Download Lesson 10 - Hypothesis testing

STATISTICS AND THE TI-83
Lesson #10. Inferential Statistics:
HYPOTHESIS TESTING
1. Inference on the Mean of a Population
Exercise 1. Test the null hypothesis that  =17 versus the alternative hypothesis that  < 17, if a
random sample of size 60 taken from the population had a mean of 14.3 and a standard deviation
of 10. Test at a 0.05 level of significance.
Ho:  = 17
H1:  < 17
left tail test
 = 0.05
Decision rule: reject Ho if p-value<0.05 or z<-1.645 (=.05)
 STAT
Tests 1 Z-Test Stats ENTER
 Enter the values  0 :  0 : 17 s : 10 x : 14.3 N : 60  :  0
 Calculate
ENTER
Conclusion: z= -2.09 < -1.645
p-value = 0.018 < 0.05.
Reject the null hypothesis because p-value < 0.05. We have statistical evidence at the
0.05 level to believe that < 17
Exercise 2. In a one-tail test, p-value is the probability that a value is as extreme as the
one observed from the sample. In exercise 1, use the normal cumulative distribution
function to find the p-value, that is, the probability that X  14.3, using n=60,
  17 and   10 seconds.
10
nd
 2 DISTR 2 normalcdf(-10^9, 14.3, 17,
60
) = 0.0182455541=p-value
Note: as expected, the answer obtained agrees with the p-value in problem 1.
Exercise 3. A random sample of 70 observations gave a mean of 40 and a standard deviation
of 15. At a 0.05 level of significance test the following hypothesis:
a) H 0 :   37 versus H1 :
b) H 0 :   37 versus H1 :
  37 right tail test
  37 two tail test
a) Decision rule: reject Ho if p-value<0.05 or z>1.645 (=.05)
 STAT
Tests 1 Z-Test Stats ENTER
 Enter the values  0 : 37
s :15
x : 40 N : 70
 Calculate
ENTER
Conclusion: z= 1.673 >1.645
p-value = 0.04713 < 0.05.
Reject the null hypothesis because p-value < 0.05. We have statistical evidence at the 0.05
level to believe that >37
b) Decision rule: reject Ho if p-value<0.05 or z>1.96 or z<-1.96 (=.05)
 STAT
Tests 1 Z-Test Stats ENTER
 Enter the values  0 : 37
: ≠0
s :15
x : 40 N : 70
 Calculate
ENTER
Conclusion: z= 1.67 < 1.96
p-value = 0.094 > 0.05.
Do not Reject the null hypothesis because p-value >0.05. We don’t have statistical evidence at
the 0.05 level to believe that ≠37
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Exercise 4. A psychologist hypothesizes that college students today sleep less than college
students 20 years ago. To test this hypothesis, he selects a sample of 25 students and asks
them to record their nightly slumbers in the same way as a student of 20 years ago. The old
study showed, college students sleeping on an average of 7.32 hours. The students in the
present study had a mean of 7.21 hours per night with a standard deviation of 1.98 hours.
Assuming normality, what conclusions can you draw at a level of significance of 0.01?
. H0
:   7.32
H1 :   7.32
left tail
Decision rule: reject Ho if p-value<0.01 or t<-2.492 (=.01 and df=24)

STAT
Tests 2 T-Test Stats
  0 : 7.32
ENTER
 x : 7.21

 S :1.98
 N : 25
  0.
 Calculate ENTER
Conclusion: t=-.277777, p-value = .3917809273>0.01 Do not accept the claim that
µ<7.32, that students sleep less today.
Note: selecting Draw instead of Calculate gives you the p-value shaded area.
Exercise 5. The time needed for a male to move his foot from the floorboard to the brake pedal
of an automobile is believe to be a normal random variable. In 36 independent tests of this
reaction time, the average time required was 0.5 seconds and the standard deviation was 0.1
seconds. Would you accept the hypothesis that his mean reaction time is greater than 0.45
seconds. Use a 0.01 level of significance. Show all steps.
H0 :   0.45
H1 :   0.45
right tail
Decision rule: reject if Prob-value < 0.01
 STAT
Tests 1 Z-Test
Stats
Enter the values  0 :
 Calculate
ENTER
0.45
ENTER
s : 0.1
x : 0.5
N : 36
 0
Conclusion: z=3
p-value = 0.0013499672
Reject the null hypothesis because p-value <0.01. We have statistical evidence at the 0.01
level to believe that his mean reaction time is greater than 0.45.
Exercise 6. In a one-tail test, p-value is the probability that a value is as extreme as the one
observed from the sample. In exercise 4, use the normal cumulative distribution function to find
the p-value, that is, the probability that X  0.5, using n=36,   0.45 and   0.1 seconds.

2nd DISTR 2 normalcdf(0.5, 10^9, 0.45,
0 .1
36
) = 0.0013499672=p-value
Note: as expected, the answer obtained agrees with the p-value in problem 4.
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2. Inference: proportion of a population
Exercise 7. Professor Fernandez asserted that no more than 60% of the 19-21 year olds
even bothered to vote in the last election. But Tony, one of his students took offense and
decided to accept the burden of proof of his conviction. In fact, he got on the phone and
called 75 of his friends (19-21) and asked them if they had voted. 57 replied that they
had. Is this sufficient proof to offer, assuming the professor would actually listen to
such evidence?. Set up and solve in the form of a hypothesis testing problem at the
0.05 level of significance.
Be sure to include all steps and to interpret your results.
H0 : P  .60
Ha :
P  .60
Right Tail Test
Decision Rule: reject the null hypothesis is p-value <0.05 or z >1.645
P
 STAT
TESTS 5
P:.60
X: 57
N:75
 Calculate ENTER
z=2.828427125 p-value = .0023389301 < 0.05
Conclusion: Tony has statistical evidence to reject the claim of the professor that no
more than 60% of the 19-21 year olds even bothered to vote in the last election
Exercise 8. A random sample of 400 citizens in a community showed that 240 favored
having the old elementary school building replaced by a new, modern building. Test the
hypothesis that 55% of the citizens in the community favor this project. Use a 5% level
of significance. Show all 8 steps. Find the p-value.
H 0 : P  0.55
H a : P  0.55
Two Tailed Test
Decision Rule: do not reject the null hypothesis is p-value <0.05
 STAT
TESTS
5
P:.55
X: 240
N:400 P 
 Calculate ENTER
z=2.01 p-value = .04442300455
Conclusion: Reject the null hypothesis, reject the claim that 55% of the citizens in the
community favor having their old school building replaced by a new, modern building.
Exercise 9. In a two-tail test, p-value is twice the probability that a value is as extreme
as the one observed from the sample. In exercise 7, use the normal cumulative
distribution function to find the p-value, that is, the probability that p̂  0.60, ,
p  0.55 and n  400 .
 2nd DISTR 2 2  normalcdf(0.6, 10^9, 0.55,
=0.04441 =p-value (see p-value in problem 7)
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(.55)(.45)
) = 2(0.02221)
400
Exercise 10. A population is known to be normally distributed with variance 9. However, there is a
dispute as to whether
H0 :   50

is 50 or 60. Suppose you have to decide by using one observation whether
is true. The following decision rule is adopted: Reject
greater than 56. Find
if the observed value is
 and  , the probabilities of Type I and Type II errors.
 = P(Type I error)=P(Rejecting H0 | H0 is true

H0
 =normalcdf(56, 1E99, 50, 3)
ENTER
)= P(x>56 given µ=50)
answer: 0.022750062
  P(Type II error)=P(Accepting H0 | H0 is false,   60) = P(x<56

  normalcdf(-1E99, 56, 60, 3)
ENTER
given µ=60)
answer: 0.0912112819
Exercise 11. Mr. Fernandez is running for election in a certain district. Suppose p represents
the proportion of voters who favor him. The following hypotheses are set up.
H : p  0.6
against H 1: p  0.6 (p  .4) . It is agreed that a sample of 8
individuals picked at random will be interviewed and if 3 or more people declare in favor of Mr.
Fernandez, then H 0 will be accepted.
Otherwise, H 0 will be rejected. Find


 = binomcfd(8, 0.6, 2)
 = 1- binomcfd(8, 0.4, 2)
ENTER answer: .04980736
ENTER answer: .68460544
Exercise 12. A population is known to be normally distributed with variance 25. However, there is a
dispute as to whether
H0 :   50

is 50 or 45. Suppose you have to decide by picking a sample whether
is accepted or rejected.
The following decision rule is adopted: Reject
average of a sample of size n=16 is less than 48. Find

 = normalcdf(-1E99, 48, 50, 5/4)
 = normalcdf(48, 1E99, 45, 5/4)
ENTER
answer: 0.0547992894
ENTER answer: 0.0081975289
.
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if the
 and  , the probabilities of Type I and
Type II errors.

H0