CHAPTER 4: SAMPLE PROBLEMS FOR HOMEWORK, CLASS OR EXAMS
These problems are designed to be done without access to a computer, but they may
require a calculator.
1. A. When the distribution of the individual observations is extremely skewed,
inferences on the ____________ may be more meaningful than inferences on the mean.
B. The degrees of freedom for the t test are based on the number of observations used in
the calculation of the ______________________ .
C. The larger the degrees of freedom for the t distribution, the more the critical values
resemble those of ____________________.
D. To demonstrate that a high proportion of observations will be close to the mean, one
can use the alternative hypothesis that the ____________________ is small.
2. You are studying educational level (measured as years of school successfully
completed) among Florida prisoners. In a sample of 40 prisoners, you find a mean of 9.4
years and a standard deviation of 2.1 years. Give a 95% confidence interval for the mean
educational level in this population, specifying the population in your sentence.
3. The U. S. Sentencing Commission reported that in 2003, the mean sentence nationally
for persons convicted of auto theft was 60.8 months. You pull records for a random
sample of 35 Florida convictions for auto theft during this same year, and find a mean of
61.9 months and a standard deviation of 4.8 months. Is there significant evidence that the
mean sentence in Florida differed from the national mean, using  = 5%?
4. In a sample of 120 households of Jacksonville Beach, 52 are rated as ‘poorly
prepared’ to evacuate on short notice. Give a 90% confidence interval for the proportion
of all households in this community that are poorly prepared to evacuate.
5. In the past, the proportion of undergraduates passing a required skills exam on the first
try has been 55%. You believe that the proportion has increased. In a sample of 60
recent first-time takers of the exam, 40 passed. Is there evidence to confirm your belief,
at  = 5%?
6. A marketing firm is planning a study among elderly persons to estimate the proportion
who do not have Internet access. They have no prior information on what this proportion
might be. Recommend a sample size, if they wish their 90% confidence interval to have a
margin of error of only 0.05.
7. A past study showed that the sentences for persons convicted of auto theft in Florida
had a standard deviation of 6.5 months. The legislature was concerned that there was too
great an inconsistency in the sentences, and approved guidelines for judges to use in
sentencing. You pull records for a random sample of 36 Florida convictions since the
institution of the guidelines and find a standard deviation of 4.8 months. Is there
significant evidence that the standard deviation has decreased, using  = 5%?
(Problem 8 is longer because it requires two hypothesis tests. It can be shortened, or
turned in to two problems.)
8. In order to qualify for automatic submission of its claims to an insurance company, a
medical billing company must pass an accuracy test. In the accuracy test, the insurance
company randomly selects and audits a sample of 30 claims from the billing company.
To pass, there must be
1) No significant evidence, at α = 10%, that the mean error differs from 0.
2) Significant evidence, at α = 1%, that the standard deviation in the errors is less than 10.
a) Explain why the first hypothesis test uses a high value of .
b) In the sample of 30, the mean error was $2.12 and the standard deviation was $8.01.
Does the company pass the test?
9. On the standard treatment, mice exposed to a toxin have a median survival time of 25
days. You have developed a new treatment, and want to show that it increases median
survival time. In a random sample of 12 mice exposed to the toxin, 10 survived more than
25 days. At  = 5%, does this constitute evidence that the median survival time has
increased?
SOLUTIONS.
1. A. median
B. standard deviation (or sums of squares or variance)
C. the standard normal
D. variance (or standard deviation)
2. With confidence 95%, the mean years of education among Florida prisoners is
between 8.728 and 10.072 years.
3.  = mean sentences for auto theft in Florida
Ho: µ = 60.8
H1:  ≠ 60.8
t = 1.356, p value = 0.1841
There is no significant evidence that the mean sentence for auto theft in Florida differs
from the national mean.
ˆ =52 ≥ 5 and
4. p̂ =52/120 We can use the normal approximation because np
ˆ)=68 ≥ 5.
n(1  p
With confidence 90%, the proportion of Jax Beach residents who are poorly prepared to
evacuate is between 0.3589 and 0.5077. (The correction due to Agresti and Coull is will
not make much difference here, since sample size is reasonably large.)
5. p = current probability a student will pass on the first try
Ho: p ≤ 0.55
H1: p > 0.55, can use normal approximation because np0 =33 ≥ 5 and
n(1  p0)=27 ≥ 5.
Z = 1.816, p value = 0.0346
There is significant evidence that the probability a student will pass on the first try has
increased.
6. n  
2
1.645 
 * .5 * .5  271
 .05 
7. Ho:  2  42.25 versus H1:  2  42.25 . SS = 35*(4.82) =806.4,
X 2  806.4 / 42.25  19.09
Using the table of the Chi-squared distribution with 35 df, reject Ho if X 2 < 22.465
There is significant evidence that the standard deviation is now below 6.5.
8 a) The hypothesis that would mean better billing practices is  = 0. Normally, we
would require the company to prove it has good billing practices, but there is no way to
make ‘=’ the alternative hypothesis. Instead, we have to refuse the company if there is
even moderate evidence that they have poor billing practices.
b) #1: Ho:  = 0, H1:  ≠ 0, t = 1.45, p value = 0.1579, There is no significant evidence
the mean differs from 0.
#2: Ho:  2 ≥ 100 versus H1:  2 < 100, SS = 1860.6429, X 2 =18.61
With 29 df, reject Ho if X 2 < 17.708. Hence, there is no significant evidence that the
standard deviation is less than 10.
The company does not pass the test.
9. M = median survival under new treatment
Ho: M ≤ 25
H1: M > 25
P value = 0.01 = probability of 10 or more surviving 25+ days under the binomial
distribution with n = 12 and p = 0.50. There is evidence that the median survival time is
now higher than 25 days.