Of 43 bank customers depositing a check, 18 received some cash back. (a) Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back. (b) Check the normality assumption.
To construct a confidence interval for the proportion of all depositors who ask for cash back, we can use the formula for a confidence interval for a proportion.
(a) Confidence Interval Calculation:
Given:
Number of depositors (n) = 43
Number of depositors who ask for cash back (x) = 18
1. Calculate the sample proportion:
p̂ = x / n
p̂ = 18 / 43 ≈ 0.4198
2. Calculate the standard error:
SE = sqrt((p̂ * (1 - p̂)) / n)
SE = sqrt((0.4198 * (1 - 0.4198)) / 43) ≈ 0.0726
3. Determine the critical value corresponding to a 90% confidence level. This can be obtained from the Z-table or using a calculator. For a 90% confidence level, the critical value is approximately 1.645.
4. Calculate the margin of error:
ME = critical value * SE
ME = 1.645 * 0.0726 ≈ 0.1195
5. Construct the confidence interval:
Confidence Interval = p̂ ± ME
Confidence Interval = 0.4198 ± 0.1195
Confidence Interval ≈ (0.3003, 0.5393)
Therefore, the 90 percent confidence interval for the proportion of all depositors who ask for cash back is approximately (0.3003, 0.5393).
(b) Checking the Normality Assumption:
To check the normality assumption, we need to ensure that both np and n(1-p) are greater than 5, where np is the number of successes and n(1-p) is the number of failures.
In this case:
np = n * p̂ = 43 * 0.4198 ≈ 18
n(1-p) = n * (1-p̂) ≈ 43 * (1-0.4198) ≈ 25
Since both np and n(1-p) are greater than 5, we can reasonably assume that the normality assumption is met for constructing the confidence interval.
Note: If either np or n(1-p) were less than 5, additional techniques, such as using the binomial distribution or applying the continuity correction, may be necessary to account for the violation of the normality assumption.