Respuesta :
Answer:
We conclude that the true average stopping distance exceeds this maximum value.
Step-by-step explanation:
We are given the following observations that are on stopping distance (ft) of a particular truck at 20 mph under specified experimental conditions.;
X = 32.1, 30.9, 31.6, 30.4, 31.0, 31.9.
Let [tex]\mu[/tex] = true average stopping distance
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 30 {means that the true average stopping distance exceeds this maximum value}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 30 {means that the true average stopping distance exceeds this maximum value}
The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean stopping distance = [tex]\frac{\sum X}{n}[/tex] = 31.32 ft
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 0.66 ft
n = sample size = 6
So, the test statistics = [tex]\frac{31.32-30}{\frac{0.66}{\sqrt{6} } }[/tex] ~ [tex]t_5[/tex]
= 4.898
The value of t-test statistics is 4.898.
Now, at 0.01 level of significance, the t table gives a critical value of 3.365 at 5 degrees of freedom for the right-tailed test.
Since the value of our test statistics is more than the critical value of t as 4.898 > 3.365, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the true average stopping distance exceeds this maximum value.
