t siven blow summarines the results of a one-way analysis of no conmpsore the pe
ID: 3045968 • Letter: T
Question
t siven blow summarines the results of a one-way analysis of no conmpsore the peefommance characteristics of four brands of vacuum he E ble is ithe asmount of time it takes to clean a specific size room with a AS BR ANDS (GROUNT . 1 488 2430 OAL t Csmplere all missing valbwex when total of 1S ohservations have been the means of four signiffcane level oro05, determine if there is any difference amongt 2 tse the omation above and determine a Tukey (g e o es) 95 percent confidence interval for n-s. The moun and sample sizes for brand I and brand 2 are as follows: *i- 8-4 ant(S marks) treblem zi A local tire dealer wants to predict the number of tires sold each month. He believes that the number of tires sold is a linear function of the amount of moncy invested in avertising He randomly selects 6 months of data consisting of tire sales (in thousands of tires) and advertising costs cin thousands of dollars) Based on the data set with 6 observations, the simpsle linear regtression model yielded the following results T24 x 124 yr = 196 1. White the estimated regression line. 2. Determine the 95 percent confidence interval for the average value of Y (number of tires sold) when advertising costs is $5,000 (X-$5,000) Using a F test, determine the significance of the regression relationship between number of tires sold and advertising costs 3. 4 Cakculate the coefficient of determination (). (8 marks)Explanation / Answer
(1) since total observation n=18 so total df=n-1=18-1=17
there are k=4 brand so brand df=k-1=4-1=13
error df=n-k=18-4=14
total SS=barnad SS + error SS i.e.
tukey HSD=(m1-m2)/sqrt(mse(1/n1 +1/n2))=(2.95-2,28)/sqrt(106.1429*(1/4 + 1/5))=0.0969
tukey q-value(0.05,4,17)=4.02 ( this is stanard value)
(1-alpha)*100% confidence interval =sample mean difference ±q*SE(difference)
95% confidence interval =(2.95-2,28)±/4.02*sqrt(106.1429*(1/4 + 1/5))=0.67±6.91=(-6.24,7.58)
(2) (a) regression equation y=b0+b1*x=3+x
(b)b0=(sum(y)*sum(x2)-sum(x)*sum(xy))/(n*sum(x2)-(sum(x))2)=(42*124-24*196)/(6*124-24*24)=3
b1=(n*sum(xy)-sum(x)*sum(y))/ (n*sum(x2)-(sum(x))2)=(6*196-42*24)/(6*124-24*24)=1
(b)when x=5000, y=3+5000=5003
Total SS=sum(y2)-sum(y)*sum(y)/n=338-42*42/6=e44
individual x and y is needed for this part and for part(c)
(c) original observation x and y are needed....
(d)correlation coefficien r=(n*sum(xy)-sum(x)*sum(y))/sqrt(n*sum(x2)-(sum(x))2)*(n*sum(y2)-sum(y))2=
=(6*196-24*42)/sqrt((6*124-24*24)*(6*338-42*42))=0.7977
and r2=0.6363
source ss df ms f= Brands 6 3 2 0.01884253 error 1486 14 106.1429 total 1492 17Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.