Question

A large​ family-held department store had the business objective of improving its response to complaints. The variable of interest was defined as the number of days between when the complaint was made and when it was resolved. Data were collected from 40 complaints that were made in the last year. Use the data to complete parts​ (a) through​ (d) below.

47 13 8 27 43 159
17
49
20
3
105
19
3
64
88
22
30
120
49
102
2
15
28
16
63
29
46
66
10
29
20
48
2
25
25
16
28
86
18
41

Click the icon to view the data table.

a. Construct a

9595​%

confidence interval estimate for the population mean number of days between the receipt of a complaint and the resolution of the complaint.The

9595​%

confidence interval estimate is from

28.728.7

days to

51.451.4

days.

​(Round to one decimal place as​ needed.)

b. What assumption must you make about the population distribution in order to construct the confidence interval estimate in​ (a)?

A.

The number of complaints per day is normally distributed.

B.

The number of days to resolve complaints follows the t distribution.

C.

The number of days to resolve complaints is normally distributed.

Your answer is correct.

D.

The number of complaints per day follows the t distribution.

c. Do you think that the assumption needed in order to construct the confidence interval estimate in​ (a) is​ valid? Explain.

A.

​No, the data suggest the population distribution is skewed to the right.

This is the correct answer.

B.

​Yes, the data suggest the population distribution approximately follows the t distribution.

C.

​Yes, the data suggest the population distribution is approximately normal.

Your answer is not correct.

D.

​No, the data suggest the population distribution is skewed to the left.

d. What effect might your conclusion in​ (c) have on the validity of the results in​ (a)?

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =40.025
standard deviation, s =35.458
sample size, n =40
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 35.458/ sqrt ( 40) )
= 5.6
II.
margin of error = t α/2 * (stanadard 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 = 39 d.f is 2.023
margin of error = 2.023 * 5.6
= 11.3
III.
CI = x ± margin of error
confidence interval = [ 40.025 ± 11.3 ]
= [ 28.7 , 51.4 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =40.025
standard deviation, s =35.458
sample size, n =40
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 2.023
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 = [ 40.025 ± t a/2 ( 35.458/ Sqrt ( 40) ]
= [ 40.025-(2.023 * 5.6) , 40.025+(2.023 * 5.6) ]
= [ 28.7 , 51.4 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 28.7 , 51.4 ] 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

b.
option:C
The number of days to resolve complaints is normally distributed
c.
the assumption needed in order to construct the confidence interval estimate in​ (a) is​ valid
option:B
​Yes, the data suggest the population distribution approximately follows the t distribution
d.
no,
effect might your conclusion in​ (c) have on the validity of the results in​ (a)