5. In a study, mosquitoes were divided into four groups of eight mosquitoes each

Question

5. In a study, mosquitoes were divided into four groups of eight mosquitoes each: infected rhesus and sporozites present (Trt 1), infective rhesus and oocysts present (Trt 2), infective rhesus and no infection developed (Trt 3), and noninfective (Trt 4). Distances flown by the mosquitoes within 24 hours were recorded. The summary data are: = 4.39, 4.52,= 5.49,= 6.36,5.19, x, = 911.91. Use the ANOVA F test at level 0.10 to decide whether there are any significant differences between true average flight times for the four treatments * Show some hand calculations in support of your final answers. (Formulas 10B, 10C, 10D.) Follow the 4-step hypothesis test outline. Give the reject region. Do not compute the p-value.

Explanation / Answer

solution;

Note:

* Notation and terminology varies from professor to professor. I will use the notation and terminology that I was taught. You should have no problem following it. :)

* Please double check my math, I'm doing this at 4am EST. :)

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Ho: mu1 = mu2 = mu3 = mu4
Ha: at least 1 mui significantly differs from the others

# of groups = g = 4
Total number of observations = N = 4(8) = 32

Overall mean = 5.19

SSTotal = sum of all x's^2 - (sum of all x's)^2 / N
--> sum of all x's = 8*4.39 + 8*4.52 + 8*5.49 + 8*6.36 = 166.08
--> SSTotal = sum of all x's^2 - (sum of all x's)^2 / N
--> SSTotal = 911.91 - 166.08^2/32
--> SSTotal = 49.9548

SSBetween = SSTreatment = SSGroup
SSB = 8(4.39-5.19)^2 + 8(4.52-5.19)^2 + 8(5.49-5.19)^2 + 8(6.36-5.19)^2
SSB = 20.3824

MSB = 20.3824/(4-1)
MSB = 6.7941

SSWithin = SSError
SSTotal = SSBetween + SSWithin
SSW = 49.9548 - 20.3824
SSW = 29.5724
MSW = SSW/(N-g)
MSW = 29.5724/(32-4)
MSW = 1.0561

F = MSB / MSW
F = 6.7941 / 1.056
F = 6.43

Note: I am using periods in the following ANOVA just to make sure the values line up.

ANOVA table

Source...........................df............SS...........MS.........F
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Between = Treatment.....3....20.3824.....6.7941......6.43
Within = Error................28....29.5724.....1.0561
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Total..............................31....49.9548

Note: The F table that I am using can be found on page 17 of the following pdf.

http://www.stat.purdue.edu/~mccabe/ips4tab/bmtables.pdf

Finding the critical value...
Using the F, go down from df numerator = 3 and dfdenominator = 28 and .05. We see that our critical value = 3.63.
--> Critical value: F = 3.63
--> Rejection region: F > = 3.63

Using the critical value approach...
Decision Rule: If the F test statistic > = 3.63, reject Ho and conclude Ha
Is F = 6.43 > 3.63???
Yes, so, we CAN reject Ho and conclude Ha
--> The population average flight times for the 4 treatments are NOT all equal

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