In: Statistics and Probability
New | Current | |
Sample Size | 8 | 8 |
Sample Mean | 558.7500 | 604.3750 |
Sample Standard Deviation | 49.1899 | 44.5162 |
df | ||
Pooled Sample Standard Deviation | 46.9113 | |
Confidence Interval (in terms of New - Current) | ||
Confidence Coefficient | 0.99 | |
Lower Limit | ||
Upper Limit | ||
Hypothesis Test (in terms of New - Current) | ||
Hypothesized Value | ||
Test Statistic | ||
p-value (Lower Tail) | 0.0361 | |
p-value (Upper Tail) | ||
p-value (Two Tail) |
New | Current | |
Sample Size | 8 | 8 |
Sample Mean | 558.7500 | 604.3750 |
Sample Standard Deviation | 49.1899 | 44.5162 |
df | 13 | |
Confidence Interval (in terms of New - Current) | ||
Confidence Coefficient | 0.99 | |
Lower Limit | ||
Upper Limit | ||
Hypothesis Test (in terms of New - Current) | ||
Hypothesized Value | ||
Test Statistic | ||
p-value (Lower Tail) | 0.0369 | |
p-value (Upper Tail) | ||
p-value (Two Tail) |
IBM is interested in comparing the average systems analyst project completion time using the current technology and using the new computer software package. So, a statistical consultant for IBM randomly selected 8 projects that used the current technology and 8 projects that used the new computer software package. The completion time (in hours) for each project was recorded and then entered into Excel. Assume the samples are independent and from normal populations with equal variances. Can IBM reject the hypothesis μNew - μCurrent = 0 at α=.05? Based on this paragraph of text, use the correct excel output above to answer the following question.
What is the 99% confidence interval for μCurrent - μNew?
a. |
(-25.0234, 116.2734) |
|
b. |
(-39.5990, 130.8490) |
|
c. |
(-4.6874, 95.9374) |
|
d. |
None of the answers is correct |
|
e. |
(-24.2025, 115.4525) |
The given interval is as:
-0.06436 < p1-p2 < 0.10778
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1
+ 1/n2))
sp = sqrt((((8 - 1)*44.5162^2 + (8 - 1)*49.1899^2)/(8 + 8 -
2))*(1/8 + 1/8))
sp = 23.4556
Given CI level is 0.99, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.977
Margin of Error
ME = tc * sp
ME = 2.977 * 23.4556
ME = 69.827
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc *
sp)
CI = (604.375 - 558.75 - 2.977 * 23.4556 , 604.375 - 558.75 - 2.977
* 23.4556
CI = (-24.2025 , 115.4525)
Option e)