Problem 4) A restaurant server recently switched restaurants, from one where cus

Question

Problem 4) A restaurant server recently switched restaurants, from one where customers tipped 20% to one that was brand new. He wondered if his average percentage tips at the new place would be different. The data for 20 randomly selected patrons are in the excel file, tab tips. d) Conduct a significance test to see if the mean percentage tip at the new restaurant would be different than 20%. Use "20", not "O.20." Use the sample mean and sample standard deviation that you calculated previously to calculate the test statistic. Be sure to state the conclusion in plain English. Calculate a 95% confidence interval for the mean tip at the new restaurant. with a quantitative response variable, a 95% confidence interval leads to a two-tailed significance test with 0.05 as our cutoff. If we reject Ho, then we would not expect Ho to be inside the confidence interval: it is not a plausible value for . Look at your confidence interval. Does it contain "20"? Look at the conclusion of the significance test. Are they consistent? e) f) Standard deviation = 1.272 Mean = 18.395 N=20

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =18.395
standard deviation, s =1.272
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.272/ sqrt ( 20) )
= 0.2844
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
margin of error = 2.093 * 0.2844
= 0.5953
III.
CI = x ± margin of error
confidence interval = [ 18.395 ± 0.5953 ]
= [ 17.7997 , 18.9903 ]
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DIRECT METHOD
given that,
sample mean, x =18.395
standard deviation, s =1.272
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 18.395 ± t a/2 ( 1.272/ Sqrt ( 20) ]
= [ 18.395-(2.093 * 0.2844) , 18.395+(2.093 * 0.2844) ]
= [ 17.7997 , 18.9903 ]
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interpretations:
1) we are 95% sure that the interval [ 17.7997 , 18.9903 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

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