sample mean for amount spent = 38.54
sample standard deviation for amount spent = 7.62
sample size is 60.
18 out of 60 chose dessert.
since we are working off the sample mean and sample standard deviation, we need to use the t-score rather than the z-score.
a. Construct a 95% confidence interval estimate for the population mean amount spent per customer in the restaurant.
standard error = standard deviation / square root of sample size = 7.62/sqrt(60) equals .983738 rounded to 6 decimal places.
critical t-score at 95% confidence interval with 59 degrees of freedom is equal to t = plus or minus 2.0.
on the high side of the 95% confidence interval, the formula of t = (x-m)/s becomes 2.0 = (x-38.54)/.983738.
solve for x to get x = 2.0 * .983738 + 38.54 = 40.5075 rounded to 4 decimal places.
on the low side of the 95% confidence interval, the formula of t = (x-m)/s becomes -2.0 = (x-38.54)/.983738.
solve for x to get x = -2.0 * .983738 + 38.54 = 36.5725 rounded to 4 decimal places.
your 95% confidence interval is from 36.5725 to 40.5075.
b. Construct a 97% confidence interval estimate for the population proportion of customers who purchase dessert.
p = 18/60 = .3
q = 1-p = .7
standard error = sqrt(p*q/n)
p = .3
q = .7
n = 60
formula becomes standard error = sqrt(.3*.7/60) = .059161 rounded to 6 decimal places.
critical t-score with 59 degrees of freedom at 97% confidence interval becomes t = plus or minus 2.2238 rounded to 4 decimal places.
on the high side of the confidence interval, t = (x-m)/s becomes 2.2238 = (x-.3) / .059161.
solve for x to get x = 2.2238 * .059161 + .3 = .4316 rounded to 4 decimal places.
on the low side of the confidence interval, t = (x-m)/s becomes -2.2238 = (x-.3) / .059161.
solve for x to get x = -2.2238 * .059161 + .3 = .1684 rounded to 4 decimal places.
your 97% confidence interval is equal to .1684 to .4316
i believe part c and d will require the z-score.
i'm beginning to think that part a and b also needed to assume the z-score.
i can redo part a and b with z-scores if that's what you need.
i'll leave part and b as is unless you tell me you need z-scores rather than t -scores.
for part c and beyond i'll assume z-scores.
here goes part c and beyond.
Jack, the owner of a competing restaurant, wants to conduct a similar survey in his restaurant. Jack does not have access to the information that Scarlett and Heather have obtained from the survey they conducted. Answer the following questions:
c. What sample size is needed to have 95% confidence of estimating the population mean amount spent in her restaurant with a margin of error of$1.50, assuming that the standard deviation is estimated to be $8?
critical z-score at 95% confidence interval = plus or minus z = 1.96.
formula used is z = (x-m)/s.
(x-m) is the margin of error which is equal to 1.5
on the high side of the confidence interval, you get z = (x-m)/s which becomes 1.96 = 1.5 / s
solve for s to get s = 1.5/1.96.
since s = standard deviation / sqrt(sample size) and since standard deviation is equal to 8, then you get 8/sqrt(sample size) = 1.5/1.96
solve for sqrt(sample size) to get sqrt(sample size) = 8 * 1.96 / 1.5 = 10.45333.....
solve for sample size to get sample size = that squared = 109.27....
since sample size has to be an integer, then round up to sample size to 110.
standard error becomes 8/sqrt(110) = .762770 rounded to 6 decimal places.
on the high side of the confidence interval, the z-score formula becomes 1.96 = (x-m) / .762770 rounded to 6 decimal places.
solve for (x-m) to get (x-m) = 1.96 * .762770 = 1.495 rounded to 3 decimal places.
that's less than 1.5 as required.
on the low side of the confidence interval, the z-score formula becomes -1.96 = (x-m) / A.
solve for (x-m) to get (x-m) = -1.96 * .762770 = -1.495 rounded to 3 decimal places.
that's greater than -1.5 as required.
the mean can be anything and you will get a margin of error of less than 1.5.
for example, assume the mean is 38.54.
your 95% confidence interval will be 38.54 - 1.495 to 38.5 + 1.495 which is equal to 37.045 to 40.035.
here's what it looks like on a z-score calculator.
d. How many customers need to be selected to have 92% confidence of estimating the population proportion of customers who purchase dessert with a margin of error 0.04?
critical z-score for 92% confidence interval is equal to z = plus or minus 1.750686071.
we'll round that to 1.7507.
z-score formula is z = (x-m)/s
(x-m) is the margin of error which is assumed to be .04.
z-score formula on the high side of the confidence interval of z = (x-m)/s which becomes 1.7507 = (x-m)/s
since (x-m) is assumed to be .04, the formula becomes 1.7507 = .04/s
solve for s to get s = .04/1.7507 = .022848 rounded to 6 decimal places.
in a proportion type study, the standard error is equal to sqrt(p * q / n)
since s is the standard error, then .022848 = sqrt(p * q / n)
since the population proportion is assumed to be .3, then we get:
.022848 = sqrt(.3*.7/n).
square both sides of that to get:
.022848^2 = .3*.7/n
solve for n to get:
n = .3*.7/.022848^2 = 402.27...
since n has to be an integer, then n is equal to the next higher integer = 403.
when n = 403, the standard error becomes sqrt(.3*.7/403) = .022827 rounded to 6 decimal places.
the high side of the confidence interval formula becomes 1.7507 = (x-m) / .022827.
solve for (x-m) to get (x-m) = 1.7507 * .022827 = .03996 which is less than .04 as required.
it won't be exactly .04 unless the sample size is not an integer.
it's the largest the margin of error can be when the margin of error has to be less than .04.
here's what it looks like on a z-score calculator.
e. Based on your answers to (c) and (d), how large a sample should Jack take?
he needs to have the margin of error less than 1.5 for part c and less than .04 for part d.
he would need to use the larger sample size of 403 for both, if he only has 1 choice.
that way, the margin of error will be less than 1.5 for part c and less than .04 for part d.