I need help regarding data mining using R-STUDIO.

it is urgent please send answer within deadline,. I need it now please check and send it to me they are about 40 questions

Question 12
Use the R output below to select the correct statement.
Call:
Im(formula = Income ~ Cards + Rating + Married, data = Credit)
Residuals:
Min
1Q Median
-39.567 -17.549 -0.192
Max
3Q
14.441
78.956
Coefficients:
Estimate Std. Error t value Pr(>It)
(Intercept) -14.668246 3.707401 -3.956 9.01e-05 ***
Cards
-1.555163 0.786828 -1.976 0.0488 *
Rating
0.180951 0.006978 25.932
< 2e-16 *** Married Yes 0.424236 2.210288 0.192 0.8479 Signif. codes: "***0.001 (**) 0.01 **) 0.05.' 0.11 Residual standard error: 21.52 on 396 degrees of freedom Multiple R-squared: 0.63, Adjusted R-squared: 0.6272 F-statistic: 224.7 on 3 and 396 DF, p-value: < 2.2e-16 Married is associated with Income (p-value < 0.0001), given that we are using Cards and Rating to predict Income Rating is associated with Income (p-value < 0.0001), given that we are using Cards and Married to predict Income We have strong statistical evidence that none of the predictor variables are related to Income (F = 224.7, p-value < 0.0001) The predictor variables, Cards, Rating, and Married, explain less than 50% of the total variability in the Income variable > A Moving to another question will save this response.
Question 13
To answer this question, use R to fit a logistic regression that predicts default using income as a predictor variable in the Default data set.
Using this model, what are the predicted probabilities of default (probability that default = “yes”) for people with an income of 100,000 and 50,000 respectively?
0.029 and 0.011
0.012 and 0.21
0.13 and 0.32
0.019 and 0.029
A Moving to another question will save this response.
A Moving to another question will save this response.
Question 14
To answer this question, use R to fit a logistic regression model that predicts default using income as a predictor variable in the Default data set.
Using this model, the odds of a person with an income of 20,000 defaulting on their loan [odds(default = “yes”)] is times greater than the odds of a person with an income of 130,000 defaulting on their loan.
4.2
2.1
6.4
2.5
Question 1
8 points
Save Answer
Using the credit data, write one expression linked by %>% operators and dplyr functions to create the summary table below. This table contains the number of customers with a credit limit greater than or equal to 3,000 that have at least 2 credit
cards, their average age and average income, by gender and marital status. Paste your final R code that produces this table as your submission to this question. As always, do not worry about rounding.
Gender Married number_of_customers avg_age avg_income
Male
No
48 55.12500 54.98554
Yes
73 55.16438 52.90101
Female No
54 57.25926 47.96756
Female Yes
88 56.39773
54.26184
Male
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Question 4
12 points
Save Answer
Using the Default data set, write one expression linked by %>% operators and dplyr and tidyr functions to create the summary table below. This table displays the average of the balance and income variables for all combinations of the default and
student variables. Paste your final R code that produces this table as your submission to this question. As always, do not worry about rounding.
Hint: First think about how to obtain the average of the balance and income variables for combinations of default and student using dplyr. Then use tidyr to reshape the result.
default student summary_statistic value
No No mean_balance
744.5044
No Yes mean_balance 948.4802
Yes No mean_balance 1678.4295
Yes Yes mean_balance 1860.3791
No No meanincome 39993.5152
No Yes
mean_income 17937.0088
Yes
No
mean_income 40625.0507
Yes Yes
mean_income 18243.5083
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Question 6
To answer this question, use R to fit a simple linear regression that predicts Income using Rating as a predictor variable in the Credit data set.
The residual standard error from this model represents of the average Income value in the Credit data set.
43.4%
24.5%
47.7%
19.4%
Question 11
8 points
Save Answer
Conduct a single partial F-test in R using multiple linear regression that answers the following question about the Credit data set:
Are either Gender or Rating helpful in predicting Income, given that we are using Cards and Age to predict Income?
Answer this question using your results. Justify your conclusion using the F statistic and p-value from your partial F-test.
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Question 8
Use the Routput below to select the correct statement.
Call:
lm(formula = Balance ~ Cards, data = Credit)
Residuals:
Min
1Q Median
-608.21 -455.24 -38.75
30 Max
350.99 1448.77
Coefficients:
Estimate Std. Error t value Pr(>It)
(Intercept) 434.29 54.57 7.958 1.83e-14 ***
Cards
28.99
16.74 1.731
0.0842
Signif. codes:
****) 0.001 (**) 0.01 **) 0.05.’ 0.11
Residual standard error: 458.6 on 398 degrees of freedom
Multiple R-squared: 0.007475, Adjusted R-squared: 0.004981
F-statistic: 2.997 on 1 and 398 DF, p-value: 0.08418
On average, for every new credit card account that customers open, we expect to see an increase of approximately $29 in their average monthly credit card balance
On average, if customers open 3 new credit card accounts, we expect to see an increase of approximately $58 in their average monthly credit card balance
On average, for every new credit card account that customers open, we expect to see an increase of approximately $434 in their average monthly credit card balance
We have strong statistical evidence (p-value < 0.0001) that the number of credit cards customers have is related to their average monthly credit card balance Question 9 To answer this question, use R to fit a multiple linear regression that predicts Income using Rating, Cards, and Age as predictor variables in the Credit data set. Using this model, the predicted Income for customers with a Rating equal to 400, Cards equal to 3, and Age equal to 40 is: 51.9 47.3 62.4 50.1 Question 7 To answer this question, use R to fit a simple linear regression that predicts Income using Rating as a predictor variable in the Credit data set. A 95% confidence interval for the true model parameter of the Rating predictor variable in this model is: (0.177, 0.204) (0.167, 0.194) (0.137, 0.214) (0.179, 0.189)