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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