Please help!!

This problem uses the teengamb data set in the faraway package. Fit a model with gamble as the response and the other variables

hichkok12 [17]

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

4 0

3 years ago

Two cards are drawn without replacement from a standard deck of 52 cards. What is the probability that one of the cards is red a

alukav5142 [94]

Total cards = 52
Total red cards = 26
Total black cards = 26

P(one red and one black) = P(red and then black)+ P(black and then red)
P(one red and one black) =(26/52)(26/51) + (26/52)(26/51)
P(one red and one black) = 26/51 (Answer A)

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Answer: 26/51 (Answer A)
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5 0

3 years ago

Heba ate 1/12 of a box of cereal. Now the box is 3/4 full

vaieri [72.5K]

Answer:

5/6

Step-by-step explanation:

1/12 + 3/4 = 9/12

9/12 + 1/12 = 10/12

simplify 10/12

= 5/6

7 0

2 years ago

Read 2 more answers

Given that R=9x+8y find x when y=8 and R=4

Ilya [14]

Answer:

Step-by-step explanation:

R = 9x + 8y

4 = 9x + 8*8

4 = 9x + 64

Subtract 64 from both sides

4 - 64 = 9x + 64 - 64

-60 = 9x

Divide both sides by 9

-60/9 = 9x/9

-20/3 = x

x = -6 2/3

8 0

3 years ago

Read 2 more answers

The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime

Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

$Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334$

X[bar] ± $Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }$

174.5 ± $2.334* \frac{6.9}{\sqrt{50} }$

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0

2 years ago