STATISTICS AND THE TI-84
Lesson #11 Inferential Statistics: Two populations
1. Inference: comparing two population means
Exercise 1. There are two populations. A sample of size 120 from one of the
populations gave a mean of 15 and a standard deviation of 1.3. A sample of
size 88 from the other population gave a mean of 13.5 and a standard deviation
of 1.5. Find a 98% confidence interval for the difference between the population
means.
Sample information:
x1  15
s1  1.3 n1  120
x2  13.5


s 2  1.5 n1  88
STAT
TESTS 9 (2-Samp Z Int) Stats
(enter the values) 1 = 1.3 2 = 1.5
x1  15
n1  120
 Calculate ENTER
Result: (1.0368, 1.9632)
x 2  13.5
ENTER
n2  88 C-Level: .98
Exercise 2: The average length of twenty trout caught in a lake was 10.8
inches with standard deviation of 2.3 inches, and the average length of fifteen
trout caught in another lake was 9.7 inches with standard deviation of 1.5 inches.
Construct a 90 percent confidence interval for the difference in the true mean
lengths of trout in the two lakes.
 STAT
TESTS 0 (2-Samp t Int) Stats ENTER

x1 : 10.8 S x1 : 2.3 n1 : 20
x 2 : 9.7 S x 2 : 1.5
n 2 : 15
C  Level : .9 Pooled : Yes
Calculate ENTER
Results: pooled: (-..0561, 2.2561)
df=33
 STAT
TESTS 0 (2-Samp t Int) Stats

x1 : 10.8 S x1 : 2.3 n1 : 20
x 2 : 9.7 S x 2 : 1.5
Sxp=2.00007576
ENTER
n 2 : 15
C  Level : .9 Pooled : No
Calculate ENTER
Results: non-pooled: (.00993, 2.1901) , df=32.48284419
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Exercise 3. A farmer tried Feed A on 256 cattle and Feed B on 144 cattle. The mean
weight of cattle given Feed A was found to be 1350 pounds with a standard deviation of
180 pounds. On the other hand, the mean weight of the cattle given Feed B was found
to be 1430 pounds with a standard deviation of 210 pounds. At the 5 percent level of
significance, is Feed B significantly better than Feed A? Find the p-value.
H0 : B   A  0
H a :  B   A  0, right tail
Reject the null hypothesis is p-value<0.05



STAT
 TESTS 3 Stats
ENTER
s1 : 210 x1 : 1430 n1 : 144
s 2 : 180 x 2 : 1350 n 2 : 256   2
Calculate
ENTER
Conclusion: z=3.845 p-value = 6.0203004E-5 <0.05
Reject the null, we have statistical evidence to believe that Feed B is significantly
better than Feed A
Exercise 4. At a certain university twelve voters were picked at random from those
who are in favor of impeachment of the president, and ten were selected at random
from those who are against. The following table give their ages.
In favor
27 34 28 30 29 50 30 44 29 32 41 35
Against
31 36 43 40 32 48 30 29 42 49
H0: 1-2=0
H1: 1-2≠0
At a 10% level of significance, is it true that the age of those in favor of impeachment
significantly differs from the age of those against. Use a two sample t-test with pooled
variance.
Decision Rule: Reject H0 if p-value <.05
 {27, 34, 28, 30, 29, 50, 30, 44, 29, 32, 41, 35}
STO L1 ENTER
 { 31, 36, 43, 40, 32, 48, 30, 29, 42, 49}
STO L2 ENTER



STAT
List 1: L1
 1: ≠2
 Calculate
 TESTS 4 DATA ENTER
List 2: L2
Freq 1: 1 Freq 2: 1
Pooled: Yes ENTER
ENTER
Conclusion: t=-1.246579 p-value =.2269 > 0.10
Do not Reject the null hypothesis, we have statistical evidence to believe that
there is no significant difference between the ages of those in favor of
impeachment and those against.
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Exercise 5; Paired t-test.
Dr. Williams claims that the special diet that he recommends significantly
reduces weight. A sample of eight persons was selected and they were put on
the diet for a period of 6 weeks. The table below shows the weights (in pounds)
of those eight person before and after dieting.
Before
182
180
195
178
177
221
198
208
After
168
183
187
169
161
204
194
196
a) Construct a 99% confidence interval for the mean difference d in weight
before and after using the dieting recommended by Dr. Williams. Use a paired
difference d  t 
2






sd
.
n
{182,180, 195,178, 177, 221, 198, 208} STO L1 ENTER
{168,183, 187,169, 161, 204, 194, 196} STO L2 ENTER
L1-L2 STO D ENTER
STAT TESTS 8 (T Interval)
DATA ENTER List: D Freq: 1 C-Level: .99
Calculate ENTER
answer: (1.3401, 17.91)
x  9.625 sx  6.696 n  8
b) Using a 1% level of significance, can you conclude that the mean weight loss
for all persons due to this special diet is greater than zero?
 STAT  TESTS 2 (T-test)
 DATA ENTER
 o: 0
List: D Freq: 1
>o ENTER
 Calculate ENTER
 t=4.0655
p-value=.002388076 x  9.625 sx  6.696 n  8
Conclusion: Reject H0 because p-value < . We have statistical evidence at a
0.01 significance level to believe that the special diet recommended by Dr.
Williams is effective.
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2. Inference: comparing two population proportions
Exercise 6. In a study to estimate the proportion of residences in a certain city and its
suburbs that subscribe to a certain magazine, it is found that 63 of 120 urban
residences subscribe while only 34 of 125 suburban residences subscribe. Find a 90%
confidence interval for the difference in the proportion of urban and suburban
residences that subscribe to writer's digest.
 STAT  TESTS B (2-PropZInt.)
 X1: 63 N1: 120 X2: 34
N2: 125
 Calculate ENTER
answer: (.15346, .35254)
C-Level: 0.90
Problem 7. A jar containing 130 mosquitoes was sprayed with an insecticide of Brand
A and it was found that 98 of them were killed. When another jar containing 150
mosquitoes of the same type was sprayed with Brand B. 120 of them were killed. At the
2 percent level of significance, do the two brands differ in their effectiveness?
H0 : p1  p2
Ha : p1  p 2 two tailed test
Decision rule: Reject the null hypothesis, if p-value<.02


STAT
 TESTS 6 X1: 98 N1: 130 X2: 120 N2:150 p1: ≠p2
Calculate ENTER answer: z= -0.9276414111 p-value =0.3535935629
Do not reject the null. We don’t have statistical evidence to believe that the brands differ
in effectiveness, p-value >.02
Problem 8. During the first exam, the average grade of 7 men in a statistic exam was
78 points with a standard deviation of 5.3. For 11 women in the same class the average
grade was 83 with a standard deviation of 6.2. At the 5% level of significance can we
conclude that women are doing better that men in the class? Use the pooled variance
with 16 degrees of freedom. What about at the 1% level of significance?
H0 : w  m  0
H a :  w   m  0, right tail
Reject the null hypothesis is p-value<0.05



STAT
 TESTS 4 Stats
ENTER
s1 : 6.2 x1 : 83 n1 : 11
s 2 : 5.3 x 2 : 78 n 2 : 7
Calculate
 2
ENTER
Conclusion: t=1.759138776 p-value = 0.048829674 <0.05
Reject the null hypothesis, we have statistical evidence to believe that women are
doing significantly better than men in this class.
p-value=0.048829674 >0.01. Do not reject the null hypothesis. We don’t have
statistical evidence at a 1% level of significance to believe that women are doing
better than men.
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