Problem

The average number of accidents at controlled intersections per year is 5.9. Is this average a different number for intersections with cameras installed? The 50 randomly observed intersections with cameras installed had an average of 5.4 accidents per year and the standard deviation was 2.29. What can be concluded at the $\alpha=0.01$ level of significance?
a. For this study, we should use $t$-test for a population mean
b. The null and alternative hypotheses would be:
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$0^{8}$
$0^{6}$
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c. The test statistic
5.9

Answer

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Answer

Final Answer: \(\boxed{\text{We fail to reject the null hypothesis}}\). There is not enough evidence to conclude that the average number of accidents at intersections with cameras installed is significantly different from the population mean at the $\alpha=0.01$ level of significance.

Steps

Step 1 :We are given the population mean (5.9), the sample mean (5.4), the sample standard deviation (2.29), and the sample size (50). We are asked to determine if the average number of accidents at intersections with cameras installed is significantly different from the population mean at the 0.01 level of significance.

Step 2 :The null hypothesis ($H_0$) is that the population mean is equal to the sample mean, i.e., the average number of accidents at intersections with cameras installed is not significantly different from the population mean. The alternative hypothesis ($H_1$) is that the population mean is not equal to the sample mean, i.e., the average number of accidents at intersections with cameras installed is significantly different from the population mean.

Step 3 :We will use a two-tailed t-test to test the hypotheses. The test statistic for a t-test is calculated as follows: $t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$ where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size.

Step 4 :Substituting the given values into the formula, we get $t = \frac{5.4 - 5.9}{2.29/\sqrt{50}} = -1.54$

Step 5 :The critical value for a two-tailed t-test at the 0.01 level of significance and 49 degrees of freedom (50-1) is approximately 2.68.

Step 6 :The calculated test statistic (-1.54) is less than the critical value (2.68). Therefore, we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the average number of accidents at intersections with cameras installed is significantly different from the population mean at the 0.01 level of significance.

Step 7 :Final Answer: \(\boxed{\text{We fail to reject the null hypothesis}}\). There is not enough evidence to conclude that the average number of accidents at intersections with cameras installed is significantly different from the population mean at the $\alpha=0.01$ level of significance.

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