Learning Team Assignment #2
Chapter 6 Case Study
#1 Use the sample to find a point estimate for the mean shoulder height of
(a) male bears.
(b) female bears.
Answer
(a) The point estimate for the mean shoulder height of male bears is 79.73
(b) The point estimate for the mean shoulder height of female bears is
75.68
#2 Find the standard deviation of the sample of shoulder heights for the
(a) male bears.
(b) female bears.
Answer
(a) The standard deviation of the sample of shoulder heights for the male
bears is 12.09
(b) The standard deviation of the sample of shoulder heights for the female
bears is 7.59
#3 Use the sample to construct a 95% confidence interval for the mean
shoulder height of
(a) male bears.
(b) female bears.
Answer
(a) The 95% confidence interval for the mean shoulder height of male
bears
12.09 
12.09 
 1.96
 3.75
 1.96
 6.32 
 40 
75.98,83.48

(b)
The 95% confidence interval for the mean shoulder height of female
bears
7.59 
7.59 
 1.96
 2.81
 1.96
5.29 
 28 
72.87, 78.49

#4 Use the sample to construct a 95% confidence interval for the mean
shoulder height of all bears in the study. How do your results differ from
those in Exercise 3? Explain.
Answer
The 95% confidence interval for the mean shoulder height of all bears
9.84 
 9.84 
 1.96
 2.338
 1.96
8.246 
 68 
75.36,80.05
The major difference is that the all the measurements are used, both male and

female to come up with the standard deviation of all the samples. Having a
bigger sample appears to decrease the margin of error as is shown in the
reading.
Chapter 7 Case Study pg 363
#1 Complete the hypothesis test for all adults (men and women by
performing the following steps. Use a level of significance of α = 0.05.
(a) Sketch the sampling distribution.
(b) Determine the critical values and add them to your sketch.
(c) Determine the regions and shade them in your sketch.
(d) Find the standardized test statistic. Add it to your sketch.
(e) Make a decision to reject or fail to reject the null hypothesis.
(f) Interpret the decision in the context of the original claim.
Answer
(a)
(b)
Significance level = 0.05 (sample mean should have under 5% chance
of taking place at 0.05 Zc = ±1.96
(c)
Rejection area is displayed in black. The two temperatures corresponding to
Zc are displayed as 98.48 and 98.73.
(d)
Z value for sample mean is
Z = Xbar – μ / (s/SQRT(N))
Z = -5.466
(e) As Z < Zc we can refuse the claim.
(f) The original claim was that the mean value of body temperature is 98.6.
Depending on the study sample we can deduce that the usual body
temperature is under 98.6.
#2 If you lower the level of significance to α = 0.01, does your decision
change? Explain your reasoning.
Answer
At significance level of 0.01 we have Zc = ±2.576.
Again as Z< Zc we can still refuse the claim.
#3 Test the hypothesis that the mean temperature of men is 98.6°F. What
can you conclude at a level of significance of α = 0.01?
Answer mean= 98.104
std dev= 0.6987
standardized test statistic= -5.715
critical values= ±2.576
As Z< Zc we can refuse the claim that the mean temperature of mean is
98.6
#4 Test the hypothesis that the mean temperature of women is 98.6°F. What
can you conclude at a level of significance of α = 0.01?
Answer mean= 98.394
std dev= 0.7434
standardized test statistic=-2.2355
critical values= ±2.576
As Z > Zc and is not in the refusal area we must accept the null hypothesis
and claim that the mean temperature of ladies is 98.6
#5 Use the sample of 130 temperatures to form a 99% confidence interval
for the mean body temperature of adult humans.
Answer E = Zc * s/(SQRT(n))
E = ±2.574 * 0.73/SQRT(130)
E = ±0.1648
With 99% confidence we might say that the mean value of human body
temperature is between 98.0852 and 98.4148
#6 The conventional “normal” body temperature was established by Carl
Wunderlich over 100 years ago. What, in Wunderlich’s sampling
procedure, do you think might have led him to an incorrect conclusion?
Answer Strictly from a sampling method point of view, from the data analysis in parts 3
and 4 we can see that in case the sample contained considerably more ladies
compared to guys, it would be more likely to create a somewhat higher mean
value of body temperature.
Chapter 7 Real Statistics / Real Decisions pg298
#1 How Would You Do It?
(a) What sampling technique would you use to select a sample for the
study? Why? What sampling technique would you use if you
wanted to select samples from four age groups: 18-29, 30-49, 5064, and 65 and over?
(b) Which technique in part (a) will give you a sample that is
representative of the population?
(c) Identify possible flaws or biases in your study.
Answer Get a random sample to obtain a varied group of people.
(a) Random Sampling would be the best method for this study. It would
produce a non-biased/broad sample necessary for an exact result.
Stratified Random Sampling would be best for having the greatest
sampling method between several age groups.
(b) Random Sampling would provide an improved representation of the
population because Stratified Random Sampling would only target
particular age groups.
(c) The age of people in the sample might produce different responses
therefore it would be significant to have an equal quantity for each
age group. Education level might also cause prejudice, however it
would be extremely tough to even this out.
#2 Testing a Proportion
Test the claim that less than 40% of people in the United States think the
Social Security system will have the money available to provide the
benefits they expect for their retirements. Use α = 0.10. Write a
paragraph that interprets the test’s decision. Does the decision support
your department’s claim?
Answer
Use a z-test for any population if
np  5
and
nq  5
z
phat  p
pq /n
z

.40  .10
(.10)(.9) /4
z 2
H()  1.9435

H()  1.394

Fail to discard
There isn’t sufficient proof to support the claim that under
40% of people in the United States believe the Social Security system will
have money available to offer the benefits they suppose for their retirements.
#3 Testing a Mean
Test the claim that the mean age of people in the United States who
would say yes to the survey question shown in the table is 60 years or
older. Use α = 0.010 and assume that the population is normally
distributed. Write a paragraph that interprets the test’s decision. Is there
enough evidence to reject your department’s claim?
Answer
Use a 2 test in case the population is usual.

2 
(n 1)s2
2
31  1.9435 2
1.5445 2
 2  49.085
2 
Discard

There is sufficient proof to discard the claim that the mean age of
people in the United States that would state yes to the survey question is 60
years or older.
#4 Your Conclusions
On the basis of your analysis of the responses to this survey question,
what would you tell your department?
Answer I would say to the department that consistent with the formulas utilized to
compute the data of the surveys, the reply would be NO. There isn’t sufficient
data to verify otherwise. You require over 40% to get a better average of the
judgment. The percentage is extremely low to defend a hard answer for the
population of the United States. The quantity of people surveyed which said
yes were 60 years old or older. A larger survey with a younger quantity of age
group would require to be incorporated.