8. A customer feels that the response time while being placed on hold awaiting a

Question

8. A customer feels that the response time while being placed on hold awaiting a customer service representative is well beyond the 10 minutes stated by an automated message indicating. The customer speaks with the representative's supervisor after conducting her business. The supervisor then reviews the logs containing the wait time for the last ten customers 16 9 3 8 18 18 5 18 12 15 6 a. Calculate the mean, median, mode, variance, and standard deviation for this data set b Estimate the true population mean that a customer must wait on hold when contacting a customer service representative with 95% confidence. c.Conduct a hypothesis test at a 0.02 level of significance to see if the customer is correct and identify the corresponding p-value for this sample data.

Explanation / Answer

8.

a.
given data,
16 , 9 , 3 , 8 , 18 , 18 , 5 , 18 , 12 , 15 , 6
arranged in ascending order
3,5,6,8,9,12,15,16,18,18,18
then median is middle most value in the given data so that it is 12
mode is most frequently repeated value in the given data
mode is 18
Mean = Sum of observations/ Count of observations
Mean = (16 + 9 + 3 + 8 + 18 + 18 + 5 + 18 + 12 + 15 + 6 / 11) = 11.6364
Variance
Step 1: Add them up
16 + 9 + 3 + 8 + 18 + 18 + 5 + 18 + 12 + 15 + 6 = 128
Step 2: Square your answer
128*128 =16384
…and divide by the number of items. We have 11 items , 16384/11 = 1489.4545
Set this number aside for a moment.
Step 3: Take your set of original numbers from Step 1, and square them individually this time
16^2 + 9^2 + 3^2 + 8^2 + 18^2 + 18^2 + 5^2 + 18^2 + 12^2 + 15^2 + 6^2 = 1812
Step 4: Subtract the amount in Step 2 from the amount in Step 3
1812 - 1489.4545 = 322.5455
Step 5: Subtract 1 from the number of items in your data set, 11 - 1 = 10
Step 6: Divide the number in Step 4 by the number in Step 5. This gives you the variance
322.5455 / 10 = 32.2545
Step 7: Take the square root of your answer from Step 6. This gives you the standard deviation
5.6793
Answers:
a.
mean = 11.6364
median = 12
mode = 18
variance = 32.2545
standard deviation = 5.6793

b.
TRADITIONAL METHOD
given that,
sample mean, x =11.6364
standard deviation, s =5.6793
sample size, n =11
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 5.6793/ sqrt ( 11) )
= 1.712
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 = 10 d.f is 2.228
margin of error = 2.228 * 1.712
= 3.815
III.
CI = x ± margin of error
confidence interval = [ 11.6364 ± 3.815 ]
= [ 7.821 , 15.452 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =11.6364
standard deviation, s =5.6793
sample size, n =11
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 10 d.f is 2.228
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 = [ 11.6364 ± t a/2 ( 5.6793/ Sqrt ( 11) ]
= [ 11.6364-(2.228 * 1.712) , 11.6364+(2.228 * 1.712) ]
= [ 7.821 , 15.452 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 7.821 , 15.452 ] 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

c.
Given that,
population mean(u)=10
sample mean, x =11.6364
standard deviation, s =5.6793
number (n)=11
null, Ho: =10
alternate, H1: >10
level of significance, = 0.02
from standard normal table,right tailed t /2 =2.359
since our test is right-tailed
reject Ho, if to > 2.359
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =11.6364-10/(5.6793/sqrt(11))
to =0.956
| to | =0.956
critical value
the value of |t | with n-1 = 10 d.f is 2.359
we got |to| =0.956 & | t | =2.359
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.9556 ) = 0.18089
hence value of p0.02 < 0.18089,here we do not reject Ho
ANSWERS
---------------
null, Ho: =10
alternate, H1: >10
test statistic: 0.956
critical value: 2.359
decision: do not reject Ho
p-value: 0.18089

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