Use knowledge of confidence intervals to find the confidence level used (90%, 95
ID: 3268568 • Letter: U
Question
Use knowledge of confidence intervals to find the confidence level used (90%, 95%, 99%) based on the margin of error and the number of individuals that participated in the poll. Show the computations involved in this process. If the poll included the confidence level, check the accuracy by using your own computations. Is the reported confidence level correct?
1.This pole is based on random sample of 74,462 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia. The margin of error associated with the poll is ±0.4 percentage point. The confidence level associated with the poll is 95% confidence level.
2.a random sample of 1,007 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia and having investments of $10,000 or more,the margin of sampling error is ±4 percentage points , 95% confidence level.
Explanation / Answer
1. Sample size = 74462 adults
Margin of error = +- 0.4 % point = 0.004
Margin of error = Critical test statistic * standard error of proportion
standard error of proportion = sqrt [p * (1-p)/N]
for 95% confidence interval (critical test statistic ) = 1.96
0.004 = 1.96 * sqrt [p * (1-p)/ 74462]
[p * (1-p)] = 0.31
p - p2= 0.31
p2- p + 0.31 = 0
this can't be true for any value of p or any proportion so the given statement is wrong that confidence level is 95% .
2. Here N = 1007
Margin of error = +- 4 % point = 0.04
Margin of error = Critical test statistic * standard error of proportion
standard error of proportion = sqrt [p * (1-p)/N]
for 95% confidence interval (critical test statistic ) = 1.96
0.04 = 1.96 * sqrt [p * (1-p)/ 1007]
[p * (1-p)] = 0.4194
p - p2= 0.42
p2- p + 0.31 = 0
this can't be true for any value of p or any proportion so the given statement is wrong that the confidence level is 95%.
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