Answer:
We cannot say that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used
Step-by-step explanation:
If we call
[tex]\mu_1[/tex]=true mean braking distance corresponding to compound 1
[tex]\mu_2[/tex]=true mean braking distance corresponding to compound 2
[tex]\bar x_1[/tex]=mean of the sample for compound 1
[tex]\bar x_2[/tex]=mean of the sample for compound 2
In this case we have the following hypothesis
[tex]H_0:\; \mu_1 - \mu_2 =0[/tex]
[tex]H_a:\; \mu_1 - \mu_2 <0[/tex]
The level of confidence is 0.05, so the region of rejection is represented in picture 1
(See picture attached).
Let's find the z-score corresponding to this data and if this value is < -1.64, then we can say that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used.
[tex]z=\frac{(\bar x_1-\bar x_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n}+\frac{s_2^2}{n}}}[/tex]
where
s1=standard deviation braking distance corresponding to the sample for compound 1
s2=standard deviation braking distance corresponding to the sample for compound 2
n=sample size
So,
[tex]z=\frac{(44-47)}{\sqrt{\frac{65.61}{51}+\frac{136.89}{51}}}=-1.5055[/tex]
Since -1.5051 is not < -1.64, we cannot refute H0 and say that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used.
