13)A sample of 20 cigarettes is tested to determine nicotine content and the ave

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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