There are 614,387 bridges in the United States. A senator sponsors a bill to rai
ID: 3325474 • Letter: T
Question
There are 614,387 bridges in the United States. A senator sponsors a bill to raise taxes to pay for repairs to those bridges. To support her bill, she claims that St ore than 5% ofbridges in the United ates are structurally deficient, and the average bridge in the United States is more than 40 years old. To support that claim, she initiates a study of the nation's bridges through the U.S. Army Corps of Engineers to test 200 bridges to see if each bridge is structurally deficient. Describe how to collect a random sample of 200 bridges from this population. Construction engineers from the Army Corps of Engineers conduct the study of the nation's bridges. Of the 200 bridges tested, 15 are deemed structurally deficient. Construct a 99% confidence interval for the population proportion. Test the claim at = 0.01. How large a sample would be required to be 99% confident that the true proportion is within 1 % of the sample proportion?Explanation / Answer
B) p = 15/200 = 0.075
At 99% Confidence interval the critical value is 2.58
The confidence interval is
p +/- z0.005 * sqrt(p * (1 - p) /n)
= 0.075 +/- sqrt(0.075 * 0.925/200)
= 0.075 +/- 0.019
= 0.056, 0.094
C) H0: P = 0.05
H1: P > 0.05
The test statistic Z = (p - P)/sqrt(P * (1 - P)/n )
= (0.075 - 0.05)/sqrt(0.05 * 0.95/200)
= 1.62
P-VALUE = P(Z > 1.62)
= 1 - P(Z < 1.62)
= 1 - 0.9474 = 0.0526
As the p-value is greater than the significance level (0.0526 > 0.01), so the null hypothesis is not rejected.
So there is not sufficient evidence to support the claim that more than 5% bridges in the united States are structurally defecient.
D) margin of error = 0.01
Or, z0.005 * sqrt (p * (1 - p )/n) = 0.01
Or, 2.58 * sqrt(0.075 * 0.925/n) = 0.01
Or, sqrt(n) = 2.58 * sqrt(0.075 * 0.925)/0.01
Or, sqrt(n) = 67.95
Or, n = 4617
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