Bio 260 (Sp17) Assignment #7 Form C, Due MAR 16 Nam Provide answers only in spac

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Bio 260 (Sp17) Assignment #7 Form C, Due MAR 16 Nam Provide answers only in space provided, do not attach additional sheets Write clearly to ensure full credit for your answers, Write the first letter of your surname in box above When describing test conclusions always include a statement about the p value that indicates the degree confidence with which you are making your conclusion For calculated values of t and F provide value to nearest 0.001 (no more and no less) or you will lose credit. (1) Recent events in Europe have brought attention to the fact that sometimes zoos euthanize animals that are not appropriate for breeding (http:llwww.cnn.com/2014/021 People aworld/europe/denmark-zoo-giraffe/ often claim that castration would be asolution. When l worked at CARE, a baboon sanctuary (http:INwww.primatecare.org), the staff there said that male baboons often lost a large amount of muscle mass after castration so this procedure isn't without risk. Those observations were anecdotal however and maybe they get bigger sometimes instead. Imagine that we conduct a study in which we randomly assign sets of male baboons into two groups (castration and control) and measure their body masses. Imagine we collect the following mass data from each set of baboons: Control Castration 92.29 89.13 90.14 93.14 88.59 89.03 88.87 90.36 89.76 92.08 87.94 93.19 91.51 90.65 87.73 91.67 88.65 92.42 87.76 90.12 91.16 91.56 87.64 89.51 87.63 1.11 88.63 90.55 93.17 90.56 90.68 90.68 (a, 2 pts) State the biological question that we will be testing with our two-tailed test, be precise. If we wished to know whether we could do a homoscedastic t test (or have no choice and must do a heteroscedastic test) we need to do a preliminary Fratio test (b, 2 pts ea) Conduct the calculation for the Fratio df(num) df(den) test and fill in the blanks to the right. Fat (a 0.05) (c, 2 pts) State the formal statistical conclusion of your F ratio test and what the practical conclusion of this test is (i.e., what t tests may be performed on this data) (d, 2 pts ea) Perform a two-tailed ter (a 0.05) heteroscedastic t test using the data above and fill in the blanks to the right. (round dif to the correct whole number) (e, 3 pts) State the biological conclusion of your t test in the box below. Use the grammar described in lecture and state with what degree of confidence you make your conclusion by providing the most specific range of p values from the table provided in lecture. You must use the phrase "significantly smaller" significantly larger or "not significantly different in your answer Note: no credit will be given for ANY text outside the box or hard to read answers

Explanation / Answer

SoutionA:

null Hypothesis:

Ho: means of bodymasses of two groups are equal.

meanof muscle mass (control)=meanof muscle mass in (castration)

Alternative

H1: Atleast one of means is different.

mean of muscle mass in (control) > mean of muscle mass (castration)

level of significance=0.05

solution b form above table

df(Numerator)=1

df(Denomiantor)=30

F cal=5.245(to 3 decimals as required)

F critical=4.171

SolutionC:

Since p value of 0.029 <0.05

Reject Null Hypothesis

Accept Alternative Hypothesis.

since p value is signifcantly smaller reject null hypothesis and accept alternative hypothesis.

ther is sufficient eviodence at 5% level of signiifcance that male babbons often lost a large amount of muscle mass after castration.Castration is not effective

Solutiond:

df=30

t calc=2.326

t crit(alpha=0.05)=2.042

SOlutione:

p value=0.013

p value for one tail

p<alpha

0.013<0.05

since p value is signifcantly smaller reject null hypothesis and accept alternative hypothesis.

ther is sufficient eviodence at 5% level of signiifcance that male babbons often lost a large amount of muscle mass after castration.Castration is not effective

Anova: Single Factor SUMMARY Groups Count Sum Average Variance control 17 1544.54 90.85529 3.122064 castration 15 1343.35 89.55667 1.922524 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 13.43877 1 13.43877 5.244849 0.029216 4.170877 Within Groups 76.86836 30 2.562279 Total 90.30712 31

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