please do the (b) and (c) Specifically In a test of braking performance, a tire

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please do the (b) and (c) Specifically

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.

(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.

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 (d-bar; mean of differences)

sd=16.66033 (sd of differences)

n=10

Explanation / Answer

b) We will conduct the hypothesis test that H0: µ1 = µ2, against the alternative hypothesis H1: µ1 µ2
    where µ1 = Mean on dry pavement
                 µ2 = Mean on wet pavement

In order to perform the hypothesis test, we use a t-statistic, for which we will calculate the degrees of freedom shortly.

Standard Error (SE) of the sampling distribution is given by SE = [(s12/n1) + (s22/n2) ]
     where, s1 = standard deviation of dry pavement
                   n1 = no. of observations on dry pavement
                              s2 = standard deviation of wet pavement
                              n2 = no. of observations on wet pavement
Therefore, SE = [9.66152/10 + 20.58692/10]               (S.D.’s are calculated using ‘stdev’ function in MS Excel)
                         = (9.3344 + 42.3822]
                         = 7.1914

The degrees of freedom (DF) is given by DF = (s12/n1 + s22/n2)2 / ( [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] )
                                                                              = (9.3344 + 42.3822)2 / (9.33442/9 + 42.38222/9)
                                                                              = 2674.6136 / 209.2650
                                                                              = 12.781
                                                                               13

The test statistic is given by t = [ (µ1 - µ2) - µ0 ] / SE
                                                     = (138.7 – 189.4) / 7.1914    (Since µ0 = 0, the hypothesized difference)
                                                     = -7.0501

The critical value of t (with DF = 13) for a two-tailed test at = 0.02 is 2.650. Since the value of the test statistic > 2.650, we reject the null hypothesis at the 0.02 level of significance. So, based on the evidence, we accept the notion that there is a difference in the mean braking performance between dry and wet roads.

c) In the context of the hypothesis testing, we could have made two types of errors:
     Type I error: The error that we reject the null hypothesis when it is actually true
     Type II error: The error that we accept the null hypothesis when it is actually false

The probability of Type I error is determined by the level of significance used in the test whereas the probability of Type II error depends on the mean of the population from which the sample is drawn.

Note: Only parts (b) and (c) have been answered, as asked in the question.

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