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%.
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