Need help in TUKEY HSD, unserue on how it\'s done. Please see below I\'m basing

Question

Need help in TUKEY HSD, unserue on how it's done. Please see below I'm basing that I do need to conduct a TUKEY since the P-Value is below 0.5

Age Happiness Engagement Mean 32.02 Mean 7.4 Mean 7.64 Standard Error 0.61378052 Standard Error 0.2 Standard Error 0.175569046 Median 31.5 Median 8 Median 8 Mode 29 Mode 8 Mode 8 Standard Deviation 4.3400837 Standard Deviation 1.414213562 Standard Deviation 1.241460628 Sample Variance 18.8363265 Sample Variance 2 Sample Variance 1.54122449 Kurtosis -1.08642015 Kurtosis -0.24983717 Kurtosis 0.859898343 Skewness 0.18868966 Skewness -0.71251831 Skewness -0.93445577 Range 15 Range 5 Range 6 Minimum 25 Minimum 4 Minimum 4 Maximum 40 Maximum 9 Maximum 10 Sum 1601 Sum 370 Sum 382 Count 50 Count 50 Count 50 Anova: Single Factor SUMMARY Groups Count Sum Average Variance Age 50 1601 32.02 18.83632653 Happiness 50 370 7.4 2 Engagement 50 382 7.64 1.54122449 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 20009.7733 2 10004.88667 1341.284396 3.94807E-95 3.057620652 Within Groups 1096.5 147 7.459183673 Total 21106.2733 149

Explanation / Answer

Here as the p - value is less than 0.05 so here

Tukey's HSD = (Mi -Mj )/sqrt[ MSw /nh ]

Where:

Here for 1 and 2

HSD = (32.02 - 7.4)/sqrt [7.4592/50]

= 24.62/0.3862 = 63.75

for 1 and 3

HSD = (32.02- 7.64)/ sqrt [7.4592/50]

= 24.38/0.3862 = 63.12

for 2 and 3

HSD = (7.64 - 7.40)/ sqrt [7.4592/50]

= 0.24/0.3862 = 0.6214

so now we will find critical value for alpha = 0.05, dF = 147 and k = 3

Q (Critical)= 3.31

so here we say that mean 1 is different from mean 2 and mean 3 but mean 2 and mean 3 are not statistically different.

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.