Answer:
We conclude that the true average stopping distance exceeds this maximum value.
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
= true average stopping distance
So, Null Hypothesis,
:
30 {means that the true average stopping distance exceeds this maximum value}
Alternate Hypothesis,
:
> 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. =
~
![t_n_-_1](https://img.qammunity.org/2021/formulas/mathematics/college/1brbzqngbb8se3mhcwr88updmia2rk3b3g.png)
where,
= sample mean stopping distance =
= 31.32 ft
s = sample standard deviation =
= 0.66 ft
n = sample size = 6
So, the test statistics =
~
![t_5](https://img.qammunity.org/2021/formulas/mathematics/college/7fvqs9ipnhlxa9ci7nxfz0yhglxbmegtod.png)
= 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.