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2. For a sample of nine automobiles, the mileage (in 1000s of miles) at which the original front brake pads were worn to 10% of their original thickness was measured, as was the mileage at which the original rear brake pads were worn to 10% of their original thickness. The results are given in the following table. Automobile Front Rear 32.8 41.2 26.6 35.2 36.6 46.1 36.4 46.0 29.2 39.9 40.9 51.7 40.9 51.6 34.8 46.1 36.6 47.3 7

(a) Construct a 90% confidence interval for the difference in the wear mileage between front and rear b) What assumptions need to be checked in order to justify the use of the statistical method? Do this and (c) Conduct a hypothesis test at the 5% significance level that tests whether the wear mileage between the brake pads. include any appropriate plots and output from software. front and rear brake pads are equal. Be sure to show the hypothesis statements. Which brakes wear faster? State your conclusion to the study appropriately.

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Answer #1

Note:- We will use R for conducting statistical test. Any other statistical tool can be used to conduct the test

Solution 1

Code Snippet

## Creating front and rear tyre wear mileage data
Front <- c(32.8,26.6,36.6,36.4,29.2,40.9,40.9,34.8,36.6)
Rear <- c(41.2,35.2,46.1,46.0,39.9,51.7,51.6,46.1,47.3)

Auto_mil<- data.frame(cbind(Front,Rear))

Auto_mil$Diff_mil<- Auto_mil$Front - Auto_mil$Rear

## Calculating mean and variance of front and rear tyre wear mileage
Auto_mil_front_means<- mean(Auto_mil$Front)
Auto_mil_front_var<- var(Auto_mil$Front)
Auto_mil_rear_means<- mean(Auto_mil$Rear)
Auto_mil_rear_var<- var(Auto_mil$Rear)

## Conducting F test to check whether variance of two samples are equal

var.test(Auto_mil$Front,Auto_mil$Rear,alternative = "two.sided")

### As p value > 0.05 we do not reject null hypothesis of equal variance

## As both sample have equal variances we use pooled variance for calculation of confidence interval

pooled_variance <- ((9-1)*Auto_mil_front_var + (9-1)*Auto_mil_rear_var)/(9+9-1-1)

confidence_interval_lower <-(Auto_mil_front_means - Auto_mil_rear_means) - qt(0.9, (9+9-2)) * sqrt(pooled_variance*(1/9 + 1/9))

confidence_interval_upper<- (Auto_mil_front_means - Auto_mil_rear_means) + qt(0.9, (9+9-2)) * sqrt(pooled_variance*(1/9 + 1/9))

Explanation

First we need to check whether variances of rear and front wear mileage is equal to know whether pooled variance or separate variance should be used for identifying confidence interval

H0 of F- test of equal variances => var(rear_tyre_mileage) = var(front_tyre_mileage)

H1 of F- test of equal variances => var(rear_tyre_mileage) not equal to var(front_tyre_mileage)  

p-value of the F test for comparing variance = 0.7535 . This implies the null hypothesis should not be rejected at 95% confidence.

Thus, we can use pooled variance to identify confidence interval

Pooled variance = ((n1-1)*s1^2 + (n2-1)*s2^2) / ((n1-1) + (n2-1))

where

first sample - front tyre wear mileage

second sample - rear tyre wear mileage

n1 = sample size of first sample = 9

n2 = sample size of second sample = 9

s1^2 = variance of first sample = 23.22

s2^2 = variance of second sample = 29.21

Pooled variance(calculated in R above ) = 26.21

At 90% cofidence interval -

(mean_sample_x1 - mean_sample_x2) +- ta/2  * sqroot(pooled variance*(1/n1 + 1/n2))

90% confidence interval for difference in wear mileage of front and rear tyres(calculated in R above) - 13.26 to -6.81

Solution2

For conducting statistical test difference in Rear and front tyre wear mileage should follow a normal distribution. We conducted jarque bera test to check normality of the difference of two sample

Output Jarque Bera Test

X-squared = 0.91893, df = 2, p-value = 0.6316

As p value is > 0.05 , the null hypothesis of normality is not rejected and we can use statistical method for the difference of two samples.

Solution3

Null hypothesis, H0 - wear mileage of front tyre = wear mileage of rear tyre

Alternate hypothesis, H1 - wear mileage of front tyre not equal to wear mileage of rear tyre

t-stat = ((mean_sample_x1 - mean_sample_x2) - (mean_population_x1 - mean_population_x2)) / sqrt(pooled variance*(1/n1+1/n2))

t-stat = ((34.9777 - 45.011) - 0 ) / sqrt(26.214*(1/9+1/9))

t-stat = -10.0333/ sqrt(5.8253) = -4.157

t critical = -1.7458

Since , t stat lies outside t critical range, we reject null hyothesis that wear mileage of front tyre = wear mileage of rear tyre

Also, as mean of wear mileage of front tyre sample is lower than mean of wear mileage of rear tyre sample, we conclude that front tyre wear faster than rear tyre at 5 % significance level.

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