Solve for 2u = 4/9 Simplify your answer as much as possible.
1 answer:
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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:
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:
The test statistic value is, 68.184.
Decision:
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.