Question

A statistical program is recommended. The data below contains mileage, age, and selling price for a sample of 33 sedans. Price Mileage Age Price Mileage Age Price Age 11,599 95,000 18,998 16,998 17,599 35,000 16,000 8 8 Mileage 20,000 | 63,000 10,499 123,000 12,599 15,998 42,000 14,599 54,000 12,599 15,998 10,599 68,000 11 10,998 70,000 88,000 86,000 60,000 61,000 102,000 109,000 16,998 11,599 13,998 11,998 20,998 12,998 14,998 50,000 11,599 17,599 11,599 12,599 13,599 9,998 40,000 7,000 81,000 70,000 86,000 92,000 120,000 45,000 95,000 60,000 12,599 88,000 58,000 27,000 32,000 9,998 11 2 14,998 11,599 16,998 36,000 7 10 15,998 4 9,599 17,599 45,000 2 Use the estimated regression equation that predicts the selling price of a used sedan given the mileage and age of the car to answer the following questions. (Let x, represent the milage of the car, x, represent the age of the car, and y represent the selling price of the car.) ý = 20385.25 - 0.03739x - 686.3367x, (a) Estimate the selling price (in $) of a five-year-old sedan with mileage of 45,000 miles. (Round your answer to the nearest dollar.) (b) Develop a 95% confidence interval for the selling price in $) of a car with the data in part (a). (Round your answers to the nearest dollar.) $ to $ (c) Develop a 95% prediction interval for the selling price in $) of a particular car having the data in part (a). (Round your answers to the nearest dollar.) to $

Answer 1

Answer:

Given that,

(a).

Given,

$\rightarrow$ The estimated regression equation for the above data,

$\hat{y}$ = 20385.25 - 0.03739 x1 - 686.3367 x2

Where,

x1 represents mileage and x2 represents the age of the car.

Here , Mileage (x1) = 45000 miles and Age (x2) = 5 years.

$\rightarrow$ Therefore , Selling price ,

$\hat{y}$ = 20385.25 - 0.03739(45000) - 686.3367(5)

$\hat{y}$= 20385.35 - 1682.55 - 3431.6835

$\hat{y}$= 15271.1165

$\hat{y}$ = $15271(approx.)

(b).

$\rightarrow$ According to the given data , we have the following values ,

Number of samples (n) = 33

Mean , ($\bar{x}$) = 13910.63636

Variance ($\sigma ^{2}$)= 8747573.364

Therefore , Standard Deviation($\sigma$) = 2957.629687

To develop a 95% confidence interval:

Therefore , we must know that z-value for 95% confidence level = 1.96

$\rightarrow$ Now , Confidence Interval is given by ,

2 xo FI u

= (13910.63636)$\pm$ (1.96 x 2957.629687)/sqrt(33)

=13910.63636 $\pm$ 1010

= ($12900 to $14900)

Thus , we have , 95% confidence interval as $12900 to $14900 .

i.e., [ $12900 , $14900 ]

(c).

$\rightarrow$Given ,

Mean $\bar{x}$ = 13910.63636

Standard Deviation ( $\sigma$ ) = 2957.629687

And z-value for 95% prediction interval = 1.96

Prediction interval given by:

$\bar{x}\mp (z\times \sigma )$

=13910.63636 $\mp$ (1.96 $\times$ 2957.629687)

=13910.63636 $\mp$ 5796.9548

= (8113.6752, 19707.5848)

=($8114, $19708)

Thus , the 95% prediction interval is given by, $ 8114 to $19708.