A Teclar, Inc. call center requires services operators to answer telephone calls
ID: 3057247 • Letter: A
Question
A Teclar, Inc. call center requires services operators to answer telephone calls from customers in an average time of 0.15 minute or less. A sample of 36 actual operator times was drawn, and the results are given in the following table. In addition, operators are expected to determine customer needs and within 0.55 minute. Another sample of 36 times was taken for this job component and is also given in the table. If these variables can be considered to be independent, is the average time take to perform each component statistically different from the standard?
component
Mean Time
Standard Deviation
Answer
0.1523
0.0183
Service
0.5790
0.0902
Is this the data?
component
Mean Time
Standard Deviation
Answer
0.1523
0.0183
Service
0.5790
0.0902
Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 0.15
Alternative hypothesis: > 0.15
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.00305
DF = n - 1 = 36 - 1
D.F = 35
t = (x - ) / SE
t = 0.75
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the t statistic having 35 degrees of freedom greater than 0.75.
Thus, the P-value = 0.23
Interpret results. Since the P-value (0.23) is greater than the significance level (0.05), we cannot reject the null hypothesis.
The average time take to answer call is not statistically different from the standard.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 0.55
Alternative hypothesis: > 0.55
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.01503
DF = n - 1 = 36 - 1
D.F = 35
t = (x - ) / SE
t = 1.93
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the t statistic having 35 degrees of freedom greater than 1.93.
Thus, the P-value = 0.031
Interpret results. Since the P-value (0.031) is less than the significance level (0.05), we have to reject the null hypothesis.
The average time take to provide service is statistically different from the standard.
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