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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


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.


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.


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.


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.


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.


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.


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


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.


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.


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)


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


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.


1.

Ho: Mu1 = mu2 = mu3


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.


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)





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





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

Chi-squared test stat = 3.467 (Not Sig difference between the observed sample data and the


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

Chi-squared test stat = 25.635 (Sig difference between the observed sample data and the


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


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.

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