A statistical program is recommended. The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experimental conditions.

32.1 30.9 31.6 30.4 31.0 31.9

The report states that under these conditions, the maximum allowable stopping distance is 30. A normal probability plot validates the assumption that stopping distance is normally distributed.

Required:
a. Does the data suggest that true average stopping distance exceeds this maximum value? Test the appropriate hypotheses using α= 0.01.
b. Calculate the test statistic and determine the P-value.
c. What can you conclude?

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 = 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.  =    ~

where,  = sample mean stopping distance = = 31.32 ft

            s = sample standard deviation = = 0.66 ft

            n = sample size = 6

So, the test statistics =    ~  

                                    =  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.