39. The diaper manufacturer of Example 4.4.4, page 149, decides that the confidence interval obtained is too wide, at 16 percentage points. He would like to narrow it to half the width, i.e. 8 percentages points. (Hint: you may want to use the results of Exercise 38.)   d) A sample of the size found in part c) is taken and it turns out that 22% of the diapers in that sample are defective. Estimate the 95% confidence level interval for the percentage of defective diapers. e) Is the result in part d) consistent with that of Example 4.4.4? a) Show that reducing the confidence level to 68%, he can get what he wants without getting a bigger sample. b) Since he does not want to compromise the confidence level of the inference, he decides to take a new random sample of double the size, i.e. 200 diapers. Assuming ˆp constant, show that this doesn’t achieve what he wants. c) How big a sample is needed to reduce the confidence interval width to 8 percentage points? (For planning purposes the manufacturer assumes ˆp = 0.2).

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39. The diaper manufacturer of Example 4.4.4, page 149, decides that the
confidence interval obtained is too wide, at 16 percentage points. He
would like to narrow it to half the width, i.e. 8 percentages points.
(Hint: you may want to use the results of Exercise 38.)

 


d) A sample of the size found in part c) is taken and it turns out that
22% of the diapers in that sample are defective. Estimate the 95%
confidence level interval for the percentage of defective diapers.
e) Is the result in part d) consistent with that of Example 4.4.4?

a) Show that reducing the confidence level to 68%, he can get what
he wants without getting a bigger sample.
b) Since he does not want to compromise the confidence level of the
inference, he decides to take a new random sample of double the
size, i.e. 200 diapers. Assuming ˆp constant, show that this doesn’t
achieve what he wants.
c) How big a sample is needed to reduce the confidence interval width
to 8 percentage points? (For planning purposes the manufacturer
assumes ˆp = 0.2).

Confidence Intervals
Suppose we perform a two-outcome random experiment n times
(n > 30) and count the number k of successes. If we denote by
k
the proportion of successes, and use the approximation
o - Vn· p· (1 – ô), we can estimate the probability p of success
as follows:
• With a 68% confidence level we can say that:
<p< p+
n
We call p
the 68% confidence interval.
-
• With a 95% confidence level we can say that:
p – 2- <p< ộ + 2–
n
We call [p – 2, ô + 2
the 95% confidence interval.
n
• With a 99.7% confidence level we can say that:
p – 3- <p< p +3-
n
We call p– 3°,p+3º the 99.7% confidence interval.
Transcribed Image Text:Confidence Intervals Suppose we perform a two-outcome random experiment n times (n > 30) and count the number k of successes. If we denote by k the proportion of successes, and use the approximation o - Vn· p· (1 – ô), we can estimate the probability p of success as follows: • With a 68% confidence level we can say that: <p< p+ n We call p the 68% confidence interval. - • With a 95% confidence level we can say that: p – 2- <p< ộ + 2– n We call [p – 2, ô + 2 the 95% confidence interval. n • With a 99.7% confidence level we can say that: p – 3- <p< p +3- n We call p– 3°,p+3º the 99.7% confidence interval.
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