In: Statistics and Probability
The managers of the center had estimated from past surveys that 28% of calls are conducted in the morning hours and had scheduled the necessary agents accordingly. However, there are concerns that this percentage has increased recently. What is your conclusion on the basis of this sample? Show the process in the excel.
Time | Region | Gender | Type of Call |
Morning | West | Female | Complaint |
Morning | West | Female | Submit request |
Afternoon | West | Female | Submit request |
Afternoon | NorthEast | Female | General Info |
Afternoon | West | Male | Submit request |
Afternoon | NorthEast | Female | Submit request |
Evening | West | Male | General Info |
Evening | South | Male | Complaint |
Evening | West | Male | General Info |
Morning | MidWest | Female | General Info |
Morning | NorthEast | Female | Submit request |
Afternoon | South | Male | Submit request |
Morning | NorthEast | Male | Complaint |
Afternoon | NorthEast | Male | General Info |
Morning | West | Female | General Info |
Afternoon | West | Male | General Info |
Afternoon | South | Female | Submit request |
Afternoon | West | Female | Submit request |
Morning | South | Male | General Info |
Ho : p = 0.28
H1 : p > 0.28
(Right tail test)
Level of Significance, α =
0.05
Number of Items of Interest, x =
7
Sample Size, n = 19
Sample Proportion , p̂ = x/n =
0.3684
Standard Error , SE = √( p(1-p)/n ) =
0.1030
Z Test Statistic = ( p̂-p)/SE = ( 0.3684
- 0.28 ) / 0.1030
= 0.8584
p-Value = 0.195337173 [Excel
function =NORMSDIST(-z)
Decision: p value>α ,do not reject null
hypothesis
There is not enough evidence that morning calls have been
increaseed considerably.
Thanks in advance!
revert back for doubt
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