Roofing injuries. According to a study conducted by the California Division of L
ID: 3149968 • Letter: R
Question
Roofing injuries. According to a study conducted by the California Division of Labor Research and Statistics, roofing is one of the most hazardous occupations. Of 2,514 worker injuries that caused absences for a full workday or shift after the injury, 23% were attributable to falls from high elevations on level surfaces, 21% to falling hand tools or other materials, 19% to overexertion, and 20% to burns or scalds. Assume that the 2,514 injuries can be regarded as a random sample from the population of all roofing injuries in California.
a. Construct a 95% confidence interval for the proportion of all injuries that are due to falls.
b. Construct a 95% confidence interval for the proportion of all injuries that are due to burns or scalds.
Please, show your work.
Explanation / Answer
a)
Note that
p^ = point estimate of the population proportion = x / n = 0.23
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.008393182
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
Margin of error = z(alpha/2)*sp = 0.016450335
lower bound = p^ - z(alpha/2) * sp = 0.213549665
upper bound = p^ + z(alpha/2) * sp = 0.246450335
Thus, the confidence interval is
( 0.213549665 , 0.246450335 ) [ANSWER]
*******************************
b)
Note that
p^ = point estimate of the population proportion = x / n = 0.2
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.007977694
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
Margin of error = z(alpha/2)*sp = 0.015635992
lower bound = p^ - z(alpha/2) * sp = 0.184364008
upper bound = p^ + z(alpha/2) * sp = 0.215635992
Thus, the confidence interval is
( 0.184364008 , 0.215635992 ) [ANSWER]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.