In Dallas some fire trucks were painted yellow (not red) to heighten their visability. During a test period, the fleet of red fire trucks made 153,348 runs and had 20 accidents, while the fleet of yellow trucks made 135,035 runs and had 4 accidents. At a= .01, did the yellow fire trucks have a significantly lower accident rate? (a) state the hypothesis (b) state the decision rule and sketch it (c) find the proportions and z test statistic (d) make a decision (e) find the p-value and interpret it (f) if statistically significant, do you think the difference is large enough to be important? If so, to whom, and why? (g)is the normally assumption fulfilled? Explain
Accident rate for Dallas Fire Trucks
Statistic Red Trucks Yellow Trucks
# of Accidentsx1=20accidents x2=4accid
# of fire runsn1=153,348 n2=135,035
Hypotheses:
Ho: pY = pR (R = red trucks; Y = yellow trucks)
Ha: pY < pR
You can use a binomial proportion 2-sample z-test for this kind of problem.
Here is one formula for this type of test:
z = (pY - pR)/√(pq(1/n1 + 1/n2)
p = (x1 + x2)/(n1 + n2)
q = 1 - p
pY = 4/135035
pR = 20/153348
p = (4 + 20)/(135035 + 153348)
q = 1 - p
Convert to decimals and substitute into the formula. Calculate z. Compare to the cutoff z at .01 for a one-tailed test (Ha shows a specific direction, so the test is one-tailed). Determine the p-value, which is the actual level of the test statistic. Determine whether to reject Ho (the null hypothesis) or fail to reject Ho. If you reject Ho, then you accept Ha and the test is statistically significant.
I hope this will help get you started.
To determine if the yellow fire trucks have a significantly lower accident rate than the red fire trucks, we need to perform a hypothesis test.
(a) State the hypothesis:
Null hypothesis (H0): The accident rate for yellow fire trucks is the same as the accident rate for red fire trucks.
Alternative hypothesis (Ha): The accident rate for yellow fire trucks is significantly lower than the accident rate for red fire trucks.
(b) State the decision rule and sketch it:
Given that the significance level (a) is 0.01, we will reject the null hypothesis if the p-value is less than 0.01.
(c) Find the proportions and z-test statistic:
To find the proportions, we calculate the accident rates (proportions) for both the red and yellow fire trucks:
Proportion for red fire trucks (p1) = x1 / n1 = 20 / 153,348
Proportion for yellow fire trucks (p2) = x2 / n2 = 4 / 135,035
To perform the z-test, we calculate the standard error (SE) and the z-test statistic:
SE = sqrt((p1(1-p1)/n1) + (p2(1-p2)/n2))
z-test statistic = (p1 - p2) / SE
(d) Make a decision:
We compare the calculated z-test statistic to the critical z-value corresponding to a significance level of 0.01. If the calculated z-test statistic is less than the critical z-value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
(e) Find the p-value and interpret it:
The p-value is the probability of observing a test statistic as extreme as the one calculated under the null hypothesis. We will compare the p-value to the significance level to determine if it is statistically significant.
To find the p-value, we calculate the area under the normal distribution curve to the left of the calculated z-test statistic. If the p-value is less than 0.01, we reject the null hypothesis.
(f) If statistically significant, do you think the difference is large enough to be important?
If the null hypothesis is rejected and there is a statistically significant difference between the accident rates of the two types of fire trucks, further analysis can be done to determine the practical significance of the difference. This can involve factors such as the magnitude of the difference, the potential impact on safety, and the cost implications of switching to yellow trucks.
(g) Is the normality assumption fulfilled? Explain:
To fulfill the normality assumption, we need to verify that the sampling distribution of the proportion (accident rate) is approximately normal. Generally, if the sample sizes are large enough (n1 and n2 > 30), the normality assumption holds. However, if the sample sizes are smaller, alternative tests such as the chi-square test or the Fisher's exact test may be used.