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In: Statistics and Probability

Question 1. Using the two-sample hypothesis test comparing old call time to new call time at...

Question 1. Using the two-sample hypothesis test comparing old call time to new call time at a .05 level of significance, discuss the hypothesis test assumptions and tests used. Provide the test statistic and p-value in your response. Evaluate the results of the hypothesis test with the scenario. Provide recommendations for the vice president.

Question 2. 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.

Old Call Time New Call Time Shift Quality Coded Quality

6.5 5.2 AM Y 0 Y= CORRECT

6.5 5.2 AM Y 0 N=INCORRECT

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

Solutions

Expert Solution

t-Test: Two-Sample Assuming Equal Variances
Variable 1 Variable 2
Mean 11.08333 8.853333
Variance 11.32103 7.327606
Observations 150 150
Pooled Variance 9.324318
Hypothesized Mean Difference 0
df 298
t Stat 6.324511
P(T<=t) one-tail 4.64E-10
t Critical one-tail 1.649983
P(T<=t) two-tail 9.28E-10
t Critical two-tail 1.967957
t-Test: Two-Sample Assuming Unequal Variances
Variable 1 Variable 2
Mean 11.08333 8.853333
Variance 11.32103 7.327606
Observations 150 150
Hypothesized Mean Difference 0
df 285
t Stat 6.324511
P(T<=t) one-tail 4.9E-10
t Critical one-tail 1.650218
P(T<=t) two-tail 9.79E-10
t Critical two-tail 1.968323

From both cases i.e. with equal variances or unequal variances the p-value<0.05, hence we reject the null hypothesis and conclude that there is a significant difference in averages of old call time to new call time.

Q2.  

Here new call time is a dependent variable and coded quality is independent variable then the regression equation of new call time on coded quality is

New call time=8.8448+0.0802 Coded quality

The detailed outputs are written below:

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.009179018
R Square 8.42544E-05
Adjusted R Square -0.006671933
Standard Error 2.715970464
Observations 150
ANOVA
df SS MS F Significance F
Regression 1 0.09199005 0.09199 0.012471 0.911234568
Residual 148 1091.721343 7.376496
Total 149 1091.813333
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 8.844776119 0.234624113 37.69764 9.08E-78 8.381130132 9.308422106 8.381130132 9.308422106
X Variable 1 0.080223881 0.718386697 0.111672 0.911235 -1.339396231 1.499843992 -1.339396231 1.499843992

R-square=0.0084% i.e. 0.0084% of total variation in new call time is explained by the regression equation and it is very poor. Hence for this data, the regression equation is fitted well and this is also seen from the ANOVA table since p-value corresponding to regression is very high i.e. 0.9112>0.05 and the regression coefficient is insignificantly different from zero since its p-value=0.7184>0.05.

Predicted call time if there is a defect=8.8448+0.0802 x1=8.925=8.93

orchestra answered 2 years ago
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