name statistics chapter 20: review a -...

Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 20-19

Name______________________________________ Statistics Chapter 20: Review A 1. The following is a confidence interval for the difference in proportions of satisfied theater

goers between those who were given a $20 discount on their tickets and those who paid full price:

95% CI for discount nodiscountp p− = (0.05, 0.14)

a. Explain what this interval means in this context. b. What assumptions or conditions needed to be met to create this confidence interval? c. Assuming this is a valid confidence interval, is the difference here statistically significant?

Explain.

2. For each example, rewrite the given statistics using appropriate symbols and detail, find the

standard error of the differences for these examples, and write the hypotheses.

a. 2000 adults were surveyed regarding their support of a new law to increase federal funding of childcare. 512 of the 832 men and 715 of the women supported the law. Is there a difference between male and female opinion?

Given information using appropriate symbols: Standard error of the difference in proportions of male and female support: Hypotheses (in words and symbols):


b. When these 2000 adults were surveyed they were also asked to rate the importance of the new law to increase federal funding of childcare. On a scale of 0 to 5, where 0 represents not important at all and 5 represents of highest importance, the average male rating (from 832 males) was 3.1 with a standard deviation of 1.2 and the average female rating (of 1168 females) was 3.9 with a standard deviation of 1.9. Is there a statistical difference between the mean ratings of males and females?

Given information written using appropriate symbols: Standard error of the difference in means of male and female ratings: Hypotheses (in words and symbols):

3. Some Canadians feel that a higher percentage of Canadians than Britons read books. A recent

Gallup Poll reported that 86% of 1004 randomly sampled Canadians read at least one book in the past year, compared to 81% of 1009 randomly sampled Britons. Do these results confirm a higher reading rate in Canada?

a. Hypotheses b. Model (including assumptions and conditions)

c. Mechanics (continue on next page)


d. Conclusion (in context) 4. A random sample of 13 men and 19 women in a college class reported their grade point

averages (GPAs). Here are histograms displaying the data:

MGPA

Fre

quen

cy

4.03.53.02.52.01.5

6

5

4

3

2

1

0

Histogram of MGPA

WGPA

Fre

quen

cy

4.003.753.503.253.002.752.50

5

4

3

2

1

0

Histogram of WGPA

Summary statistics for these data are:

y s Men 2.898 0.583Women 3.330 0.395

A woman in the class says that she believes that college women tend to have higher GPAs than do college men. a. Justify that a t-model is appropriate, and then find a 95% confidence interval for the difference

in mean GPAs between men and women b. Interpret your confidence interval in this context. c. Do these data support her claim? Explain.


Statistics Chapter 20: Review A – KEY 1. The following is a confidence interval for the difference in proportions of satisfied theater

goers between those who were given a $20 discount on their tickets and those who paid full price:

95% CI for discount nodiscountp p− = (0.05, 0.14)

a. Explain what this interval means in this context.

I am 95% confident that the proportion of theater goers receiving a discount who are satisfied with their theater experience is between 5% and 14% higher than those theater goers not receiving a discount.

b. What assumptions or conditions needed to be met to create this confidence interval?

The independence assumption, random sampling condition, 10% condition, independent groups assumption, and success/failure condition must be met.

c. Assuming this is a valid confidence interval, is the difference here statistically significant?

Explain.

Yes. I am 95% confident that there is a difference because zero is not in the interval. 2. For each example, rewrite the given statistics using appropriate symbols and detail, find the

standard error of the differences for these examples, and write the hypotheses.

a. 2000 adults were surveyed regarding their support of a new law to increase federal funding of childcare. 512 of the 832 men and 715 of the women supported the law. Is there a difference between male and female opinion?

Given information using appropriate symbols:

ˆ832 512 0.6154M M Mn x p= = = ˆ1168 715 0.6122F F Fn x p= = =

Standard error of the difference in proportions of male and female support:

ˆ ˆ ˆ ˆˆ ˆ( ) M M F F

M FM F

p q p qSE p p

n n− = + (0.6154)(0.3846) (0.6122)(0.3878)

0.0221832 1168

= + ≈

Hypotheses (in words and symbols):

The null hypothesis is that the proportion of males surveyed who support federally funded childcare is no different than the proportion of females who support federally funded childcare. The alternative hypothesis is that the proportion of women in support of federally funded childcare is different than the proportion of men. In symbols: 0 : 0 : 0M F A M FH p p H p p− = − < .

jbowman

Pencil


b. When these 2000 adults were surveyed they were also asked to rate the importance of the new law to increase federal funding of childcare. On a scale of 0 to 5, where 0 represents not important at all and 5 represents of highest importance, the average male rating (from 832 males) was 3.1 with a standard deviation of 1.2 and the average female rating (of 1168 females) was 3.9 with a standard deviation of 1.9. Is there a statistical difference between the mean ratings of males and females?

Given information written using appropriate symbols:

832 3.1 1.2M M Mn x s= = = 1168 3.9 1.9F F Fn x s= = =

Standard error of the difference in means of male and female ratings:

2 2 2 21.2 1.9( ) 0.069

832 1168M F

M FM F

s sSE y y

n n

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− = + = + ≈

⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Hypotheses (in words and symbols):

The null hypothesis is that mean importance men place on increasing funding for childcare is no different than that of women. The alternative hypothesis is that the mean rating of importance that women place on increased funding for childcare is different than that of men. In symbols: 0 : 0 : 0M F A M FH Hµ µ µ µ− = − < .

3. Some Canadians feel that a higher percentage of Canadians than Britons read books. A recent Gallup Poll reported that 86% of 1004 randomly sampled Canadians read at least one book in the past year, compared to 81% of 1009 randomly sampled Britons. Do these results confirm a higher reading rate in Canada?

a. Hypotheses

0H : 0B Cp p− = , AH : 0B Cp p− < , where B = Britons and C = Canadians.

b. Model (including assumptions and conditions)

Randomization Condition: The Britons and Canadians were randomly sampled by Gallup. 10% Condition: The number of Britons and Canadians is greater than 10,090 (10 x 1009) and 10,040 (10 x 1004), respectively.

Independent samples condition: The two groups are clearly independent of each other. Success/Failure Condition: Of the Britons, approximately 817 read at least one book and 192 did not; of the Canadians, approximately 863 read at least one book and 141 did not. The observed number of both successes and failures in both groups is larger than 10. Because the conditions are satisfied, we can model the sampling distribution of the difference in proportions with a Normal model. We can perform a two-proportion z-test.

c. Mechanics 1009Bn = , ˆ 0.81Bp = , 1004Cn = , ˆ 0.86Cp = .

( ) ( )( ) ( )( )0.81 0.19 0.86 0.14ˆ ˆˆ ˆˆ ˆ 0.0165

1009 1004C CB B

B CB C

p qp qSE p p

n n− = + = + =

( )ˆ ˆ 0 0.05 0

3.03ˆ ˆ( ) 0.0165

B C

B C

p pz

SE p p

− − − −= = = −−

, so the -value ( 3.03) 0.0012P P z= < − =

(continue on next page)

jbowman

Pencil


d. Conclusion (in context)

The P-value of 0.0012 is low, so we reject the null hypothesis. There is strong evidence that the percentage of Britons who read at least one book in the past year is less than the percentage of Canadians who read at lease one book in the past year.

4. A random sample of 13 men and 19 women in a college class reported their grade point averages (GPAs). Here are histograms displaying the data:

MGPA

Fre

quen

cy

4.03.53.02.52.01.5

6

5

4

3

2

1

0

Histogram of MGPA

WGPA

Fre

quen

cy4.003.753.503.253.002.752.50

5

4

3

2

1

0

Histogram of WGPA

Summary statistics for these data are:

y s Men 2.898 0.583Women 3.330 0.395

A woman in the class says that she believes that college women tend to have higher GPAs than do college men. 1. Justify that a t-model is appropriate, and then find a 95% confidence interval for the difference

in mean GPAs between men and women

Independent group assumption: Men and women are independent groups. Randomization condition: We are told that these are random samples of men and women in the class.

10% condition: These samples represent less than 10% of all possible men and less than 10% of all possible women taking the course. Nearly Normal condition: The samples are pretty small, but the histograms given don’t seem to deviate too badly from being unimodal and symmetric.

Because the conditions are satisfied, it is appropriate to proceed with a two-sample t-interval.

2 2 2 20.395 0.583

( ) 0.185419 13

W MW M

W M

s sSE y y

n n

⎛ ⎞ ⎛ ⎞− = + = + ≈⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

( )2 2

* *19.44 19.44

0.395 0.583( ) (3.330 2.898) (0.0446,0.8194)

19 13W M W MCI y y t SE y y t⎛ ⎞⎜ ⎟= − ± × − = − ± + =⎜ ⎟⎝ ⎠

(Note: The number of degrees of freedom and t* come from the calculator’s 2-SampTInt function)

2. Interpret your confidence interval in this context.

We are 95% confident that the mean GPA for college women is 0.045 to 0.819 points higher than the mean GPA for college men.

3. Do these data support her claim? Explain.

Yes. Zero is not in the interval. This means that this sample provides evidence that college women have a higher mean GPA than college men.


Name______________________________________ Statistics Chapter 20: Review B

1. A National Cancer Institute study published in 1991 examined the incidence of cancer in dogs. Of 827 dogs whose owners used the weed killer 2-4-D on their lawns or gardens, 473 were found to have cancer. Only 19 of the 130 dogs that had not been exposed to this herbicide had cancer.

a. Find the standard error for the difference in proportions of cancer between the exposed and unexposed groups.

b. Address the assumptions and conditions to construct a 95% confidence interval for

the difference in pets’ cancer risk.

c. Find the 95% confidence interval and interpret your interval in this context.

2. Write the hypotheses for a testing whether the proportion of high school football games won by teams whose home fields are artificial turf are better than teams whose home fields are natural grass.


3. An agronomist hopes that a new fertilizer she has developed will enable grape growers to increase the yield of each grapevine. To test this fertilizer she applied it to 44 vines and used the traditional growing strategies on 47 other vines. The fertilized vines produced a mean of 53.4 pounds of grapes with standard deviation 3.7 pounds, while the unfertilized vines yielded an average of 52.1 pounds with standard deviation 3.4 pounds of grapes. Do these experimental results confirm the agronomist’s expectations?

a. Find the standard error for the difference in means of grapes produced between the fertilized and unfertilized groups.

b. Hypotheses

c. Model (including assumptions and conditions)

d. Mechanics

e. Conclusion


4. Some of the cigarettes sold in the US claim to be “low tar”. How much less tar would a smoker get by smoking low tar brands instead of regular cigarettes? Samples of 15 brands of each type were randomly chosen from the 1206 varieties (no kidding) that are marketed. Their tar contents (mg/cig) are listed in the table below.

Type Milligrams of tar per cigarette Regular 18 10 14 15 15 12 17 11 14 17 12 14 15 15 12 Low Tar 9 5 10 4 8 9 9 3 7 12 6 10 8 11 8

a. Check assumptions and conditions to use these data for inference.

b. Find a 95 % confidence interval for the difference between regular and low tar brands.

c. Interpret your interval in the context of the problem.

d. Explain to someone who doesn’t understand much about statistics that why the difference in tar between regular and low tar brands is considered to be “statistically significant”.


Statistics Chapter 20: Review B – KEY 1. A National Cancer Institute study published in 1991 examined the incidence of cancer in dogs.

Of 827 dogs whose owners used the weed killer 2-4-D on their lawns or gardens, 473 were found to have cancer. Only 19 of the 130 dogs that had not been exposed to this herbicide had cancer.

a. Find the standard error for the difference in proportions of cancer between the exposed and unexposed groups.

ˆ ˆˆ ˆˆ ˆ( ) U UE E

E UE U

p qp qSE p p

n n− = +

473 354 19 111827 827 130 130

0.0354827 130

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠= + ≈

b. Address the assumptions and conditions to construct a 95% confidence interval for the difference in pets’ cancer risk.

Randomization Condition: The dogs were not chosen randomly. 10% Condition: The number of dogs exposed to the weed killer and those not exposed to the weed killer in these samples is certainly less than 10% of all dogs. Independent samples condition: The two samples should be independent as long as not all the dogs came from the same community where they could be exposed to (or not) to the weed killer. Success/Failure Condition: 473 and 19 are both larger than 10.

c. Find the 95% confidence interval and interpret your interval in this context.

( ) ( )* 473 19ˆ ˆ ˆ ˆ( ) 2.000 0.0354 (0.355,0.497)

827 130E U E Up p z SE p p⎛ ⎞− ± × − = − ± =⎜ ⎟⎝ ⎠

I am 95% confident that the cancer rate for dogs exposed to weed killer was between 35.5% to 49.7% higher than the cancer rate of dogs not exposed to the weed killer.

2. Write the hypotheses for a testing whether the proportion of high school football games won by teams whose home fields are artificial turf are better than teams whose home fields are natural grass.

The null hypothesis is that the proportion of games won on artificial turf is no different than the proportion of games won on natural grass. The alternative hypothesis is that the proportion of won on artificial turf is higher than the proportion of games won on natural grass. In symbols: 0 : 0 : 0ARTIFICIAL NATURAL A ARTIFICIAL NATURALH p p H p p− = − > .


3. An agronomist hopes that a new fertilizer she has developed will enable grape growers to increase the yield of each grapevine. To test this fertilizer she applied it to 44 vines and used the traditional growing strategies on 47 other vines. The fertilized vines produced a mean of 53.4 pounds of grapes with standard deviation 3.7 pounds, while the unfertilized vines yielded an average of 52.1 pounds with standard deviation 3.4 pounds of grapes. Do these experimental results confirm the agronomist’s expectations?

a. Find the standard error for the difference in means of grapes produced between the fertilized and unfertilized groups.

2 2 2 23.7 3.4( ) 0.7464

44 47F T

F TF T

s sSE y y

n n

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− = + = + ≈⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

b. Hypotheses

The null hypothesis is that mean yield of the vines that received the fertilizer treatment is 5 pounds greater than the mean yield of the vines that received the traditional growing strategies. The alternative hypothesis is that the mean yield of the fertilized vines is 5 pounds higher the mean yield of those not receiving the fertilizer. In symbols:

0 : 0 : 0F T A F TH Hµ µ µ µ− = − > , where F = new fertilizer and T = traditional strategies.

c. Model (including assumptions and conditions)

Independent group assumption: The two groups may not be independent (no random assignment). Randomization condition: We are not told that the treatments were randomly assigned. 10% condition: These samples most likely represent less than 10% of all vines at this vineyard. Nearly Normal condition: The samples are rather large, so the CLT will guarantee a t-model for the sampling model, regardless of the shape of the populations. Because the conditions are satisfied, it is appropriate to proceed with a two-sample t-interval.

d. Mechanics 44Fn = 53.4Fy = 3.7Fs = 47Tn = 52.1Ty = 3.4Ts =

6.3F Ty y− = 87.02df = (from technology)

2 2 2 2

53.4 52.1 1.743.7 3.444 47

F T

F T

F T

y yt

s sn n

=− −= ≈

++

87.02( 1.74) 0.043P P t= > =

e. Conclusion

Since the P-value is low, we reject the null hypothesis. There is evidence to suggest that mean yield of vines with the new fertilizer is more than 5 pounds greater than the mean yield of vines with the traditional fertilizer.


4. Some of the cigarettes sold in the US claim to be “low tar”. How much less tar would a smoker get by smoking low tar brands instead of regular cigarettes? Samples of 15 brands of each type were randomly chosen from the 1206 varieties (no kidding) that are marketed. Their tar contents (mg/cig) are listed in the table below.

Type Milligrams of tar per cigarette Regular 18 10 14 15 15 12 17 11 14 17 12 14 15 15 12 Low Tar 9 5 10 4 8 9 9 3 7 12 6 10 8 11 8

a. Check assumptions and conditions to use these data for inference.

Independent group assumption: Regular and low tar are definitely independent groups. Randomization condition: We are told that these are random samples of cigarettes. 10% condition: These samples do represent less than 10% of the possible 1206 possible regular and low tar cigarettes in the United States. Nearly Normal condition: The histograms appear to be fairly unimodal and symmetric.

Because the conditions are satisfied, it is appropriate to proceed with a two-sample t-interval.

b. Find a 95 % confidence interval for the difference between regular and low tar brands.

2 2 2 22.5486 2.3135

( ) 0.888715 15

LT RLT R

LT R

s sSE y y

n n

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− = + = + ≈⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

( )2 2

* *27.74 27.74

2.5486 2.3135( ) (7.9333 14.0667) ( 7.955, 4.312)

15 15LT R LT Ry y t SE y y t⎛ ⎞⎜ ⎟− ± × − = − ± + = − −⎜ ⎟⎝ ⎠

(Note: The number of degrees of freedom and t* come from the calculator’s 2-SampTInt function)

c. Interpret your interval in the context of the problem.

I am 95% confident that the mean tar content in low tar cigarettes is 4.312 to 7.955 mg lower than the mean tar content in regular cigarettes.

d. Explain to someone who doesn’t understand much about statistics that why the difference in tar between regular and low tar brands is considered to be “statistically significant”.

I am very confident that there is a mean difference in tar content between the two types of cigarettes because if there were no evidence of a difference, then zero would fall in my interval. Because zero is not in my interval, I am quite confident that there is a mean difference in tar content. This makes my results statistically significant.

19171513119

4

3

2

1

0

Tar (mg)

Fre

quen

cy

Regular Cigarettes

12108642

6

5

4

3

2

1

0

Tar (mg)

Fre

quen

cyLow-Tar


Name______________________________________ Statistics Chapter 20: Review C

1. Write the hypotheses for the following tests.

a. Do high school football teams whose home fields are artificial turf win more than teams than teams whose home fields are natural grass?

b. Do people who regularly eat a high protein breakfast eat fewer calories during the day than those who eat regularly eat a high carbohydrate breakfast?

2. Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects with allergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and the current Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reported drowsiness, and 22 of the 80 patients with Brand C reported drowsiness.

a. Compute a 95% confidence interval for the difference in proportions of subjects reporting drowsiness.

b. Does the interval in question a provide evidence that the side effect of drowsiness is different with the new medication? Explain.


c. If, instead of creating a confidence interval, you had conducted a hypothesis test, would you have rejected the null hypothesis? Explain.

3. Every year favorite songs compete to be on a Top 200 list based upon sales and rankings by the experts in the music industry. These songs have many characteristics, such as song length and beats per minute, which vary from category to category in the music industry. A disc jockey wondered if the number of beats per minute in songs classified as dance music were lower than the beats per minute in the songs that are ranked on a Top 200 list from 2001. A random sample of songs from each group was selected and the beats per minute are listed in the chart at the right. Does this sample indicate that songs classified as dance music have lower beats per minute than the songs ranked on a Top 200 list?

a. Test an appropriate hypothesis and state your conclusion.

b. Create and interpret a 90% confidence interval.

Beats per Minute

Dance Songs Top 200 Songs121 119 122 120 122 121 121 118 117 122 121 121 120 119 122 123 120 119 121 118 121 118 119 120 118 120 120 124 120 123 119 117 118


Statistics Chapter 20: Review C – KEY

1. Write the hypotheses for the following tests.

a. Do high school football teams whose home fields are artificial turf win more than teams than teams whose home fields are natural grass?

The null hypothesis is that the proportion of games won on artificial turf is no different than the proportion of games won on natural grass. The alternative hypothesis is that the proportion of won on artificial turf is higher than the proportion of games won on natural grass. In symbols: 0 : 0 : 0ARTIFICIAL NATURAL A ARTIFICIAL NATURALH p p H p p− = − > .

b. Do people who regularly eat a high protein breakfast eat fewer calories during the day than those who eat regularly eat a high carbohydrate breakfast?

The null hypothesis is that the mean number of calories consumed during the day for those who eat a high protein breakfast is no different than the mean number of calories consumed during the day for those who eat a high carbohydrate breakfast. The alternative hypothesis is that the that the mean number of calories consumed during the day for those who eat a high protein breakfast is less than the mean number of calories consumed during the day for those who eat a high carbohydrate breakfast. In symbols: 0 : 0 : 0protein carbohydrate A protein carbohydrateH Hµ µ µ µ− = − < .

2. Researchers conduct a study to test a potential side effect of a new allergy medication. A random sample of 160 subjects with allergies was selected for the study. The new “improved” Brand I medication was randomly assigned to 80 subjects, and the current Brand C medication was randomly assigned to the other 80 subjects. 14 of the 80 patients with Brand I reported drowsiness, and 22 of the 80 patients with Brand C reported drowsiness.

a. Compute a 95% confidence interval for the difference in proportions of subjects reporting drowsiness.

Model: Randomization Condition: The treatments were randomly assigned to subjects.

10% Condition: The subjects were randomly selected. We assume it is from a large populations of allergy sufferers. Independent samples condition: The two groups are independent of each other because the treatments were assigned at random. Success/Failure Condition: For Brand C, 22 were drowsy and 58 were not. For Brand I, 14 were drowsy and 66 were not. The observed number of both successes and failures in both groups is larger than 10. Because the conditions are satisfied, we can model the sampling distribution of the difference in proportions with a Normal model.

Mechanics: 80In = , ˆ 0.175Ip = , 80Cn = , ˆ 0.275Cp =

( )( ) ( )( )0.275 0.725 0.175 0.825ˆ ˆ ˆ ˆ

ˆ ˆ( ) 0.06680 80

C C I IC I

C I

p q p qSE p p

n n− = + = + ≈

( )1 2 1 2

ˆ ˆ ˆ ˆ2 ( ) 0.10 2 0.066 ( 0.032,0.232)p p SE p p− ± − = ± = −i i

b. Does the interval in question a provide evidence that the side effect of drowsiness is different with the new medication? Explain.

There is not sufficient evidence because 0 is contained in the interval. There may be no difference in the proportion of drowsiness reported.


c. If, instead of creating a confidence interval, you had conducted a hypothesis test, would you have rejected the null hypothesis? Explain.

A hypothesis test at a .05 level should reach the same conclusion if a two-sided test is used. If the alternate hypothesis were “less drowsy”, the significance level would need to be changed, or a different conclusion could occur.

3. Every year favorite songs compete to be on a Top 200 list based upon sales and rankings by the experts in the music industry. These songs have many characteristics, such as song length and beats per minute, which vary from category to category in the music industry. A disc jockey wondered if the number of beats per minute in songs classified as dance music were lower than the beats per minute in the songs that are ranked on a Top 200 list from 2001. A random sample of songs from each group was selected and the beats per minute are listed in the chart at the right. Does this sample indicate that songs classified as dance music have lower beats per minute than the songs ranked on a Top 200 list?

a. Test an appropriate hypothesis and state your conclusion.

Hypotheses: Let: mean of dance songs = Dµ and mean of Top 200 songs = Tµ .

Then, 0H : 0T Dµ µ− = AH : 0T Dµ µ− >

Model: Independent group assumption: Random samples of dance songs

and Top 200 songs were taken. These samples were taken from song lists with separate categories.

Randomization condition: Samples were taken at random. 10% condition: The sample represents less than 10% of all possible

songs classified as dance and less than 10% of songs who made the Top 200 songs in 2001.

Nearly Normal condition: The histograms show that both samples are unimodal and roughly symmetric. Because the conditions are satisfied, it is appropriate to model the sampling distribution of the difference in the means with a Student’s t-model. We will perform a two-sample t-test. Mechanics:

18Dn = 119.7Dy = 1.74Ds = 15Tn = 120.6Ty = 1.72Ts =

0.9T Dy y− =

( )2 2 2 2

0 0.9 01.449

1.74 1.7218 15

T D

D T

D T

y yt

s s

n n

− − −= = =++

( 1.449) 0.079P P t= > =

Conclusion: Since the P-value of 0.079 is fairly high, we fail to reject the null hypothesis. There is little evidence of a difference between the mean beats per minute of dance songs and the mean beats per minute of Top 200 songs.

b. Create and interpret a 90% confidence interval. The conditions have been met, so we can create a two-sample t-interval for difference in means, with 90% confidence.

( )2 2

* *30.06 30.05

1.74 1.72( ) (119.7 120.6) ( 1.906,0.1500)

18 15D T D Ty y t SE y y t⎛ ⎞

− ± × − = − ± + = −⎜ ⎟⎝ ⎠

I am 90% confident that the mean beats per minute for a dance song is between 1.9 beats below to 0.1 beats higher than the mean beats per minute of a Top 200 song.

jbowman

Pencil

name statistics chapter 20: review a -...

Documents