Last year\'s records of auto accidents occurring on a given section of highway w
ID: 3156979 • Letter: L
Question
Last year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was $1000 or more and to whether a physical injury resulted from the accident. The data follows. (a) Estimate the true proportion of accidents involving injuries when the damage was $1000 or more for similar sections of highway. (Round your answer to three decimal places.) Find the 95% margin of error. (Round your answer to three decimal places.) (b) Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under $1000 and those with damage of $1000 or more. Use a 95% confidence interval. (Round your answers to three decimal places.) toExplanation / Answer
a)
Here,
p^ = x2/n2 = 25/49 = 0.510204082 [ANSWER]
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Note that
p^ = point estimate of the population proportion = x / n = 0.510204082
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.071413695
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
Margin of error = z(alpha/2)*sp = 0.13996827 [ANSWER]
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b)
Getting p1^ and p2^,
p1^ = x1/n1 = 0.242424242
p2 = x2/n2 = 0.510204082
Also, the standard error of the difference is
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] = 0.103272507
For the 95% confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = 1.959963985
Margin of error = z(alpha/2)*sd = 0.202410394
lower bound = p1^ - p2^ - z(alpha/2) * sd = -0.470190233
upper bound = p1^ - p2^ + z(alpha/2) * sd = -0.065369445
Thus, the confidence interval is
( -0.470190233 , -0.065369445 ) [ANSWER]
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