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 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. 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 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 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. 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
In: Math
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
In: Math
In: Math
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 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 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 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