Assignment: 1-1
Problem Definition: Which t-calc formula do we use?
Hypotheses: H0: σTSA2 σ401K2
H1: σTSA2 σ401K2
Decision Rule: If Critical Ratio of F is greater than the Critical Value of F (2.48), reject the null.
Test and CI for Two Variances: TSA, 401K
Statistics
Variable N StDev Variance
TSA 15 709.723 503707.095
401K 15 593.924 352746.067
Ratio of standard deviations = 1.195
Ratio of variances = 1.428
95% One-Sided Confidence Intervals
Lower Bound Lower Bound
Distribution for StDev for Variance
of Data Ratio Ratio
Normal 0.758 0.575
Continuous 0.803 0.644
Tests
Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 14 14 1.43 0.257
Levene's Test (any continuous) 1 28 0.96 0.168
Conclusion:
The F Critical Ratio of 1.43 is less than the F critical value of 2.48, fail to reject the null. Type 2 error may have been happened.
Interpretation:
The variances for the TSA and 401K are equal, so maybe use pooled the variance t test.
Assignment: 1-2
Problem Definition: Is there any difference between the two retirement programs TSA and 401K
Hypotheses: H0: µTSA = µ401K
H1: µTSA µ401K
Decision Rule: If t Critical Ratio is less than t critical value of -2.0484, or greater than t critical value of 2.0484, reject the null.
Test:
Two-Sample T-Test and CI: TSA, 401K
Two-sample T for TSA vs 401K
N Mean StDev SE Mean
TSA 15 2120 710 183
401K 15 1778 594 153
Difference = mu (TSA) - mu (401K)
Estimate for difference: 342
95% CI for difference: (-148, 831)
T-Test of difference = 0 (vs not =): T-Value = 1.43 P-Value = 0.164 DF = 28
Both use Pooled StDev = 654.3902
Conclusion:
1) The T Critical ratio of 1.43 is not greater than the T critical value of 2.0484. Fail to reject the null. Type 2 error may have been happened.
2) P-value of 0.164 is greater than alpha of 0.05, fail to reject the null.
3) Confidence Interval: difference of 342 is within the confidence interval of (-148, 831)
Interpretation:
There is no significant difference in the amount of contributions to the two retirement programs TSA and 401K.
Boxplot of TSA, 401K
Assumption:
The boxplot indicates that the graphed medians falls at the center of the distribution and the arithmetic means are close to the medians for each distribution. Thus, the populations from which the samples were taken are approximately normally distributed. The whiskers are approximately equal in length indicating equal variances for both populations. As a result, the data can be pooled.
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