Answer:
Explained below.
Step-by-step explanation:
The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².
(1)
The hypothesis for both the test can be defined as:
<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.
<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.
(2)
A Chi-square test will be used to perform the test.
The significance level of the test is, <em>α</em> = 0.05.
The degrees of freedom of the test is,
df = n - 1 = 55 - 1 = 54
Compute the critical value as follows:
![\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D_%7B%5Calpha%2C%20%28n-1%29%7D%3D%5Cchi%5E%7B2%7D_%7B0.05%2C%2054%7D%3D72.153)
Decision rule:
If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.
(3)
Compute the test statistic as follows:
![\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D%3D%5Cfrac%7B%28n-1%29%5Ctimes%20s%5E%7B2%7D%7D%7B%5Csigma%5E%7B2%7D%7D)
![=\frac{(55-1)\times 94.7}{75}\\\\=68.184](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%2855-1%29%5Ctimes%2094.7%7D%7B75%7D%5C%5C%5C%5C%3D68.184)
The test statistic value is, 68.184.
Decision:
![cal.\chi^{2}=68.184](https://tex.z-dn.net/?f=cal.%5Cchi%5E%7B2%7D%3D68.184%3C%5Cchi%5E%7B2%7D_%7B0.05%2C54%7D%3D72.153)
The null hypothesis will not be rejected at 5% level of significance.
Conclusion:
The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.