SOLUTION: Of 41 bank customers depositing a check, 25 received some cash back.
a)Construct a 90 percent confidence interval for a proportion of all depositors who ask for cash back. (Round
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Question 1197184: Of 41 bank customers depositing a check, 25 received some cash back.
a)Construct a 90 percent confidence interval for a proportion of all depositors who ask for cash back. (Round your answers to 4 decimal places.)
b) May normality of p be assumed?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Part (a)
At 90% confidence, the z critical value is roughly z = 1.645
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 90% confidence level.
You can also use the NORMINV function in a spreadsheet to calculate the critical z value.
On TI83 and TI84 calculators, the function you'll use is called invNorm to calculate the critical z value.
It is found by pressing the button labeled "2nd" in the top left corner. Then hit the VARS key. The invNorm function is the third item.
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x = number of people who got cash back = 25
n = sample size = 41
phat = sample proportion of people who got cash back
phat = x/n
phat = 25/41
phat = 0.6097561 approximately
E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 1.645*sqrt(0.6097561*(1-0.6097561)/41)
E = 0.12531992
L = lower boundary of the confidence interval
L = phat - E
L = 0.6097561 - 0.12531992
L = 0.48443618
L = 0.4844
U = upper boundary of the confidence interval
U = phat + E
U = 0.6097561 + 0.12531992
U = 0.73507602
U = 0.7351
The confidence interval for p in the format (L, U) is (0.4844, 0.7351) when rounding to four decimal places.
The confidence interval in the format L < p < U is 0.4844 < p < 0.7351
An alternative format is which in this case is roughly when rounding to four decimal places.
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Part (b)
x = number of people who got cash back = 25
n = 41 = sample size
phat = x/n = 25/41
n*phat = 41*(25/41) = 25
n*(1-phat) = 41*(1-25/41) = 16
Both results are larger than 5, so normality of p can be assumed.
Some textbooks require the threshold of 5 to be 10 instead. In this case, both 25 and 16 are larger than 10, so they are both over the threshold here as well.
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