Using the New Call Center Data, provide a summary report for the vice president

Question

Using the New Call Center Data, provide a summary report for the vice president including the following information

Using new call time and coded quality, develop a prediction equation for new call time. Evaluate the model and discuss the coefficient of determination, significance, and use the prediction equation to predict a call time if there is a defect.

Evaluate whether the new call time meets customer specification. As stated in a previous lesson, customers indicated they did not want a call time longer than 7.5 minutes. Assume a standard deviation of 2 min is acceptable. Is the call center now meeting the customer specifications? If not where is the specification not being met? Explain your answers.

Old Call Time New Call Time Shift Quality Coded Quality 6.5 5.2 AM Y 0 6.5 5.2 AM Y 0 6.5 5.2 AM Y 0 6.5 5.2 AM Y 0 7 5.6 AM Y 0 7 5.6 AM Y 0 7 5.6 AM Y 0 7 5.6 AM N 1 7 5.6 AM Y 0 8 6.4 AM Y 0 8 6.4 AM Y 0 8.5 6.8 AM Y 0 8.5 6.8 AM Y 0 9 7.2 AM Y 0 9 7.2 AM Y 0 9 7.2 AM N 1 9 7.2 AM N 1 9.5 7.6 AM N 1 9.5 7.6 AM Y 0 9.5 7.6 AM Y 0 10 8 AM Y 0 10 8 AM Y 0 10 8 AM Y 0 10 8 AM Y 0 10 8 AM Y 0 10.5 8.4 AM Y 0 10.5 8.4 AM Y 0 10.5 8.4 AM Y 0 10.5 8.4 AM Y 0 10.5 8.4 AM Y 0 10.5 8.4 AM Y 0 10.5 8.4 AM Y 0 10.5 8.4 AM Y 0 11.5 9.2 AM Y 0 11.5 9.2 AM Y 0 11.5 9.2 AM Y 0 12 9.6 AM Y 0 12 9.6 AM Y 0 12 9.6 AM N 1 12 9.6 AM Y 0 12.5 10 AM Y 0 12.5 10 AM N 1 13 10.4 AM Y 0 13 10.4 AM Y 0 13.5 10.8 AM Y 0 15.5 12.4 AM Y 0 16 12.8 AM Y 0 16.5 13.2 AM Y 0 17 13.6 AM Y 0 18 14.4 AM Y 0 6 4.8 PM Y 0 9 7.2 PM N 1 9.5 7.6 PM N 1 10 8 PM Y 0 10.5 8.4 PM Y 0 10.5 8.4 PM Y 0 11 8.8 PM Y 0 11 8.8 PM Y 0 11 8.8 PM Y 0 11 8.8 PM Y 0 11.5 9.2 PM Y 0 11.5 9.2 PM Y 0 11.5 9.2 PM Y 0 12 9.6 PM Y 0 12 9.6 PM Y 0 12 9.6 PM Y 0 12 9.6 PM Y 0 12 9.6 PM Y 0 12 9.6 PM Y 0 12 9.6 PM Y 0 12.5 10 PM Y 0 12.5 10 PM Y 0 12.5 10 PM Y 0 12.5 10 PM Y 0 13 10.4 PM N 1 13 10.4 PM N 1 13.5 10.8 PM Y 0 13.5 10.8 PM Y 0 14 11.2 PM Y 0 14 11.2 PM Y 0 14 11.2 PM Y 0 14 11.2 PM N 1 14 11.2 PM Y 0 14.5 11.6 PM Y 0 14.5 11.6 PM Y 0 14.5 11.6 PM N 1 15 12 PM N 1 15 12 PM Y 0 15.5 12.4 PM N 1 16 12.8 PM Y 0 16.5 13.2 PM Y 0 17 13.6 PM Y 0 17.5 14 PM Y 0 18 14.4 PM Y 0 18 14.4 PM Y 0 18 14.4 PM Y 0 18.5 14.8 PM Y 0 19 15.2 PM Y 0 19.5 15.6 PM Y 0 19.5 15.6 PM Y 0 5.25 4.2 AM Y 0 5.25 4.2 PM Y 0 5.25 4.2 AM Y 0 5.25 4.2 PM Y 0 5.75 4.6 AM Y 0 5.75 4.6 PM Y 0 5.75 4.6 AM Y 0 5.75 4.6 PM Y 0 5.75 4.6 AM Y 0 6.75 5.4 PM Y 0 6.75 5.4 AM Y 0 7.25 5.8 PM Y 0 7.25 5.8 AM Y 0 7.75 6.2 PM Y 0 7.75 6.2 AM Y 0 7.75 6.2 PM N 1 7.75 6.2 AM Y 0 8.25 6.6 PM Y 0 8.25 6.6 AM N 1 8.25 6.6 PM Y 0 8.75 7 AM Y 0 8.75 7 PM Y 0 8.75 7 AM Y 0 8.75 7 PM Y 0 8.75 7 AM Y 0 9.25 7.4 PM Y 0 9.25 7.4 AM Y 0 9.25 7.4 PM Y 0 9.25 7.4 AM Y 0 9.25 7.4 PM Y 0 9.25 7.4 AM Y 0 9.25 7.4 PM Y 0 9.25 7.4 AM Y 0 10.25 8.2 PM Y 0 10.25 8.2 AM Y 0 10.25 8.2 PM Y 0 10.75 8.6 AM Y 0 10.75 8.6 PM Y 0 10.75 8.6 AM Y 0 10.75 8.6 PM Y 0 11.25 9 AM Y 0 11.25 9 PM Y 0 11.75 9.4 AM Y 0 11.75 9.4 PM Y 0 12.25 9.8 AM Y 0 14.25 11.4 PM Y 0 14.75 11.8 AM Y 0 15.25 12.2 PM Y 0 15.75 12.6 AM Y 0 16.75 13.4 PM Y 0

Explanation / Answer

First copy given data in Excel as it is. Then copy only that part in which the lables are not included.

Then run following R- code.

a=read.table("clipboard",header=F)
attach(a)
l=lm(V2~V5)
summary(l)

And the output is:

> summary(l)

Call:
lm(formula = V2 ~ V5)

Residuals:
Min 1Q Median 3Q Max
-4.6448 -1.8448 -0.3448 1.4750 6.7552

Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) 8.84478 0.23462 37.698 <2e-16 ***
V5 0.08022 0.71839 0.112 0.911   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.716 on 148 degrees of freedom
Multiple R-squared: 8.425e-05, Adjusted R-squared: -0.006672
F-statistic: 0.01247 on 1 and 148 DF, p-value: 0.9112

Here the equation is:

New Call Time = (0.08022)*Coded Quality + 8.84478

Multiple R-squared: 8.425e-05

p-value: 0.9112

Thus the model is not significant.

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