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.

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step-by-step explanation:

1+2+3+4+5+6+7+8+9+10 = 55

2*10 = 20

55-20 = 35

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column