1.Below are 2 sample t test and test for equal variance for an experiment lookin

Question

1.Below are 2 sample t test and test for equal variance for an experiment looking at an alternative process condition.

A)What can you conclude about the mean and the variation between the two process conditions?

Two-Sample T-Test and CI: C1, C2 Two- sample T for C1 vs C2 SE N Mean St Dev Mean 580 415 47 C1 78 29 C2 4 462.6 58.7 mu (C2) mu (C1) Difference Estimate for difference 117.8 95 CI for difference (4.7, 230.9) T-Test of difference 0 (vs not T-Value 2.13 P-Value 0.042 DE 30 Test for Equal Variances for C1, C2 F-Test Test Statistic 49.90 P-Value 0.008 Levene's Test Test Statistic 5.08 P-Value 0.027 95% Bonferroni Confidence Intervals for StDevs 1000 1750 Data

Explanation / Answer

A) Based on mean and standard deviation, we should calculate coeffiecient of variation for both C1 and C2

CV of C1 = [SD / mean]* 100 = [415/580]*100 = 71.55%

CV of C2 = [SD / mean]* 100 = [58.7/462.6]*100 = 12.69%

   Here we can infer that Second process is more consistent(less variation) than process one.

B)Given that p value = 0.042

Based on this we can infer that there is significant difference between two process at 5%,

since p value = 0.042 < 0.05

But we can infer there is no significant difference between two process at 1% and 2%,

Hence statistical statements are different with different significant levels and general decisions.

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