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Of 54 bank customers depositing a check, 17 received some cash back. Construct a 90% confidence...

Question:

Of 54 bank customers depositing a check, 17 received some cash back.

Construct a 90% confidence interval for the proportion of all depositors who ask for cash back. (Round answers to 4 decimal places.)

Confidence Interval for Proportion:

Confidence intervals are a method of estimation of a parameter.

The confidence interval for p is given by:

{eq}\\ \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\ \\ Where\\\\ \hat{p}=\frac{x}{n}\\ \alpha= Level~ of~Confidence {/eq}

A confidence interval of a proportion gives certain confidence that the population proportion will lie within this interval.

Answer and Explanation:

According to question,

p: Proportion of depositors who ask for cashback.

{eq}\\ \hat{p}=\frac{x}{n}\\ \\ \hat{p}=\frac{17}{54}\\ \\ =0.3148\\ {/eq}

The 90% confidence interval for p is:

{eq}\\ \alpha=0.1\\ Z_{\alpha}=1.645\\ Therefore\\ CI~ is\\ \\ \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\ \\ 0.3148 \pm 1.645\sqrt{\frac{0.3148(1-0.3148)}{54}}\\ \\ \mathbf{(0.2108 , .4188)} {/eq}


Learn more about this topic:

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Finding Confidence Intervals for Proportions: Formula & Example

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 8
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