By hand please, no R In a test of braking performance, a tire manufacturer measu

ID: 3230780 • Letter: B

Question

By hand please, no R

In a test of braking performance, a tire manufacturer measured the stopping distance for one of its tire models. On a test track, a car made repeated stops from 60 mph. The company tested the tires on 10 different cars, recording the stopping distance for each car on both dry and wet pavement, with the following results. car=1:10 dist=c(150,147,136,134,130,134,134,128,136,158,201,220,192, 146,182,173,202,180,192,206) pavement=c(rep('Dry',each=10),rep('Wet',each=10)) stop=data.frame(car,dist,pavement) dbar = -50.7 sd = 16.66033 (sd of differences) n = 10 2 (a) Estimate the true average difference of braking performance on wet vs. dry roads with 98% confidence. Interpret. (b) Is there a difference in the mean braking performance on wet vs. dry roads? Conduct hypothesis test. Let = 0.02 (c) State the kind of error could have been made in context of the problem. (d) Now do part b again in R

Explanation / Answer

The 98% confidence interval for the true average difference of braking performace on wet vs. dry road assuming the population variances not equal is given by : (x1 – x2) ± t* ((s12 /n1 + s22 / n2))

Where

x1 is the mean of braking performance on dry road

x2 is the mean of braking performance on wet road

s12 is the sample variance of braking performance on dry road

s22 is the sample variance of braking performance on wet road

n1 is the sample size of braking performance on dry road

n2 is the sample size of braking performance on wet road

t* is ( t1* s12/n1 + t2* s22/n2 )/( s12/n1 + s22/n2)

t1 and t2 are the table values of t at a prefixed level of significance with (n1 – 1) and (n2 – 1) d.f. respectively.

cars

Dry

wet

difference

150

201

-51

147

220

-73

136

192

-56

134

146

-12

130

182

-52

134

173

-39

134

202

-68

128

180

-52

136

192

-56

158

206

-48

mean

138.7

189.4

-50.7

variance

93.34444

423.8222

t values:

t1 @ =0.02 and 9 d.f. = 2.821

t2 @ =0.02 and 9 d.f. = 2.821

t* = (2.821*93.3444/10+ 2.821*423.8222/10)/ (93.3444/10 + 423.8222/10) = 2.821

98% confidence interval

= (138.7 – 189.4) ± 2.821*((93.34444/10 + 423.82222/10))

= -50.7 ± 20.28702

98% confidence interval = (-70.98702, -30.4129)

To test the difference in the mean braking performance on wet vs. dry roads:

H0 : There is no difference between the mean braking performance on wet vs. dry roads. µ1 = µ2

H1 : There is difference between the mean braking performance on wet vs. dry roads. µ1 µ2

t = x1 - x2 / ((s12 /n1 + s22 / n2))

= -50.7/ 7.19143

= -7.05

t* = 2.821 ( Refer the calculation given above)

Since t < -2.821, we reject the null hypothesis. Hence we can conclude that there is difference between the mean braking performance on wet vs. dry roads.