Question

Of 58 bank customers depositing a check, 15 received some cash back. (a) Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back. (Round your answers to 4 decimal places.)

Answer 1

Answer:

0.1640 ≤ p ≤0.3532

Explanation:

Sample size (N) = 58 customers

Z-score for 90% interval (Z) = 1.645

The proportion of customers that received some cash back is given by:

$p=\frac{15}{58}\\p=0.2586$

The confidence interval for the proportion of all depositors who ask for cash back is determined by the following relationship:

$p\pm Z*\sqrt{\frac{p*(1-p)}{N}}$

Applying the given data:

$0.2586\pm 1.645*\sqrt{\frac{0.2586*(1-0.2586)}{58}}\\0.2586\pm0.0946$

The upper and lower limits are:

$L=0.2586-0.0946\\L =0.1640 \\L=0.2586-0.0946\\L = 0.3532$

Using a 90% confidence level, the interval for the proportion (p) is:

0.1640 ≤ p ≤0.3532