As district manager part of your responsibility is to identify reasons for chang

Question

As district manager part of your responsibility is to identify reasons for changes in sales by store. In one of the district stores you have identified that sales have dropped by 15% and you believe the reason is that not enough time is spent on cleaning and maintenance of inventory displayed. Cleaning and maintenance hours are supposed to be reported as after store closing hours. Random samples of cleaning and maintenance hours were surveyed to determine the amount of time spent daily on cleaning and maintenance. The times are shown below. If the times are normally distributed with a variance of 1.20 hours, estimate with 99% confidence the mean daily time spent cleaning and maintaining inventory by store personnel.

Standard Deviation =

Sample Mean =

Stantard Error of the Mean =

Margin of Error =

Upper Confidence Limit =

Lower Confidence Limit =

Explanation / Answer

TRADITIONAL METHOD
given that,
standard deviation, = sqrt(1.20) = 1.095
sample mean, x =5.1111
population size (n)=9
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1.095/ sqrt ( 9) )
= 0.365
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.365
= 0.94
III.
CI = x ± margin of error
confidence interval = [ 5.1111 ± 0.94 ]
= [ 4.171,6.051 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =1.095
sample mean, x =5.1111
population size (n)=9
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 5.1111 ± Z a/2 ( 1.095/ Sqrt ( 9) ) ]
= [ 5.1111 - 2.576 * (0.365) , 5.1111 + 2.576 * (0.365) ]
= [ 4.171,6.051 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [4.171 , 6.051 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 5.1111
standard error =0.365
z table value = 2.576
margin of error = 0.94
confidence interval = [ 4.171 , 6.051 ]

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