Answer:
A. 95% confidence interval of gamble amount is (18.78277, 37.70227)
B. The 95% confidence interval of gamble amount is (42.23237, 100.3835)
C. 95% confidence interval of sqrt(gamble) is (3.180676, 4.918371)
D. The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information
Step-by-step explanation:
to)
We will see a code with which it can be predicted that an average man with income and verbal score maintains an appropriate 95% CI.
attach (teengamb)
model = lm (bet ~ sex + status + income + verbal)
newdata = data.frame (sex = 0, state = mean (state), income = mean (income), verbal = mean (verbal))
predict (model, new data, interval = "predict")
lwr upr setting
28.24252 -18.51536 75.00039
we can deduce that an average man, with income and verbal score can play 28.24252 times
using the following formula you can obtain the confidence interval for the bet amount of 95%
predict (model, new data, range = "confidence")
lwr upr setting
28.24252 18.78277 37.70227
as a result, the confidence interval of 95% of the bet amount is (18.78277, 37.70227)
b)
Run the following command to predict a man with maximum values for status, income, and verbal score.
newdata1 = data.frame (sex = 0, state = max (state), income = max (income), verbal = max (verbal))
predict (model, new data1, interval = "confidence")
lwr upr setting
71.30794 42.23237 100.3835
we can deduce that a man with the maximum state, income and verbal punctuation is going to bet 71.30794
The 95% confidence interval of the bet amount is (42.23237, 100.3835)
it is observed that the confidence interval is wider for a man in maximum state than for an average man, it is an expected data because the bet value will be higher than the person with maximum state that the average what you carried s that simultaneously The, the standard error and the width of the confidence interval is wider for maximum data values.
(C)
Run the following code for the new model and predict the answer.
model1 = lm (sqrt (bet) ~ sex + status + income + verbal)
we replace:
predict (model1, new data, range = "confidence")
lwr upr setting
4,049523 3,180676 4.918371
The predicted sqrt (bet) is 4.049523. which is equal to the bet amount is 16.39864.
The 95% confidence interval of sqrt (wager) is (3.180676, 4.918371)
(d)
We will see the code to predict women with status = 20, income = 1, verbal = 10.
newdata2 = data.frame (sex = 1, state = 20, income = 1, verbal = 10)
predict (model1, new data2, interval = "confidence")
lwr upr setting
-2.08648 -4.445937 0.272978
The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information
is the square of
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