Math 140 Hypothesis Test Activities
Answer Keys
Hyp Test Act 2 Answers
1.
Ho: p = 0.93
Ha: p < 0.93 (claim)
Left Tailed Test
2.
: 1.25 (claim)
: 1.25
Ho
Ha
Two Tailed Test
3.
Ho: p = 0.74
Ha: p > 0.74 (claim)
Right Tailed Test
4.
: 98.6
: 98.6
Ho
Ha claim
Left Tailed Test
5.
0 : 2.9
: 2.9 ( )A
H
H claim
Two Tailed Test
6.
0 : 0.1 (claim)
H : 0.1A
H p
p
Left Tailed Test
7.
(If women are P1 and men are P2)
0 1 2
1 2
:
: ( )A
H p p
H p p claim
Left Tailed Test
(If men are P1 and women are P2)
0 1 2
1 2
:
: ( )A
H p p
H p p claim
Right Tailed Test
8.
0 1 2
1 2
: ( )
: A
H claim
H
Two Tailed Test
9.
Ho: p = 0.5
Ha: p > 0.5 (claim)
Right Tailed Test
Hyp Test Act 3 Answers
1. z = +1.48
The sample percentage is only 1.48 standard deviations above the population value, so it is not
significant.
2. z = - 0.74
The sample percentage is only 0.74 standard deviations below the population value, so it is not
significant.
3. z = -3.70
The sample percentage is 3.70 standard deviations below the population value, so it is very
significant.
4. z = +1.84
The sample percentage is 1.84 standard deviations above the population value. This is on the
borderline of significant and not significant. It depends on the significance level chosen.
Hyp Test Act 5 Answers
1.
Ho: p = 0.93
Ha: p < 0.93 (claim)
If Ho is true and 93% of Americans own a traditional phone, then there is 2.69% chance of
getting 454 or less out of 500 in the sample.
Since p-value 0.0269 < sig level 0.05, we reject the null hypothesis.
The sample data was significantly lower than the population value 93%. It was highly unlikely
that the sample data happened by random chance.
2.
: 1.25 (claim)
: 1.25
Ho
Ha
If Ho is true and people do spend 1.25 hours a day eating and drinking, then there is a 24.8%
chance of getting a sample mean of 1.22 hours or less.
Since p-value 0.248 > 0.10, we fail to reject the null hypothesis.
The sample data was not significantly lower than the population value. The sample data could of
happened by random chance.
3.
Ho: p = 0.74
Ha: p >0.74 (claim)
If Ho is true and 74% of Americans own a credit card, then we have a 8.57% chance of getting a
sample percent of 76% or more.
Since the p-value 0.0857 >0.05, we fail to reject the null hypothesis
The sample data was not significantly lower than the population value. The sample data could of
happened by random chance.
4.
: 98.6
: 98.6
Ho
Ha claim
If Ho is true and normal body temperature really is 98.6 degrees Fahrenheit, then we have a
0.23% chance of getting a sample mean of 98.2 or less.
Since p-value 0.0023<sig level 0.01, we will reject the null hypothesis.
The sample data was significantly lower than the population value. It was highly unlikely that
the sample data happened by random chance.
Hyp Test Act 6 Answers
1.
: 0.04
: 0.04
Ho p
Ha p claim
(reject Ho) There is significant sample evidence to support the claim that less than 4% of people
on this medicine had side effects.
(fail to reject Ho) There is not significant sample evidence to support the claim that less than 4%
of people on this medicine had side effects.
2.
: 63.5
: 63.5( )
Ho
Ha claim
(reject Ho) There is significant sample evidence to support the claim that the average height of
women is more than 63.5 inches
(fail to reject Ho) There is not significant sample evidence to support the claim that the average
height of women is more than 63.5 inches
3.
: 0.54( )
: 0.54
Ho p claim
Ha p
(reject Ho) : There is significant sample evidence to reject the claim that the republican
candidate will receive 54% of the vote.
(fail to reject Ho) : There is not significant sample evidence to reject the claim that the
republican candidate will receive 54% of the vote.
4.
: 2000
: 2000( )
Ho
Ha claim
(reject Ho) There is significant sample evidence to support the claim that the average weight of
electrically powered cars is less than 2000 pounds.
(fail to reject Ho) There is not significant sample evidence to support the claim that the average
weight of electrically powered cars is less than 2000 pounds.
5.
: 0.5
: 0.5
Ho p
Ha p claim
(reject Ho) There is significant sample evidence to support the claim that the majority of
patients taking Toprol have seen improvement in their migraine symptoms.
(fail to reject Ho) There is not significant sample evidence to support the claim that the majority
of patients taking Toprol have seen improvement in their migraine symptoms.
Hyp Test Act 7 Answers
3.
:
:
Ho 30 claim
Ha 30
Two tailed test.
Data was random. The shape was not bell shaped (skewed) , but sample size was over 30
(65). So it does pass the assumption of 30 or normal.
Test stat = -1.55 (Might not be significant enough) Sentence: The sample value was 1.55
standard errors below population value.
P-value = 0.126 If Ho is true, there is a 12.6% chance of getting the sample data or more
extreme.
(Sample data could of happened by random chance i.e. sampling variability.)
Pvalue 0.126 > sig level 0.1 Fail to reject Ho.
Conclusion: There is not significant sample evidence to reject the claim that the average
age of students at UCLA is 30 years old.
7.
: 64
: 64
Ho claim
Ha
Assumptions: Data was not random, however it was a census (better than random).
Sample size was over 30 (324). The shape of the data was nearly normal.
Test Stat = 9.506 (very significant) The sample value was 9.506 standard errors more than
the population value.
Pvalue < 0.0001 (close to zero). If Ho is true, then there is less than 0.0001 chance of getting
the sample data or more extreme.
Sample data probably did not happen by random chance.
Pvalue < sig level 0.05 Reject Ho.
There is significant sample evidence to reject the claim that the average height of math 140
students is 64 inches.
Hyp Test Act 8 Answers
5.
: p .
: p .
Ha 0 5 claim
Ho 0 5
Assumptions: Data was not random, however it was a census (better than random). There
was at least 10 success (female) and 10 failures (male).
Test stat = 3.33 (significant!) The sample value was 3.33 standard errors above population
value.
P-value = 0.0004 (probably did not happen by random chance.) Sentence: If Ho is true,
then there was a 0.0004 (0.04%) chance of getting the sample data or more extreme.
Pvalue < sig level (0.05) Reject Ho.
Conclusion: There is significant sample evidence to support the claim that more than 50%
of math 140 students are female.
Hyp Test Act 9 Answers
1.
0 1 2
1 2
: (Will approve sale of medicine)
: (Will not approve sale of medicine)A
H p p
H p p
Type 1: Reject H0 by mistake. You say you have evidence to support that the percent of women helped
by medicine is lower than the percent of men, when you really do not have evidence. FDA will not
approve medicine and they say they have evidence to back it up. People (especially men) will not have
access to a good stress medicine. The company that makes the medicine will lose money. FDA may be
liable for a law suit.
Type II: Fail to reject H0 by mistake. The FDA would say that they do not have sufficient evidence to
show that the percent of women that the medicine helps is lower than the percent of men, when they
really do have enough evidence. FDA will approve the sale of the medicine by mistake. Women taking
the medicine for stress may find that the medicine does not work. If it should come to light that the
medicine does not work, the company and the FDA may be liable.
2.
0 : 0.02 (Recall)
: 0.02 (No Recall)A
H p
H p
Type I: Reject H0 by mistake. Support HA that the percent of airbag malfunctions is less than 2% when
it really is not. Acura will not recall the cars when they really should have. There will be cars with bad
airbags on the road. Acura could maybe be sued if a person is killed or injured in a car accident.
Type II: Fail to reject H0 by mistake. Acura says that they do not have significant evidence to support
HA when they really do. Acura will recall the cars when they really didn’t have to. Acura is losing money
and reputation.
3.
0 : 0.04 (# of People drinking soda has not increased)
: 0.04 (# of People drinking soda has increased)A
H p
H p
Type I: Reject the null and supporting HA by mistake. Mike tells his boss that he has data showing that
the percent of people drinking the new flavor of Pepsi has increased, when he really does not have
evidence. Pepsi may invest in a new flavor that people may not like and lose money in the process.
Pepsi might feel that Mike lied to them or is incompetent, so he might get into trouble and maybe get
fired.
Type II: Fail to reject the null by mistake. Mike tells his boss that he does not have significant evidence
to show that the percent of people that like the new flavor has increased, when it really has increased.
Pepsi may not invest in this flavor. If the flavor was popular, they may lose money on the investment or
Coke beats them to it.
4.
In order to decrease the chances that a type I error occurring, we could decrease the significance level.
5.
In order to decrease the chances that a type II error occurring, we could increase the sample size.
6.
If we increase the significance level, then we have a greater chance of having a type I error, but we will
also have smaller chance of having a type II error.
7.
If we decrease the significance level, then we have a smaller chance of having a type I error, but we will
also have a greater chance of having a type II error.
8.
Most Statisticians agree that a 5% significance level maintains a good balance between type I and type II
errors. The probabilities for type I and type II errors will both be relatively small.
Hyp Test Act 11 Answers
1.
1 : ACT scores after class
2 = ACT scores before class
1 2
1 2
:
:
: 0
: 0
d
d
OR
Ho
Ha claim
Ho
Ha claim
Right tail test
Paired data
Assumptions? It was random. The sample sizes were both 20 (under 30) however the data
was bell shaped. So it does meet the 30 or normal requirement.
Test Stat = 2.9166 Sentence: Sample 1 (after) was 2.9166 standard errors above sample 2
(before). (significant)
Pvalue = 0.0044 Sentence: If Ho is true (two groups same), then there was a 0.0044
chance of getting the sample difference or more extreme. (very unlikely that it happened
by random chance.)
Pvalue < sig level (0.05) Reject the Ho!!
Conclusion: There is significant sample evidence to support the claim that the class is
effective in raising ACT scores.
2.
Both samples are independent and random. Since the sample sizes are both 30 or bell shaped,
it does meet all asumptions for a 2 mean hypothesis test.
Pop 1 : Do not live with smokers
Pop 2: Live with smokers
0 : 1 2
: 1 2( )
H
HA claim
The test statistic was t = -9.758. This means that the average cotinine level for those that don’t
live with smokers was 9.758 standard errors below the cotinine level for those that live with
smokers.
The Pvalue < 0.0001 . It is extremely unlikely that this data happened by random chance.
Sentence: If the cotinine levels for both groups are the same there was less than 0.0001 chance
of getting the sample data or more extreme.
Pvalue < sig level 0.01. Reject H0.
Conclusion: There is significant sample evidence to support the claim that the cotinine levels for
people that don’t live with smokers is lower than the cotinine level of the those who live with
smokers.
3.
Mu1 = systolic BP
Mu2 = diastolic BP
Ha: µ1 > µ2 (claim)
Ho: µ1 < or = µ2
Or
Ha: µ1 - µ2 > 0 (claim)
Ho: µ1- µ2 < or = 0
Or
Ha: µd > 0 (claim)
Ho: µd < or = 0
Assumptons: Was random. N=40 > 30. Data looks right skewed. Passes the 30 or normal
requirement.
Test Stat = 25.52 (very extremely significant). Systolic BP was 25.52 standard errors above
diastolic.
Pvalue < 0.0001 (zero) (Very unlikely to happen by random chance) Sentence: If Ho is
true, we had less than 0.0001 chance of getting the sample data or more extreme.
Reject Ho.
Conclusion: There is significant sample evidence to support the claim that systolic BP is
higher than diastolic BP.
4.
The sample data was randomly collected. The sample sizes were both over 30 or bell shaped.
The data sets are independent since a cholesterol level for a man and a woman are not related.
Data does meet the assumptions to do a 2 mean test.
Pop 1 : Cholesterol for Women
Pop 2: Cholesterol for Men
0 : 1 2
: 1 2( )
H
HA claim
The test statistic was t = -2.817. This tells us that the average cholesterol for the women was
2.817 standard errors below the average cholesterol for men. (This looks significant.)
The Pvalue was 0.0064. (Highly unlikely that this happened by random chance.) Sentence: If
cholesterol of men and women are the same, then there was only a 0.0064 chance of getting
the sample data or more extreme.
So Pvalue < sig level 0.05. Reject H0.
Conclusion: There is significant sample evidence to support the claim that the cholesterol levels
of men and women are different.
Hyp Test Act 12 Answers
1.
P1: percent of teen pregnancy 2012 (u.s.a.)
P2: percent of teen pregnancy 2008 (u.s.a.)
1 2
1 2
: p p
: p p
Ho
Ha claim
Assumptions: Data was random. 2008 data had at least 10 success and at least 10 failures.
2012 data had at least 10 success and at least 10 failures. Passes all the assumptions.
Test stat = 0.21 pregnancy rate in 2012 data was only 0.2 standard errors above preg rate
in 2008. (Not significant – close)
P-value = 0.4168 If the Ho is true, then there was a 41.7% chance of getting the sample
data or more extreme. (could of happened by random chance)
Fail to reject Ho
Conclusion: There is not significant sample evidence to support the claim that 2012 has a
higher teen pregnancy rate than 2008.
2.
P1: marijuana users
P2: non-marijuana users
1 2
1 2
: p p
: p p
Ho
Ha claim
Right Tailed Test
Assumptions: Data was random, At least 10 success and failures in marijuana group. At
least 10 success and failures in non marijuana group.
Test Stat = 6.85 Sentence: Group 1 are 6.85 standard errors above Group 2. (Very
Significant)
Pvalue < 0.0001 Sentence: If Ho is true, then there was less than 0.0001 chance of getting
the sample data or more extreme. (Probably did not happen by random chance)
Pvalue < sig level (0.05) Reject Ho.
Conclusion: There is significant sample evidence to support the claim that marjijuana
users use other drugs a lot more than non-marijuana users, i.e. gateway drug
8.
P1: percent of female math 140 students
P2: percent of male math 140 students
1 2
1 2
: p p
: p p
Ho
Ha claim
Assumptions: Not random, but it was a census (better than random). At least 10 success
and failures for both groups (females and males). Passes assumptions
Test Stat = 4.71 (significant) Group 1 (female) was 4.71 standard errors above Group 2
(males)
P-value < 0.0001 (did not happen by random chance)
Sentence: If Ho is true, then there was less than 0.0001 chance of getting the sample data
or more extreme.
Reject Ho
There is significant evidence to support that there is a higher percent of females in math
140 than males.
10.
P1: percent of Instagram
P2: percent of Facebook
1 2
1 2
: p p
: p < p
Ho
Ha claim
Assumption: Not random, but was a census. At least 10 people that use Instagram and at
least 10 that do not use Instagram. At least 10 that use facebook and at least 10 that do not
use facebook.
Test Stat: 4.16 (Instagram was 4.16 standard errors above facebook) (Significantly higher
NOT lower!!)
Pvalue = 1 If the null is true, then there is a 100% chance of getting the sample data or
more extreme.
Fail to reject Ho.
There is not significant evident to support that percent of Instagram is lower than the
percent of facebook.
Hypothesis Test Review Sheet #1 Answers
1. Simulation is an important part of inferential statistics. The idea is to assume that the
population value in the null hypothesis is correct and simulate what we would expect
random samples from that population value to look like. Now look at a real sample data set
and compare it to the simulation. We can look at how many of the simulated sample values
were the same or more extreme as the real sample data value. The probability of this is
an approximate P-value. If the real sample data value happened a lot in the simulation,
then we are pretty sure that the real sample value could have happened by random chance.
Therefore the real sample value is not significantly different than the population value
that the simulation is based on.
2. A test statistic tells us how many standard errors the sample data is above or below
the population value used in the null hypothesis.
3. There are several ways to know that the sample data is significantly different. A test
statistic that is very unusual or a very small P-value can indicate this. Also a value that is
very rare in the simulation can also.
4. If the population value in the null hypothesis is correct, then the P-value is the
probability of getting the sample value or more extreme.
5. There are several ways to know if the sample data could have happened by random
chance and does not necessarily contradict the population value. A test statistic that is
very small or a very large P-value can indicate this. Also a value that is happens often in
the simulation can also.
6. If P-value is small (less than or equal to the significance level), reject the null
hypothesis.
If P-value is large (greater than the significance level), fail to reject the null hypothesis.
7. If you reject the null hypothesis, start conclusion with “there is significant sample
evidence”.
If you fail to reject the null hypothesis, start conclusion with “there is not significant
sample evidence”.
If the claim is the null hypothesis, finish the conclusion with “to reject the claim”
If the claim is the alternative hypothesis, finish the conclusion with “to support the claim”
8. The assumptions are as follows:
To test a hypothesis test about a population mean, the sample needs to be collected
randomly and be either nearly normal or at least 30.
To test a hypothesis test about a population percentage (proportion), the sample needs to
be collected randomly and n (p) and n (1-p) must both be at least 10.
9.
0 : 180 (claim)
: 180A
H
H
Simulating a population mean of 180 pounds gave the following printout from StatKey.
Answers may vary since it is a randomized simulation. The sample mean in the health data
was 172.55 pounds. Simulating 2000 times, that sample mean of 172.55 or more extreme
occurred a total of 138 times (two tails). This indicates that the chances of the sample
data or more extreme data happening from a population of 180 was 0.069 (6.9%). This is
an estimated P-value. Since we are using a significance level of 6.9%, we will fail to reject
the null hypothesis. There is evidence, but it is not quite significant enough to reject the
claim that the average weight of men is 180 pounds. The sample data was unlikely to
happen from the population value of 180, but it was not quite unlikely enough. The sample
mean 172.55 was different than the population value of 180, but it was not quite
significant enough. Conclusion: There is not significant enough evidence to reject the
claim that the mean average weight of men is 180 pounds.
10.
0 : 25
: 25 (claim)A
H
H
Right Tailed Test. Test Stat T = 2.204 The sample mean of $26.82 was 2.204 standard
errors above the population value of $25. There is a significant difference between the
sample value and the population value. The sample value was significantly higher than the
population value. P-Value = 0.0181 If the average salary of nurses is $25, then there was
only a 1.81% chance of getting the sample data or more extreme. If the average salary
really is $25, then it was very unlikely that the sample data happened by random chance
(1.8%). Reject the null hypothesis. There is significant evidence to support the claim that
the average salary of nurses is greater than $25.
11.
0 : p 0.1
: p 0.1 (claim)A
H
H
Right Tailed Test. Answers may vary in simulation. In 5000 simulations we only got a
sample percent of 13% or higher only 1 time. Estimated P-value = 0.0004 Reject Ho. The
sample value of 13% is significantly higher than the population value of 10%. It is very
unlikely (0.0004) that the sample value happened by random chance. There is significant
sample evidence to support the claim that the more than 10% of women have at least one
tattoo.
12.
0 : p 0.25 (claim)
: p 0.25 A
H
H
Two tailed test. The sample percent was 0.259 and the Z test Stat = 0.197 . So the
sample percent was only 0.197 standard errors above the population value. This is very
close. They are not significantly different. The Pvalue = 0.8435 So if the population
value is 25%, there was an 84.4% chance of getting the sample data or more extreme.
The sample data could definitely happen by random chance. It had an 84% chance of
happening. We fail to reject H0. There is not significant sample evidence to reject the
claim that 25% of drowning deaths happen at the beach.
13. P1 : percent of high level preforming people that drink Gatoraid.
P2 : percent of high level preforming people that drink water.
0 1 2
1 2
: (gatoraide is not better than water)
: (gatoraide is better than water)A
H p p
H p p
Type I: Reject H0 by mistake and support that people perform better when drinking
Gatoraide when they reall don’t. This would result in people drinking Gatoraid thinking it
will improve performance when it really does not. Gatoraid may be liable for false
advertising since they said they have significant data when they really do not have
evidence. To decrease the chances of making a type I error, we could lower the
significance level.
Type II: Fail to Reject H0 by mistake and tell people that those that drink Gatoraid
perform no better than those that drink water, when they really do. People may not buy
Gatoraid since they do not think it works. Gatoraid may lose money in sales. To decrease
the chances of making a type II error, we should increase the sample size.
14.
0 : 2% (Put out recall)
: 2% (Will not put out recall)A
H p
H p
Type I: Reject H0 by mistake and support that the bracket is defective less than 2% of
the time when it is actually defective much more. The car company will not put out a recall
when they should. If a tire falls off the truck on the road it may cause an accident and
injuries. The company would be liable for damages and may be sued.
Type II: Fail to Reject H0 by mistake, so we will tell the company to put out a general
recall when they really don’t need to. This will cost the company money to replace
brackets that are probably not defective. The company may also lose some reputation
with its customers.
Hypothesis Test Act 13 answers
1.
0 : 1 2 3
: at least one is not equal
H p p p
HA
Expected values are n/k = 60/3=20. Notice all expected values >5. The data was randomly collected
also. So it does meet the assumptions necessary for a goodness of fit test. Our test statistic was Chi
squared = 11.7. Our P-value was 0.0029. Since our p-value was less than our sig level of 0.05 we reject
the null hypothesis. There is significant sample evidence to reject the claim that the percent of people
from each political part are the same.
2.
H0: p1=.43 , p2=.23 , p3=0.2 , p4 = 0.08 , p5 = 0.06
HA: Distribution is different than the null hypothesis (at least one is not equal) (claim)
The data was randomly selected and all the expected values from Statcato were greater than 5. The test
statistic was chi squared = 17.3766 and the p-value was 0.0016. We reject the null hypothesis since
pvalue is less than sig level 0.05. There is sufficient sample evidence to support the claim that the
percentages of COC students is different than what was given in the magazine.
3.
H0: Pm=Pt=Pw=Pth=Pf=Psa=Psu (claim)
HA: At least one day has a different probability
The expected values from Statcato are all greater than five. The data was randomly selected so it meets
all the assumptions for a goodness of fit test.
The test statistic was chi squared = 7.5478 and the p-value was 0.2731. Since the Pvalue was greater
than the sig level of 0.01, we fail to reject the null. There is not sufficient sample evidence to reject the
claim that the probability of dying in a car accident is the same on each day of the week.
4.
H0: p1 = 0.57 , p2 = 0.31 , p3 = 0.03 , p4 = 0.06 , p5 = 0.03
Ha: at least one is not equal (claim)
Expected Values are 1178.76 , 641.08 , 62.04 , 124.08 , 62.04
Degrees of freedom = 4
Test Statistic = 121.3667
P-value = 0
There is strong sample evidence to support the claim that the distribution does not match the NHTSA.
The largest discrepancy was in deaths by head injury. We expected 641 deaths , but the actual sample
data was 864. It is important to wear a helmet.
Hypothesis Test Act 14 answers
3. b)
P(business / male) = 112/357 = 0.314
P(business / female) = 89/335 = 0.266
Guess that gender and major are probably related (dependent)
Hypothesis Test Act 15 answers
5.
Ho: Distributions are same
Ha: Distributions are different (claim)
less than 3
3-5 months over 5 months Total
Intended 593 (574.15)
(0.62)
26 (36.14)
(2.85)
33 (41.71)
(1.82)
652
unintended 64 (73.09)
(1.13)
8 (4.6)
(2.51)
11 (5.31)
(6.1)
83
mistimed 169 (178.76)
(0.53)
18 (11.25)
(4.04)
16 (12.99)
(0.7)
203
Total 826 52 60 938
Chi-Square test:
Statistic DF Value P-value
Chi-square 4 20.301606 0.0004
Assumptions: Random. Expected value = 574, 36, 41.7, 73.1, 4.6 , 5.31 , 178.8 , 11.25 , 12.99 .
We did have one expected value that fell below 5. Meaning the data is not large enough to
handle this test.
Chi-squared test stat = 20.3 (significant)
P-value = 0.0004 (unlikely to happen by random chance)
Sentence: If the Ho is true (distributions are same) then there was a 0.0004 chance of getting
the sample data or more extreme.
Pvalue < sig level 0.05
Reject Ho
If this data had satisfied the assumptions then the conclusion would have been “there is
significant sample evidence to support the claim that the distributions are different between
the groups.” However this problem did not meet assumptions. Specifically the data set was
too small.
Hypothesis Test Activity 16 Answers
1.
Ho: Mu1 = mu2 = mu3
Ha: at least one is not equal (CLAIM)
F test stat = 5.627
Significant???
Original Data F value was in tail and looked significantly different than most of simulated
dots.
Pvalue = 0.013 was very small. Very unlikely that the groups could be the same and F
probably did not happen by random chance.
Pvalue < sig level 5%
Reject Ho
There is significant sample evidence to support the claim that the average amount of ants
will be different depending on the type of sandwich.
2.
Ho: Mu1 = mu2 = mu3 (CLAIM)
Ha: at least one is not equal
F test stat = 7.104
Significant???
Original Data F value was in tail and looked significantly different than most of simulated
dots.
Pvalue = 0.00067 was very small. Very unlikely that the groups could be the same and F
probably did not happen by random chance.
Pvalue < sig level 5%
Reject Ho
There is significant sample evidence to reject the claim that the average pulse rates are the
same for the 3 groups.
Hypothesis Test Activity 17 Answers
1.
Ho: Mu1 = mu2 = mu3
Ha: at least one is not equal (CLAIM)
Side by side boxplot showed that the sample means did look different. Bears later in the
year had a higher mean average. Also the boxplots showed that some groups had a lot
more spread than others. (might fail the similar variance assumption)
Summary statistics:
Column n Mean Variance
Apr-July Bear Weight 13 151.38462 504.58974
Aug-Sept Bear Weight 24 182.125 2305.6793
Oct-Nov Bear weight 17 228.11765 1662.1103
Assumptions:
Quantitative, all measuring weights of bears
Random
All of the data sets were less than 30. However all the data sets had an almost bell
shaped histogram. So the data does pass the 30 or nearly normal assumption.
Fail the similar variance assumption. One variance was more than 4 times as large
as one of the others.
F test stat = 13.55 (significant difference between the variance between and the variance
within.)
Variance between is 13.55 times greater than the variance within.
Degrees of freedom between = 3 groups – 1 = 2
Degrees of freedom within groups = (13-1) + (24-1) + (17-1) = 51
Pvalue < 0.0001 (If the groups were the same there was less than 0.0001 chance of getting
the sample data or more extreme by random chance) (Highly unlikely)
Pvalue < sig level
Reject Ho
If this data had met all the assumptions then the conclusion would be this:
There is significant sample evidence to support the claim that the average weight of bears is
different depending on the time of year they were measured.
Hypothesis Test Activity 19 Answers
1.
Ho: rho = 0 (no correlation)
Ha: rho > 0 (is positive correlation) - CLAIM
Assumptions
2 quantitative data sets
Random
Weak linear pattern. Passes
Fails outlier assumption. There are a few influential outliers
Fails nearly normal requirement. Histogram of residuals is skewed right
Histogram of residuals is centered close to zero
Residual plot shows even spread pattern so passes the homoscedasticity requirement.
r = 0.364
Pvalue = 0.0208
If the null is true and there is no correlation, then there was 0.0208 chance of getting the
sample data or more extreme by random chance.
Pvalue < sig level
Reject Ho
If the problem meets the assumptions, then there would be significant evidence to support the
claim that there is positive correlation between the height and weight of women.
Hyp Test Act 19 /#6
Ho: rho = 0 (no correlation) claim
Ha: rho not = 0 (is correlation)
Assumptions
2 quantitative data sets
Random
Moderate linear pattern. However, we see a nonlinear curved pattern in the data
indicating the line is probably not the best model for this data set.
Don’t see any influential outliers
Histogram of residuals is bell shaped
Histogram of residuals is centered close to zero
Residual plot shows a distinct nonlinear curve pattern. The left side is more spread out
than the right side, so it fails homoscedasticity
r = 0.719
Pvalue < 0.0001
If the null is true and there is no correlation, then there was less than 0.0001 chance of getting
the sample data or more extreme by random chance.
Reject Ho
If the problem meets the assumptions, then there would be significant evidence to reject the
claim that there is no correlation between the length and age of the bear.
Joe is right. The data fails the linear trend assumption and the homoscedasticity assumption.
Hypothesis Test Review Sheet #2 Answers
1.
a) Two population Proportion (percentages)
Categorical data
0 1 2
1 2
: p
: p (claim)A
H p
H p
or 0 1 2
1 2
: p 0
: p 0 (claim)A
H p
H p
Assumptions: Random, at least 10 success, at least 10 failures in both
data sets.
Z-test statistic : The number of standard errors that sample percent 1 is (above
or below) the sample percent 2
b) Two population mean (not related or matched pairs)
Quantitative data
2 Types: If two data sets not related = 2 sample t –test
If two data sets related = t –test (paired)
0 1 2
1 2
:
: (claim)A
H
H
or
0 1 2
1 2
: 0
: 0 (claim)A
H
H
or
0 : 0
: 0 (claim)
d
A d
H
H
Assumptions: Random quantitative data sets, not related or matched pair,
At least 30 or nearly normal in both data sets
T-test statistic : The number of standard errors that sample mean 1 is (above or
below) the sample mean 2
c) Chi squared goodness of fit test (checking the same percent in 3 or more groups)
Categorical data
H0: p1 = p2 = p3 = p4
Ha: at least one not =
Assumptions: Random categorical data, All Expected values at least 5
Chi Squared Test stat sentence? The sum of the averages of the squares of the
differences between the observed sample data and the expected values from
the null hypothesis.
d) Chi squared test for Independence
Categorical Data
Ho: The two categorical variables are independent (not related)
Ha: The two categorical variables are dependent (related)
Assumptions: Random categorical data, All Expected values at least 5
Chi Squared Test stat sentence? The sum of the averages of the squares of the
differences between the observed sample data and the expected values from
the null hypothesis.
e) Chi squared test of Homogeneity
Categorical Data
Ho: The distributions for the groups are the same
Ha: The distributions for the groups are different
Assumptions: Random categorical data, All Expected values at least 5
Chi Squared Test stat sentence? The sum of the averages of the squares of the
differences between the observed sample data and the expected values from
the null hypothesis.
f) ANOVA (Analysis of Variance)
quantitative data 3 or more groups
Ho: mu1 = mu2 = mu3 = mu4 = mu5 (claim)
Ha: at least one not =
Assumptions: Random quantitative data (same units between groups), Samples
sizes at least 30 or nearly normal, Groups are not related, No groups has
variance more than twice as large as any other groups variance
F test statistic sentence? The ratio of the variance between the groups to the
variance within the groups.
g) Correlation Hypothesis Test
Two different quantitative variables (ordered pair)
Ho: rho = 0 (no correlation)
Ha: rho not = 0 (is correlation)
Assumptions for Correlation Rho Test: Random ordered paired data , sample
sizes at least 30, Scatterplot shows a linear shape (No pattern in the scatterplot
or in the verses fit residual plot), Residuals are nearly normal, Residuals are
centered close to zero, No fan shape in the residual plot vs x value
(Homoscedasticity), No outliers that are overly influential
T-Test statistic: The number of standard errors that the slope of the regression
line is above or below zero.
2. P1: percent of babies exposed to cocaine that passed the test P2: percent of babies not exposed to cocaine that passed the test H0: P1 = p2 HA: P1 < p2 (claim) Assumptions: Passed the assumptions. The data was random, There was at least 10 successes and failures in both groups. ( x1 = 139 > 10 , n1-x1 = 51 > 10 , x2 = 153 > 10 , n2 – x2 = 33 > 10 ) Test Statistic: z = -2.12 . This means that the sample percent of cocaine babies is 2.12 standard deviations below the sample percent of babies not exposed to cocaine. There is a significant difference between the groups. P value = 0.0164. This means that if the null hypothesis is true and the percent of children exposed to cocaine is the same as the percent of children not exposed, then there was a 1.64% chance of getting the sample values or more extreme. It is very unlikely that this data would happen by random chance from equal populations. Reject H0. Conclusion: There is sufficient evidence to support the claim that the passage rates
for babies that were exposed to cocaine is lower than the passage rates for babies not exposed to cocaine on the test of object assembly. 3.
1
2
: Ave weight of male German Shepherds
: Ave weight of male Dobermans
The problem does meet the assumptions. Both data sets were random and despite the data
sets being too small (n1=20 and n2=14) , they were normally distributed.
The data sets are independent. There is no relationship between the weight of a random
German Shepherd and the weight of a random Doberman. (Unless they have the same owner
which was very unlikely.) We will perform a regular 2 pop T-test. (Not paired)
0 1 2
2
:
: (claim)A
H
H
Test Stat t = 0.558 This means that the sample mean weight of the German Shepherds in the
data is only 0.558 standard errors above the sample mean weight of the Dobermans. There
was not a significant difference between the sample means. They were very close.
P-value = 0.2906 . This means that if the null hypothesis is true and the mean average weight of German Shepherds and Dobermans are the same, then there was a 29.06% chance of getting the sample values or more extreme. This data could of happen by random chance from equal population means. Fail to Reject Ho. Conclusion: There is not sufficient sample evidence to support that the
average weight of male German Shepherds is more than the average weight of male Doberman
Pinchers.
4. Population 1: mom’s height
Population 2: daughters height
The problem does meet the assumptions to create a confidence interval. Both data sets were
random. Both data sets were too small (n1=20 and n2=20). To check if they were normally
distributed, we created histograms of each. The histogram for the mom was bell shaped and
the histogram for the daughters was more uniform but not radically skewed. So both data sets
meet the “nearly normal” assumption.
The data sets are definitely matched pairs. Mom’s and daughters are related to each other
genetically and will probably play a role in their heights.
0 : 0
: 0
d
A d
H
H
Test Stat t = -1.519 This means that on average, the heights of moms were 1.519 standard
errors below their daughters. There was a difference between moms and daughters, but it may
not be significant enough.
P-value = 0.0726 This means that if the null hypothesis is true and moms and daughters have
the same height, then there was only a 7.26% chance of getting this sample data or more
extreme. It was unlikely that this happened by random chance, however this data may not be
significant enough evidence.
Fail to Reject Ho. Conclusion: There is not sufficient sample evidence to support that the
mothers are shorter than their daughters on average.
5.
Ho: p1=p2=p3=p4=p5=p6 (claim)
Ha: at least one is not equal
Assumptions: All Expected values = 20 > 5. Data was also random. So passes all the
assumptions
Chi-squared test stat = 35.8 (Sig difference between the observed sample data and the
expected values from Ho)
Pvalue < 0.0001 Unlikely to happen by random chance.
P-value < sig level
Reject Ho
There is significant sample evidence to reject the claim that the various sports have the same
percentage of knee injuries.
6.
Ho: The distributions for the groups are the same (claim)
Ha: The distributions are different
Assumptions: All Expected values = 20 > 5. Data was also random. So passes all the
assumptions
Chi-squared test stat = 3.467 (Not Sig difference between the observed sample data and the
expected values from Ho)
Pvalue = 0.1767 Likely to happen by random chance.
P-value > sig level
Fail to Reject Ho
There is NOT significant sample evidence to reject the claim that the various animals have the
same distributions of rabies.
7.
Ho: Age and music are independent (not related)
Ha: Age and music are dependent (related) CLAIM
Assumptions: All Expected values = 20 > 5. Data was also random. So passes all the
assumptions
Chi-squared test stat = 25.635 (Sig difference between the observed sample data and the
expected values from Ho)
Pvalue = 0.0003 Unlikely to happen by random chance.
P-value < sig level
Reject Ho
There is significant sample evidence to support the claim that age and favorite type of music
are related.
8.
Ho: Rho = 0 (no correlation)
Ha: Rho > 0 (is positive correlation) CLAIM
Assumptions:
Two quantitative ordered pair data sets
Random
N =40 > 30
Histogram of residuals was nearly normal
Histogram centered close to zero
Residual plot showed an even spread pattern
One possible outlier. Do not think it is influential though because the r value was still strong
(close to 1)
Data did have a linear trend (r =0.8)
P-value < 0.0001 Unlikely to happen by random chance
Pvalue < sig level
Reject Ho
There is sig sample evidence to support the claim that there is positive correlation between the
weight and BMI of men.
9.
Ho: mu1 = mu2 = mu3
Ha: at least one is not equal (CLAIM)
Quantitative data
Random
n>30 or nearly normal
similar variances
Groups are supposed to be independent (fails this one)
F test stat = 4.7 (variance between groups is sig higher than the variance within groups)
Pvalue = 0.012 (unlikely to happen by random chance)
Reject Ho because P-value < sig level
If this problem had met assumptions, then we would have sig evidence to support the claim
that the average height of daughters, mom and grandmothers are different.