Math 104 Fall 14Lab Assignment #4Math 104 Fall 14Lab Assignmen.docx

Math 104 Fall '14Lab Assignment #4
[1] Lipitor is a drug used to control cholesterol. In clinical
trials of Lipitor, 94 subjects were treated with Lipitor and 270
subjects were given a placebo. Among those who were treated
with Lipitor, 7 developed infections. Among those given a
placebo, 27 developed infections. Use a 0.05 significance level
to test the claim that the rate of infections was the same for
those treated with Lipitor and those given a placebo.
Z Test for Differences in Two Proportions
Data
Hypothesized Difference
0
Level of Significance
0.05
Group 1
Number of Items of Interest
7
Sample Size
94
Group 2
Number of Items of Interest
27
Sample Size
270
Intermediate Calculations
Group 1 Proportion
0.074468085

Group 2 Proportion
0.1
Difference in Two Proportions
-0.02553191
Average Proportion
0.0934
Z Test Statistic
-0.7326
Two-Tail Test
Lower Critical Value
-1.9600
Upper Critical Value
1.9600
p-Value
0.4638
Do not reject the null hypothesis
[2] Simple random samples of high-interest (5.36%) mortgages
and low-interest (3.77%) mortgage were obtained. For the 40
high-interest mortgages, the borrowers had a mean FICO credit
score of 594.8 and standard deviation of 12.2. For the 40 low-
interest mortgages, the borrowers had a mean FICO credit score
of 785.2 and standard deviation of 16.3.
a) Use a 0.01 significance level to test the claim that the mean
FICO score of borrowers with high-interest mortgage is lower
than the mean FICO score of borrowers with low-interest
mortgage.
b) Does the FICO credit rating score appear to affect mortgage
payments? If so, how?
The test of interest here is,

As here the sample size for both the samples are large enough
so we can assume the normality using CLT and thus a Z-test
would be appropriate. And the hypotheses also show that it’s a
left tail test.
The output is given below,
Z Test for Differences in Two Means
Data
0
0.01
Population 1 Sample
Sample Size
40
Sample Mean
594.8
Population Standard Deviation
12.2
Population 2 Sample
Sample Size
40
Sample Mean
785.2
Population Standard Deviation
16.3
Difference in Sample Means
-190.4
Standard Error of the Difference in Means
3.2192

Z Test Statistic
-59.1451
Lower-Tail Test
-2.3263
p-Value
0.0000
Reject the null hypothesis
Here the lower p-value suggests that we should reject the null
hypothesis and conclude that the mean FICO score of borrowers
with high-interest mortgage is lower than the mean FICO score
of borrowers with low-interest mortgage.
Note that the above analysis showed that the interest rate is
higher for the group with lower FICO scores. Thus the FICO
credit rating scores appear to affect mortgage payments. If the
FICO score is high then the interest rate is expected to be low
so the mortgage payments would be low. And for low FICO
score the opposite is expected to be observed.[3] A random
sample of the birth weights of 186 babies has a mean of 3103g
and a standard deviation of 696g (based on data from
“Cognitive Outcomes of Preschool Children with Prenatal
Cocaine Exposure,” by Singer et al., Journal of the American
Medical Association, Vol. 291, No. 20). These babies were
born to mothers who did not use cocaine during their
pregnancies. Further, a random sample of the birth weights of
190 babies born to mothers who used cocaine during their
pregnancies has a mean of 2700g and a standard deviation of
645g.
a) The birth weights of babies are known normally distributed.
Use a 0.05 significance level to test the claim that both samples
are from populations having the same standard deviation.
b) Using your finding in (a), construct a 95% confidence

interval estimate of the difference between the mean birth
weight of a baby born to mothers who did not use cocaine and
that of a baby born to mothers who used cocaine during their
pregnancies.
c) Using your finding in (b), does cocaine use appear to affect
the birth weight of a baby? Substantiate you conclusion.
a) Here we need to test for the equality of the standard
deviation (same as variance) so the hypotheses of interest are,
As this is two sample test for standard deviation or variance so
the test can be performed using a F test. The obtained output is
given below.
F Test for Differences in Two Variances
Data
0.05
Larger-Variance Sample
Sample Size
186
Sample Variance
484416
Smaller-Variance Sample
Sample Size
190
Sample Variance
416025
F Test Statistic

1.1644
Population 1 Sample Degrees of Freedom
185
189
Two-Tail Test
1.3329
p-Value
0.2986
Do not reject the null hypothesis
Here the p-value is larger than the significance level so we are
failing to reject the null hypothesis and concluding that the data
does not suggest that both samples are from populations having
different standard deviation.
b) Here we found out that the variations are not different in two
groups so we should use a pooled variance t-test.
Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
0
0.05
Population 1 Sample
Sample Size
186

Sample Mean
3103
Sample Standard Deviation
696
Population 2 Sample
Sample Size
190
Sample Mean
2700
645
185
189
Total Degrees of Freedom
374
Pooled Variance
449854.7727
Standard Error
69.1826
403.0000
t Test Statistic
5.8252
Two-Tail Test
-1.9663

1.9663
p-Value
0.0000
Confidence Interval Estimate
for the Difference Between Two Means
Data
Confidence Level
95%
Degrees of Freedom
374
t Value
1.9663
Interval Half Width
136.0356
Confidence Interval
Interval Lower Limit
266.9644
Interval Upper Limit
539.0356
From the values from above table,
95% confidence interval for difference =
c) From the above calculated confidence interval we can see
that the confidence interval does not contain the value 0. So
based on the findings we can conclude that cocaine use appears

to affect the birth weight of a baby.[4] A large discount chain
compares the performance if its credit managers in Ohio and
Illinois by comparing the mean dollars amounts owed by
customers with delinquent charge accounts in these two states.
Here a small mean dollar amount owed is desirable because it
indicates that bad credit risks are not being extended large
amounts of credit. Two independent, random samples of
delinquent accounts are selected from the populations of
delinquent accounts in Ohio and Illinois, respectively. The first
sample, which consists of 15 randomly selected delinquent
accounts in Ohio, givens a mean dollar amount of $524 with a
standard deviation of $38. The second sample, which consists
of 20 randomly selected delinquent accounts in Illinois, gives a
mean dollar amount of $473 with a standard deviation of $22.
a) Assuming that the normality assumption, test to determine if
Here we need to test for the equality of the variance so the
hypotheses of interest are,
As this is two sample test for variance and normality
assumption is valid so the test can be performed using a F test.
The obtained output is given below.
F Test for Differences in Two Variances
Data
0.05
Larger-Variance Sample
Sample Size
15
Sample Variance
1444
Smaller-Variance Sample

Sample Size
20
Sample Variance
484
F Test Statistic
2.9835
14
19
Two-Tail Test
2.6469
p-Value
0.0283
Here the p-value is smaller than the significance level so we are
rejecting the null hypothesis and concluding that the data
suggests that the population variances are not equal.
population mean dollar amounts owed by consumers with
delinquent charge accounts in Ohio and Illinois.
Here the population variances are not equal so we should run a
separate variance t test. The results are as follows,
Separate-Variances t Test for the Difference Between Two

Means
(assumes unequal population variances)
Data
0
0.05
Population 1 Sample
Sample Size
15
Sample Mean
524
38.0000
Population 2 Sample
Sample Size
20
Sample Mean
473
22.0000
Numerator of Degrees of Freedom
14512.2178
Denominator of Degrees of Freedom
692.7711
Total Degrees of Freedom
20.9481
Degrees of Freedom
20
Standard Error
10.9757

51.0000
Separate-Variance t Test Statistic
4.6466
Two-Tail Test
-2.0860
2.0860
p-Value
0.0002
Here based on the result we are rejecting the null hypothesis
and thus the conclusion is based in the data there is a difference
between the population mean dollar amounts owed by
consumers with delinquent charge accounts in Ohio and Illinois
at 0.05 significance level.
c) Assuming that the normality assumption and the condition
you checked in a) hold, calculate a 95 percent confidence
interval for the difference between the mean dollar amounts
owed in Ohio and Illinois. Based on this interval, do you think
that these mean dollar amounts differ in a practically important
way?
Here the 95% confidence interval is =
As we can see the computed confidence interval has values not
so high as lowest is 28.10 so based on this interval, I don’t
think that these mean dollar amounts differ in a practically
important way.

Math 104 Fall 14Lab Assignment #4Math 104 Fall 14Lab Assignmen.docx

Math 104 Fall 14Lab Assignment #4Math 104 Fall 14Lab Assignmen.docx

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Math 104 Fall 14Lab Assignment #4Math 104 Fall 14Lab Assignmen.docx