we use t distiribution.
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 = 1.2
t(alpha/2) = critical t for the confidence interval = 2.860934606
s = sample standard deviation = 0.2
n = sample size = 20
df = n - 1 = 19
Thus,
Margin of Error E = 0.127944885
Lower bound = 1.072055115
Upper bound = 1.327944885
Thus, the confidence interval is
( 1.072055115 , 1.327944885 ) [ANSWER]
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15.
We can take any value above 1.3279. Hence, we can use
c = 1.35 [ANSWER]
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