4. The Federal Trade Commission (FTC) has, for several years, studied the accura

Question

4. The Federal Trade Commission (FTC) has, for several years, studied the accuracy of price scanners used in many retail stores. Grocery stores were found, in general, to have a smaller incidence of scanner errors than non-grocery retail stores. The manager of a Richmond retailer (non-grocery) wishes to investigate scanner errors at his store. A random sample of 101 scanner transactions from last week’s purchases is obtained, and the average transaction error is + $4.32 with a standard deviation of $1.52. In other words, the customers making these 101 transactions were over charged, on average, by amount $4.32.

A.Use the sample data to calculate and interpret a 90%confidence interval for the mean transaction error in the population of scanner transactions that occurred at this store last week. (4+2=6 pts)

B. Determine the margin of error for the interval estimate in a, and explain what it represents about scanner errors. (2 pts)

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =4.32
standard deviation, s =1.52
sample size, n =101
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.52/ sqrt ( 101) )
= 0.151
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 100 d.f is 1.66
margin of error = 1.66 * 0.151
= 0.251
III.
CI = x ± margin of error
confidence interval = [ 4.32 ± 0.251 ]
= [ 4.069 , 4.571 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =4.32
standard deviation, s =1.52
sample size, n =101
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 100 d.f is 1.66
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 = [ 4.32 ± t a/2 ( 1.52/ Sqrt ( 101) ]
= [ 4.32-(1.66 * 0.151) , 4.32+(1.66 * 0.151) ]
= [ 4.069 , 4.571 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 4.069 , 4.571 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

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