THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
A sales representative has 425 accounts.From a random sample of 60 of these accounts,she found that the average yearly order was $1,250 with a standard deviation of $70.She also found that 28 out of 60 in the sample have placed an order in the last four weeks.
-Find the 90% confidence interval for the proportion of all accounts that placed an order within the last four weeks.
A)0.467 ± 0.412
B)0.467 ± 0.266
C)0.467 ± 0.838
D)0.467 ± 0.120

Question

Quizplus · Accepted Answer

To find the confidence interval, we need to use the formula: 
Margin of error = z* (standard error)
Where z* is the z-score of the confidence level, and the standard error can be found using the following formula: 
Standard error = sqrt(p̂(1-p̂)/n)
where p̂ is the sample proportion, and n is the sample size.

Here, we are given p̂ = 28/60 = 0.467 and n = 60. 
Using a 90% confidence level, we can find the z-score from the standard normal distribution table or calculator to be 1.645.

Plugging in the values, we get: 
Standard error = sqrt(0.467(1-0.467)/60) = 0.069
Margin of error = 1.645*0.069 = 0.114

Therefore, the 90% confidence interval for the proportion of all accounts that placed an order within the last four weeks is:
0.467 ± 0.114

Simplifying, we get 
Lower bound = 0.467 - 0.114 = 0.353
Upper bound = 0.467 + 0.114 = 0.581

So, the best choice is B, which corresponds to the above interval.