Of 41 bank customers depositing a check, 22 received some cash back. Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back. (Round your answers to 4 decimal places.)

Respuesta :

Answer:

CI: {0.4085; 0.6647}

Step-by-step explanation:

The confidence interval for a proportion (p) is given by:

[tex]p \pm z*\sqrt{\frac{(1-p)*p}{n} }[/tex]

Where n is the sample size, and z is the z-score for the desired confidence interval. The score for a 90% confidence interval is 1.645. The proportion of depositors who ask for cash back is:

[tex]p=\frac{22}{41}=0.536585[/tex]

Thus the confidence interval is:

[tex]0.536585 \pm 1.645*\sqrt{\frac{(1-0.536585)*0.536585}{41}}\\0.536585 \pm 0.128109\\L=0.4085\\U=0.6647[/tex]

The confidence interval for the proportion of all depositors who ask for cash back is CI: {0.4085; 0.6647}