Part a) Find the 68% confidence interval for the percentage of leaky diapers in Example 4.4.4. Note that it is 8 percentage points long. Part b) Find the 95% confidence interval for the percentage of leaky diapers in Example 4.4.4 assuming 200 diapers were tested and 20% were leaky. Note that it is not 8 percentage points long. Part c) Find the 95% confidence interval for the percentage of leaky diapers in Example 4.4.4 assuming 400 diapers were tested and 20% were leaky. Note that it is 8 percentage points long. Also, for exercise 40, fair means that heads happens with probability 0.5. Asking if a coin is fair is asking whether 0.5 is inside the confidence interval. 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.) 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 ˆ p 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 ˆ p = 0.2). 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? 38. This problem refers to Formula (4.3), page 149, for the standard error. Throughout the problem we assume that p, and with it ˆ p, is constant. a) What effect has an increase of n on the standard error? b) By what factor is the standard error changing if the sample size is changed from n to 2n? c) When changing the sample size from n to kn, what value of k reduces the standard error by a factor of 1/2

Part a) Find the 68% confidence interval for the percentage of leaky diapers in Example 4.4.4. Note that it is 8 percentage points long. Part b) Find the 95% confidence interval for the percentage of leaky diapers in Example 4.4.4 assuming 200 diapers were tested and 20% were leaky. Note that it is not 8 percentage points long. Part c) Find the 95% confidence interval for the percentage of leaky diapers in Example 4.4.4 assuming 400 diapers were tested and 20% were leaky. Note that it is 8 percentage points long. Also, for exercise 40, fair means that heads happens with probability 0.5. Asking if a coin is fair is asking whether 0.5 is inside the confidence interval. 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.) 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 ˆ p 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 ˆ p = 0.2). 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? 38. This problem refers to Formula (4.3), page 149, for the standard error. Throughout the problem we assume that p, and with it ˆ p, is constant. a) What effect has an increase of n on the standard error? b) By what factor is the standard error changing if the sample size is changed from n to 2n? c) When changing the sample size from n to kn, what value of k reduces the standard error by a factor of 1/2

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Question

Part a) Find the 68% confidence interval for the percentage of leaky diapers in Example 4.4.4. Note that it is 8 percentage points long.

Part b) Find the 95% confidence interval for the percentage of leaky diapers in Example 4.4.4 assuming 200 diapers were tested and 20% were leaky. Note that it is not 8 percentage points long.

Part c) Find the 95% confidence interval for the percentage of leaky diapers in Example 4.4.4 assuming 400 diapers were tested and 20% were leaky.

Note that it is 8 percentage points long.
Also, for exercise 40, fair means that heads happens with probability 0.5. Asking if a coin is fair is asking whether 0.5 is inside the confidence interval.

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.)

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 ˆ
p 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 ˆ
p = 0.2).

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?

38. This problem refers to Formula (4.3), page 149, for the standard error.
Throughout the problem we assume that p, and with it ˆ
p, is constant.

a) What effect has an increase of n on the standard error? b) By what factor is the standard error changing if the sample size
is changed from n to 2n? c) When changing the sample size from n to kn, what value of k
reduces the standard error by a factor of 1/2

Example 4.4.3. We suspect that a coin has been doctored, and we perform
a run of n = 100 flips to determine if that is the case or not. Of the flips,
k = 58 come up heads. Estimate the probability p of heads for this coin.
What can we conclude about the coin?
A quick naive answer would be that the coin has been doctored, because
we did not get k = 50 heads. But we know better by now. We need to find
out if the probability p of heads is 50% or not.
By the binomial principle k has a near-normal distribution. So we can say
with 68% certainty that -o< k <u+o. A little algebraic manipulation
yields:
-a< -k <a
くk+の
Pn <k+ a
k +a
k - a
p -
In the last line we have p in the middle, which is the quantity we are interested
in, and p occurs on the sides. We began with u - a <k<u + o, which
means that 4 - << + . or p - < p sp+. Thus, we have
essentially reversed our original equation, reflecting that we now know p and
want to estimate p instead of the other way around.
CHAPTER 4. STATISTICS
149
Take a look at the last line in the equation above. We know the values
of p and n. Unfortunately, we do not know the value of a. By the binomial
principle o = Vn.p. (1 - p), but this requires knowing p to get a. Fortu-
nately, we can get a very good approximation of a by using p = " instead
of p. Doing that, we get
58
a* Vn p. (1– p) = /100 x
100
58
1-
100
%3!
= 4.94.
58 – 4.94
58 + 4.94
and with 68% confidence we can say
that is
100
100
0.5306 <p<0.6294.
Looks like a doctored coin! But we have a low confidence in this result, only
68% confidence level.
We can increase the confidence of our result if we use 2 standard de-
viations instead of 1. Repeating the above calculations, now with a 95%
confidence we can say that 0.4812 <p< 0.6788. This time the possibility of
p = 0.5 is there.
In conclusion, we can say that the coin looks suspicious, but we cannot
conclude that it has been doctored; not with a decent confidence level.
We summarize the main points of the previous example in the box on
раge 150.
Note the expression " in all of the confidence intervals. This is the
standard deviation expressed as a percentage of the sample size. We call this
the standard error.
p. (1 - p)
st.err. = -
(4.3)

Example 4.4.4. A diaper manufacturer wants to determine the percentage
of leaky diapers in a newly installed production line. He selects a random
sample of 100 diapers, and tests them. It turns out that 20 of the diapers in
the sample are defective. Estimate the percentage of defective diapers in the
production line.
CHAPTER 4. STATISTICS
150
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
n
o x Vn· p· (1 – p), we can estimate the probability p of success
as follows:
• With a 68% confidence level we can say that:
<p< p+
We call p
the 68% confidence interval.
п
n
• With a 95% confidence level we can say that:
p – 2 <p< ô+ 2
n
n
We call p- 2°, p +2
the 95% confidence interval.
• With a 99.7% confidence level we can say that:
p – 3- <p< p + 3-
n
We call p-
- 3-,
- p+3
the 99.7% confidence interval.
n
The first thing we need to decide is the confidence level we are going to work
with. In most situations a confidence level of 95% is appropriate, and that
is the case here. That means we obtain the confidence interval by going 2
standard errors away from p. From Example 4.4.1 we have n = 100, k = 20
so p = 0.2 and o 2 4. The standard error is st.err. =
0.04. Therefore
the confidence interval is [0.12, 0.28], i.e. 0.12 < p < 0.28. So the best we
can say is that the percentage of defective diapers in the new production line

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 15 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Point Estimation, Limit Theorems, Approximations, and Bounds$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.