Is x/5 = y/3 a linear function
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Answer:
[-0.65;2.54]km
Step-by-step explanation:
Hello!
So, you need to find a 99%CI for the difference in mean life of two brands of radial tires. Since he assigned one tire of each brand at random to the two rear wheels of each car, in other words, every car tested had one rear tire of each brand at the same time, this test can be considered to be of paired samples.
Brand 1 Brand 2 X₁-X₂
car 1: 36,925 ; 34,318 ; 2.607
car 2: 45,300 ; 42,280 ; 3.020
car 3: 36,240 ; 35,500 ; 0.740
car 4: 32,100 ; 31,950 ; 0.150
car 5: 37,210 ; 38,015 ; -.0805
car 6: 48,360 ; 47,800 ; 1.160
car 7: 38,200 ; 37,810 ; 0.390
car 8: 33,500 ; 33,215 ; 0.285
n= 8
With this in mind, we define the study variable as Xd= X₁-X₂
Where X₁ corresponds to the lifespan, in km, of a tire from Brand 1
and X₂ corresponds to the lifespan, in km, of a tire from Brand 2
so Xd will be the difference between the lifespan of the tires from Brand 1 and Brand 2.
This variable Xd~N(μd;δd²) (p-value for normality test is 0.4640)
To calculate the CI the best statistic is the Student's t with the following formula:
t= (xd[bar] - μd)/(Sd/√n) ~t₍ₙ₋₁₎
sample mean: xd[bar]= 0.94
standard deviation: Sd= 1.29
= 3.355
xd[bar] ± *(Sd/√n)
⇒ 0.94 ± 3.355*(1.29/√8)
[-0.65;2.54]km
The Confidence Interval can be compared to a pair of bilateral hypothesis. If we were to determine the following hypothesis
H₀:μd=0
H₁:μd≠0
Using the level of significance of 0.01 (complementary to the confidence level)
As the calculated confidence interval contains zero, we do not reject the null hypothesis, that is, there is no significant difference between the two tire brands.
I hope you have a SUPER day!