Answer Happy

Does the Carotenoid Pigment of the Egyptian Vulture Vary by Region? 25.3) The bright yellow head of the adult male An Egyptian Vulture, Neophron percnopterus Egyptian vulture requires carotenoid pigments. The brighter the yellow pigment, the more likely that the male will attract a mate. These pigments can not be synthesized by the vultures, so they must be obtained through their diet. Unfortunately, carotenoids are scarce in rotten flesh and bones, but they are readily available in the dung of ungulates. Which is why Egyptian vultures are frequently seen eating the droppings of cows, goats, and sheep in Spain, where they have been studied. image by San Diego Zoo Safari Park Ungulates are common in some areas but not in others. Could the variation in yellow pigment coloring of the Egyptian vulture be due to a variation in the availability in ungulate dung? Negro et al. (2002) measured plasma carotenoids in wild Egyptian vultures caught at four regions in Spain as part of a study to determine the causes of variation among the sites in carotenoid availability. Their data (altered slightly for randomization and preliminary checks) is summarized in the table below. Use a one way ANOVA hypothesis at a 10% LOS with preliminary checks to determine if the mean serum pigment wg/ml) of the Egyptian vulture is significantly different for the four selected regions in Spain.

Site N plasma carotenoids concentration (g/ml) 8.6 6.8 8.4 14.5 Standard deviation 2.2 2.7 2.8 2.8 24 69 76 WN 2 3 4 11 25.3) Step 1a: Preliminary Checks . The data sets are collected from independent and random samples. • With summarized data, we assume that all data sets pass a normality test by the Anderson-Darling normality test, as we are using mean values.

25.3) Step 1b: Check for Homoscedastity (equal variances): ke Because of unequal sample sizes we must first calculate a harmonic mean for the grouped sample size, n'. n = 1 + 1 +...+ 2 n'a Use at least second decimal accuracy Rounded up to the next integer for f-table use. no Establish the decision line for Hartley's Equal variance test. Recognize that this is NOT the decision line for the ANOVA test, rather it is the decision line for the preliminary check of equal variances, without which we can not proceed with the ANOVA test. For Hartley's test: df = (k, n'-1) =( Do Not Reject Reject Fcritical

25.3) Step 1c: The statistical conclusion for Hartley's F test is: At the 5% LOS We Do Not Reject v the null hypothesis of H., because: F-Notational Support: Fsample ✓ Fcritical < Numeric Validation: < P-Notational Support: p> να Numeric Validation: (p>0.20) 0.05 With the preliminary checks of independent and random samples, a normal distribution and equal variances, all met, we can proceed with a one-way ANOVA test.

25.3) Step 2: Declare the variables for this hypothesis test: English text equivalence for the Variable and Hypothesis Options: Multi-parameter Hypotheses: Notel Because of the limitations in Variable English A ki = = t; = id software most special characters Equivalence B cannot be included in a 'pop-up X X X z # X4 CAt least one mean is not equal answer format and yet they are X; X-bar-i D needed for the hypothesis formation. the = Hly = ls = 14 Hi The pop-up selections will be in the mu-i E H & M ₂ & 4 F text-form for the special characters $i = {z = {z = X embedded within the pop-up answer. X X-tilde-i G & # % & # x4 Use the table at the right as a guide if H At least one median is not equal mi eta-i needed. Na = n2 = na na D M = 12 na na 25.3) Step 2: Declare the variables for this hypothesis test: mu-i The mean plasma concentration of carotenoids in ug/ml) for a wild Egyptian vulture from the ih region in Spain 1

25.3) Step 3) Choose the correct hypothesis statement to test the claim that the mean plasma concentration of carotenoids in (ug/ml) for a wild Egyptian vulture varies by region. Test the hypothesis at a 10% LOS. H:D Ha:c a = 0.10 25.3) Step 4) Complete the partially filled ANOVA table as an aid in forming the conclusion for the ANOVA hypothesis. ANOVA TABLE Source Sum of Squares (SS) Degrees of freedom: df Mean Squares (MS) F-sample SSB = 578.675 dfo MSB Fsample Between Groups Within Groups SSW =1273.4 df MSW- Bracketed p-value: Total SST dfe (p <0.0001)

25.3), Step 5) Write a statistical conclusion for the ANOVA hypothesis. At the 10% LOS we Reject the null hypothesis of H., because: df = (dfx, dfw)=(0,0 df and F Notation Support: Fsample > v Fcritical Numeric Validation: p-Notational Support: peva Numeric p = (p <0.0001) 0.10 Validation 25.3), Step 6) Write an English sentence conclusion The evidence supports the case that mean plasma carotenoid concentration in pg/ml of Egyptian vultures varies according to region