Questions
Consider the following information about travelers on vacation: 40% check work email, 30% use a cell...

Consider the following information about travelers on vacation: 40% check work email, 30% use a cell phone to stay connected to work, 25% bring a laptop with them, 16% both check work email and use a cell phone to stay connected, and 44% neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop. (a) What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected? Incorrect: Your answer is incorrect. (b) What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? (c) If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected? (Round your answer to four decimal places.)

In: Math

This problem is going to use the data set in R called "ChickWeight" that has 4...

This problem is going to use the data set in R called "ChickWeight" that has 4 variables, as described below.

ChickWeight:
A data frame with 578 observations on 4 variables.
1) weight: a numeric vector giving the body weight of the chick (gm).
2) Time: a numeric vector giving the number of days since birth when the measurement was made.
3) Chick: an ordered factor with levels 18 < ... < 48 giving a unique identifier for the chick. The ordering of the levels groups chicks on the same diet together and orders them according to their final weight (lightest to heaviest) within diet.
4) Diet: a factor with levels 1, ..., 4 indicating which experimental diet the chick received.

Using a significance level of 0.05, is there evidence to support that the weight can be determined by the Time, Diet, and the interaction between the two? (Appears as Time:Diet in RStudio)

Fill in the R code below.

dat.aov = aov( ~ factor( ) *  , data= )
summary( )

Fill in the ANOVA table below.
Type the values into the table EXACTLY as they appear in your output in RStudio.

df SS MS F Pr(>F)
factor(Time) 2e-16
Diet 2e-16
factor(Time):Diet 0.00017
Residuals

Is there evidence to support a significant interaction between Time and Diet?
1. ?0:H0: No AB interaction vs ??:Ha: Factors A and B interact
2. ?=0.05α=0.05
3. F =  
4. ??Fα =  
5. Conclusion:
Reject H0
Fail to reject H0
Interpretation:
There is sufficient evidence to support that the interaction between Time and Diet is significant.
There is not sufficient evidence to support that the interaction between Time and Diet is significant.

In: Math

A defective car has a probability of 2/3 upon turning on the ignition in each attempt....

A defective car has a probability of 2/3 upon turning on the ignition in each attempt. Assume attempts

are independent.

• (a) What is the probability that exactly 3 attempts are needed until the car starts?

• (b) What is the probability that 3 or 4 attempts are needed?

• (c) What is the probability of success in 4 or more trials?

In: Math

This problem is going to use the data set in R called "ChickWeight" that has 4...

This problem is going to use the data set in R called "ChickWeight" that has 4 variables, as described below.

ChickWeight:
A data frame with 578 observations on 4 variables.
1) weight: a numeric vector giving the body weight of the chick (gm).
2) Time: a numeric vector giving the number of days since birth when the measurement was made.
3) Chick: an ordered factor with levels 18 < ... < 48 giving a unique identifier for the chick. The ordering of the levels groups chicks on the same diet together and orders them according to their final weight (lightest to heaviest) within diet.
4) Diet: a factor with levels 1, ..., 4 indicating which experimental diet the chick received.

Using a significance level of 0.05, is there evidence to support that the weight can be determined by the Time (treatment) and Diet (block)?

Fill in the R code below.

dat.aov=aov( ~ factor( ) +  ,data= )
summary( )

Fill in the ANOVA table below.
Type the values into the table EXACTLY as they appear in your output in R.

df SS MS F Pr(>F)
factor(Time) 2e-16
Diet 2e-16
Residuals

Is there evidence to support that the treatment variable Time is significant?
1. ?0:?1=?2=...=?12H0:μ1=μ2=...=μ12 vs ??:????Ha:ALOI
2. ?=0.01α=0.01
3. F =  
4. ??Fα =  
5. Conclusion:
Reject H0
Fail to reject H0
Interpretation:
There is sufficient evidence to support that the variable Time is significant.
There is not sufficient evidence to support that the variable Time is significant.

Is there evidence to support that the block variable Diet is significant?
1. ?0:H0: No block effect vs ??:Ha: There is a block effect
2. ?=0.01α=0.01
3. F =  
4. ??Fα =  
5. Conclusion:
Reject H0
Fail to reject H0
Interpretation:
There is sufficient evidence to support that the variable Diet is significant.
There is not sufficient evidence to support that the variable Diet is significant.

In: Math

There are 6 purple balls, 5 blue balls, and 3 green balls in a box. 5...

There are 6 purple balls, 5 blue balls, and 3 green balls in a box. 5 balls were randomly chosen (without replacing them). Find the probability that

(a) Exactly 3 blue balls were chosen.

(b) 2 purple balls, 1 blue ball, and 2 green balls were chosen.

In: Math

1. STATISTICAL RESULTS: Report and statistically interpret the results with an appropriate tabular format and text....

1. STATISTICAL RESULTS: Report and statistically interpret the results with an appropriate tabular format and text. See the Output B. [6 pt]

2. DISCUSSION: Discuss what the results imply. [2 pt]

Output A. Descriptive statistics

Statistics

Exam Performance (%)

Exam Anxiety

N

Valid

103

103

Missing

0

0

Mean

56.57

74.3437

Median

60.00

79.0440

Std. Deviation

25.941

17.18186

Minimum

2

1.00

Maximum

100

100.00

Output B. Simple Linear Regression Results

Correlations

Exam Performance (%)

Exam Anxiety

Pearson Correlation

Exam Performance (%)

1.000

-.441

Exam Anxiety

-.441

1.000

Sig. (1-tailed)

Exam Performance (%)

.

.000

Exam Anxiety

.000

.

N

Exam Performance (%)

103

103

Exam Anxiety

103

103

Model Summary

Model

R

R Square

Adjusted R Square

Std. Error of the Estimate

1

.441a

.194

.186

23.397

a. Predictors: (Constant), Exam Anxiety

Coefficientsa

Model

Unstandardized Coefficients

Standardized Coefficients

t

Sig.

95.0% Confidence Interval for B

B

Std. Error

Beta

Lower Bound

Upper Bound

1

(Constant)

106.071

10.285

10.313

.000

85.667

126.474

Exam Anxiety

-.666

.135

-.441

-4.938

.000

-.933

-.398

a. Dependent Variable: Exam Performance (%)

In: Math

Consider a binomial experiment with n=5 and p=0.20 What is Var(x)?

Consider a binomial experiment with n=5 and p=0.20 What is Var(x)?

In: Math

Figure shows a contact lens table tht contains information about contact lens prescriptions (hard, soft and...

Figure shows a contact lens table tht contains information about contact lens prescriptions (hard, soft and no contact lens) From the table derive quantitative association rules by mapping tables to Boolean association rules.

ID

Age

Spectacle

Astigmatic

Tear Production

Contact lens

1

21

Myope

No

Reduced

None

2

24

Myope

No

Normal

Soft

3

20

Myope

Yes

Reduced

None

4

26

Myope

Yes

Normal

Hard

5

27

Hypermyope

No

Reduced

None

6

22

Hypermyope

No

Normal

Soft

7

28

Hypermyope

Yes

Reduced

None

8

27

Hypermyope

Yes

Normal

Hard

9

38

Myope

No

Reduced

None

10

32

Myope

No

Normal

Soft

11

36

Myope

Yes

Reduced

None

12

37

Myope

Yes

Normal

Hard

13

33

Hypermyope

No

Reduced

None

14

32

Hypermyope

No

Normal

Soft

15

39

Hypermyope

Yes

Reduced

None

16

34

Hypermyope

Yes

Normal

None

17

52

Myope

No

Reduced

None

18

51

Myope

No

Normal

None

19

50

Myope

Yes

Reduced

None

20

54

Myope

Yes

Normal

Hard

21

52

Hypermyope

No

Reduced

None

22

55

Hypermyope

No

Normal

Soft

23

58

Hypermyope

Yes

Reduced

None

24

54

Hypermyope

Yes

Normal

None

In: Math

Prove Var(|X|) <= Var(X)

Prove Var(|X|) <= Var(X)

In: Math

For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015....

  1. For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015. Assume the standard deviation is $3,540 and that debt amounts are normally distributed.
  1. What is the probability that the debt for a borrower with good credit is more than $18,000?
  2. What is the probability that the debt for a borrower with good credit is less than $10,000?
  3. What is the probability that the debt for a borrower with good credit is between $12,000 and $18,000?
  4. What is the probability that the debt for a borrower with good credit is no more than $14,000?

In: Math

For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015....

  1. For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015. Assume the standard deviation is $3,540 and that debt amounts are normally distributed.
  1. What is the probability that the debt for a borrower with good credit is more than $18,000?
  2. What is the probability that the debt for a borrower with good credit is less than $10,000?
  3. What is the probability that the debt for a borrower with good credit is between $12,000 and $18,000?
  4. What is the probability that the debt for a borrower with good credit is no more than $14,000?

In: Math

Regression analysis is often used to provide a means to express the relationship between one or...

Regression analysis is often used to provide a means to express the relationship between one or more input variables and a result. It is easy to plot in Excel (“add trendline”) so is found frequently in business presentations. Your company has made a model with 10 different factors measured from past years’ and states based upon the model, the company expects to make a 23 million dollar profit next year. Discuss possible concerns with banking on the 23 million dollar prediction, including concepts of

a. correlation

b. causation

c. single point prediction (that is, just plugging the values into the equation and saying that single number is the prediction of future performance)

d. confidence interval.  

e. prediction interval.

In: Math

You would like to study the height of students at your university. Suppose the average for...

You would like to study the height of students at your university. Suppose the average for all university students is 68 inches with a SD of 20 inches, and that you take a sample of 17 students from your university.

a) What is the probability that the sample has a mean of 64 or more inches? probability = .204793 (is this answer correct or no? and I need help with part b too.)

b) What is the probability that the sample has a mean between 63 and 68 inches?

In: Math

Safeco company produces two types of​ chainsaws: The Safecut and the Safecut Deluxe. The Safecut model...

Safeco company produces two types of​ chainsaws: The Safecut and the Safecut Deluxe. The Safecut model requires 2 hours to assemble and 1 hour to​ paint, and the Deluxe model requires 4 hours to assemble and one half
hour to paint. The daily maximum number of hours available for assembly is 32​, and the daily maximum number of hours available for painting is 10. If the profit is ​$26 per unit on the Safecut model and ​$40 per unit on the Deluxe​ model, how many units of each type will maximize the daily profit and what will that profit​ be?

In: Math

You have 120 mice lacking insulin receptors in their brain tissue. On average 15.2% of these...

You have 120 mice lacking insulin receptors in their brain tissue. On average 15.2% of these types of mice will die within a month. What is the probability at least 100 mice live until next month? (use binomial approximation of normal)

In: Math