SOLUTION: Of 41 bank customers depositing a check, 20 received some cash back. (a) Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back

Question 1198867: Of 41 bank customers depositing a check, 20 received some cash back.
(a) Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**1. Find the sample proportion (p̂)**
* p̂ = (Number of depositors who received cash back) / (Total number of depositors)
* p̂ = 20 / 41
* p̂ ≈ 0.4878
**2. Find the standard error (SE)**
* SE = √(p̂ * (1 - p̂) / n)
* SE = √(0.4878 * (1 - 0.4878) / 41)
* SE ≈ 0.0764
**3. Find the critical value (z*) for a 90% confidence level**
* For a 90% confidence level, the area in each tail of the standard normal distribution is (100% - 90%) / 2 = 5%.
* Using a standard normal distribution table or calculator, the critical value (z*) for 5% in the tail is 1.645.
**4. Calculate the margin of error**
* Margin of Error = z* * SE
* Margin of Error = 1.645 * 0.0764
* Margin of Error ≈ 0.1257
**5. Construct the 90% confidence interval**
* Lower limit = p̂ - Margin of Error = 0.4878 - 0.1257 = 0.3621
* Upper limit = p̂ + Margin of Error = 0.4878 + 0.1257 = 0.6135
**Therefore, the 90% confidence interval for the proportion of all depositors who ask for cash back is approximately (0.3621, 0.6135).**
This means we are 90% confident that the true proportion of all depositors who ask for cash back lies between 36.21% and 61.35%.

RELATED QUESTIONS
Of 41 bank customers depositing a check, 25 received some cash back. a)Construct a 90... (answered by math_tutor2020)

The First National Bank of Wilson has 700 checking account customers. A recent sample of... (answered by ewatrrr)

The first national bank of wilson has 650 (population) checking account customers. A... (answered by robertb)

The First National Bank of Wilson has 700 checking account customers. A recent sample of... (answered by oscargut)

The U.S. Customs Service conducted a random check of Miami longshoremen and found that 36 (answered by stanbon)

Smallville Bank has 650 checking accounts. A recent sample of 50 of these customers... (answered by stanbon)

A random sample of 20 acres gave a mean yield of sugarcane equal to 41.2 tons per acre... (answered by Boreal)

The U.S. Customs Service conducted a random check of Miami longshoremen and found that 36 (answered by stanbon)

Can someone please check to see if this is done correctly? 1. A random sample of (answered by stanbon)