Math 3339
Homework 5 (Chapter 7)
Name:__________________________________ PeopleSoft ID:_______________
Instructions:
• Homework will NOT be accepted through email or in person. Homework must be submitted through
CourseWare BEFORE the deadline.
• Print out this file use or software and complete the problems.
• Write in black ink or dark pencil or type your solutions in the space provided. You must show all
work for full credit.
• Submit this assignment at http://www.casa.uh.edu under “Assignments" and choose hw5.
• Total possible points: 15
• You can use RStudio for any of these problems unless otherwise indicated.
1. Section 7.3.4, problem 1
A sample of size 36 from a normally distributed population variable with population
standard deviation 20 had a sample mean of 88. Find a 90% confidence interval for the
population mean
The population standard deviation is given thus it is a z-confidence interval:
88+c(-1,1)*qnorm(1.9/2)*20/sqrt(36)
[1] 82.51715 93.48285
We are 90% confident that the population mean is between 82.52 and 93.48.
2. Section 7.3.4, problem 2
A sample of size 90 from a population variable had a sample mean of 4.74 and a sample standard deviation of 0.71. Find a 95%
confidence interval for the population mean.
The standard deviation is from the sample thus it is a t-confidence interval:
4.74+c(-1,1)*qt(1.95/2,89)*0.71/sqrt(90)
[1] 4.591293 4.888707
We are 95% confidence interval that the population mean is between 4.59 and 4.89.
3. Section 7.3.4, problem 3
Import the ”reacttimes” data set and consider the 50 observations of the variable ”Times” to
be a sample from a larger population. Find a 99% confidence interval for the population
mean. Construct a normal quantile plot and comment on the appropriateness of the
procedure.
> t.test(Times,conf.level = 0.99)
One Sample t-test
data: Times
t = 11.511, df = 49, p-value = 1.538e-15
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
1.336739 2.148061
sample estimates:
mean of x
1.7424
Confidence interval = (1.34, 2.15)
We are 99% confident that the mean reaction times is between 1.34 and 2.15.
Looking at this q-q plot since the points do not all follow the line, this confidence interval may not be
the best to use to get an estimate for the population mean.
4. Section 7.3.4, problem 5
We wish to estimate the population mean of a variable that has standard deviation 70.5. We
want to estimate it with an error no greater than 5 units with probability 0.99. How big a
sample should we take from the population? What happens if the standard deviation and the
margin of error are both doubled?
Z* = 2.5758
n > (2.5758*70.5/5)2 = 1319.04, need at least a sample of 1320.
If both are doubled, the sample size will not change.
5. Section 7.4.3, problem 1
The Food and Drug Administration monitors the production line of a breakfast cereal
company to determine what proportion of its boxes of cereal contain insect parts. The FDA
would like to know that proportion to within 5 percentage points with 95% confidence. How
many boxes of cereal should they sample?
Z* = 1.96, p* = 0.5
n > 0.5*0.5*(1.96/0.05)^2 = 384.16, we need at least 385 samples.
6. Section 7.4.3, problem 2
Suppose the company has been inspected before and that previously the proportion of cereal
boxes with insect parts was 0.15. The FDA wants to be as unobtrusive as possible. How
many boxes of cereal should they sample?
Z* = 1.96, p* = 0.15
n > 0.15*0.85*(1.96/0.05)^2 = 195.9216, we need at least 196 samples.
7. Section 7.4.3, problem 4
The FDA sampled 60 boxes of cereal and found 12 with insect parts. Find a 95% confidence
for the true proportion using all three methods described above. Use R to answer question 3
with Method 3.
> prop.test(n=60,x=12,correct = F)
1-sample proportions test without continuity
correction
data: 12 out of 60, null probability 0.5
X-squared = 21.6, df = 1, p-value = 3.359e-06
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.1182851 0.3178181
sample estimates:
p
0.2
Confidence Interval: (0.118, 0.318), We are 95% confident that the proportion of boxes with insect
parts is between 11.8% and 31.8%.
8. Section 7.6.1, problem 1
A sample of size 20 from a normal distribution has sample mean 3.5 and sample variance 14.75. Find 90%
confidence intervals for the variance and standard deviation of the distribution.
Lower limit = 19*14.75/qchisq(.95,19) = 9.2972
Upper limit = 19*14.75/qchisq(.05,19) = 27.701
Confidence interval for variance: (9.2972, 27.701)
Confidence interval for standard deviation (take the square root of the limits from the confidence interval of
variance):
sqrt(c(9.2972,27.701))
[1] 3.049131 5.263174
(3.049, 5.263)
9. Section 7.6.1, problem 2
The variable ”Temp” in the airquality data set is approximately normally distributed.
Treating the data as a sample from a larger normal population, find 95% confidence intervals
for the variance and standard deviation of temperatures.
> lcl=152*var(Temp)/qchisq(.975,152)
> ucl=152*var(Temp)/qchisq(.025,152)
> c(lcl,ucl)
[1] 72.42541 113.71100 #Confidence interval for variance
> sqrt(c(lcl,ucl))
[1] 8.510312 10.663536 #Confidence interval for standard deviation
10. Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the
alcohol content of each bottle is determined. Let µ denote the average alcohol content for the
population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval
is (7.8, 9.4).
a. Would a 90% confidence interval calculated from this same sample have been narrower or wider
than the given interval? Explain your reasoning.
b. Consider the following statement: There is a 95% chance that µ is between 7.8 and 9.4. Is this
statement correct? Why or why not?
c. Consider the following statement: We can be highly confident that 95% of all bottles of this type
of cough syrup have an alcohol content that is between 7.8 and 9.4. Is this statement correct?
Why or why not?
d. Consider the following statement: If the process of selecting a sample of size 50 and then
computing the corresponding 95% interval is repeated 100 times, 95 of the resulting intervals will
include µ. Is this statement correct? Why or why not?
a. It will decrease. As the confidence level decreases, the width of the interval will also decrease.
b. This is false. It is not a chance. We are 95% confident that the mean is between 7.8 and 9.4.
c. This is false. The interval is about the mean of the population not the number.
d. This statement is not true. In the long run it will be 95% of the intervals will capture the true
population mean. Does not mean that exactly 95 out of 100 will capture the true mean.
