In: Math
Most companies have increased their dependence on computers and software. As a result, more employee time is spent on the telephone with technical support for the software. A sample of 8 times spent on the phone with technical support yielded the following data:
| 
 Time spent on phone (in minutes)  | 
| 
 11  | 
| 
 9  | 
| 
 9  | 
| 
 8  | 
| 
 12  | 
| 
 13  | 
| 
 11  | 
| 
 14  | 
Construct a 98 percent confidence interval estimate of the true mean population time that is spent by employees on the telephone with technical support for the software. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Negative values should be indicated by a minus sign. Report your answers to 2 decimal places, using conventional rounding rules.
Answer: $ _____≤ (Click to select) ≤ $____
Solution:
Given: A sample of 8 times spent on the phone with technical support yielded the following data:
| 
 Time spent on phone (in minutes)  | 
| 
 11  | 
| 
 9  | 
| 
 9  | 
| 
 8  | 
| 
 12  | 
| 
 13  | 
| 
 11  | 
| 
 14  | 
We have to construct a 98 percent confidence interval estimate of the true mean population time that is spent by employees on the telephone with technical support for the software.
Formula:
where



Thus we need to make following table:
| x : Time spent on phone (in minutes) | x^2 | 
| 11 | 121 | 
| 9 | 81 | 
| 9 | 81 | 
| 8 | 64 | 
| 12 | 144 | 
| 13 | 169 | 
| 11 | 121 | 
| 14 | 196 | 
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![]()  | 
Thus







and
tc is t critical value for c = 98% confidence level
Thus find two tail area = 1 - c = 1 - 0.98 = 0.02
df = n - 1 = 8 - 1 = 7
Thus from t table for df = 7 and two tail area = 0.02
tc = 2.998
Thus margin of error = E is



Thus




Thus a 98% confidence interval estimate of the true mean
population time that is spent by employees on the telephone with
technical support for the software is between