Of 57 bank customers depositing a check, 25 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.) The 90% confidence interval is from to

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Find the confidence interval
Of 57 bank customers depositing a check, 25 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.)
The 90% confidence interval is from
to
(b) May normality of p be assumed?
Yes
No
Transcribed Image Text:Of 57 bank customers depositing a check, 25 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.) The 90% confidence interval is from to (b) May normality of p be assumed? Yes No
Expert Solution
Step 1

Given, Total number of persons, n = 57

  number of persons who received cashback is = 25

The proportions of persons who received cashback is, p^ = 25/57  = 0.4386

Since np >10 sampling distribution of sampling proportion is approximately normal. Therefore, a 90% confidence interval for the true proportion of persons who received cashback is given by

             p^ ± 1.645p^(1-p^)/n= 0.4386 ± 1.6450.4386×0.5614/57= 0.4386 ± 0.1081= 0.3305 to 0.5467

 

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