Question

Dr. Mack Lemore, an expert in consumer behavior, wants to estimate the average amount of money that people spend in thrift shops. He takes a small sample of 8 individuals and asks them to report how much money they had in their pockets the last time they went shopping at a thrift store. Here are the data:


19, 21, 14, 19, 20, 17, 25, 15.

Find the upper bound of a 95% confidence interval for the true mean amount of money individuals carry with them to thrift stores, to two decimal places. Take all calculations toward the final answer to three decimal places.

Answer 1

TRADITIONAL METHOD
given that,
sample mean, x =18.75
standard deviation, s =3.4949
sample size, n =8
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 3.4949/ sqrt ( 8) )
= 1.236
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 7 d.f is 2.365
margin of error = 2.365 * 1.236
= 2.922
III.
CI = x ± margin of error
confidence interval = [ 18.75 ± 2.922 ]
= [ 15.828 , 21.672 ]
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DIRECT METHOD
given that,
sample mean, x =18.75
standard deviation, s =3.4949
sample size, n =8
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 7 d.f is 2.365
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 = [ 18.75 ± t a/2 ( 3.4949/ Sqrt ( 8) ]
= [ 18.75-(2.365 * 1.236) , 18.75+(2.365 * 1.236) ]
= [ 15.828 , 21.672 ]
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interpretations:
1) we are 95% sure that the interval [ 15.828 , 21.672 ] 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
Answer:
upper bound confidence interval is 21.67