Female primates visibly displa
Female primates visibly display their fertile window, often withred or pink coloration. Do humans also do this? A study looked atwhether human females are more likely to wear red or pink duringtheir fertile window (days 6-14 of their cycle). They collecteddata on 24 female undergraduates at the University of BritishColumbia, and asked each how many days it had been since her lastperiod, and observed the color of her shirt. Of the 10 females intheir fertile window, 4 were wearing red or pink shirts. Of the 14females not in their fertile window, only 1 was wearing a red orpink shirt.
(a) State the null and alternative hypotheses. (2 points)
(b) Calculate the relevant sample statistic, ?̂? −?̂??, for thedifference in proportion wearing a pink or red shirt between thefertile and not fertile groups. (2 points)
(c) For the 1000 statistics obtained from the simulatedrandomization samples, only 6 different values of the statistic,?̂? − ?̂?? are possible. The table below shows the number of timeseach difference occurred among the 1000 randomizations. Calculatethe p-value. (2 points) Randomization distribution for differencein proportion wearing red or pink in 1000 samples ?̂? −?̂?? −0.357−0.186 −0.014 0.157 0.329 0.500 Count 39 223 401 264 68 5
(d) Interpret the p-value. (1 point)
Just need help with part D.
observed the color of her shirt. Of the 10 females in their fertilewindow, 4 were wearing red or pink shirts. Of the 14 females not intheir fertile window, only 1 was wearing a red or pink shirt.
Letgroup 1 be females wearing a red or pink shirt in the fertilegroup; and group 2 be females wearing a red or pink shirt in thenot fertile group.
nullhypotheses :H0 : p1 = p2
nulland alternative hypotheses : Ha : p1 >p2
(b) We have to find
p1^-p2^, for the difference in proportion wearing a pink or redshirt between the fertile and not fertile groups.
p^1 – p^2 = 4/10 – 1/14 = 0.3286
(c) Given data
1000statistics obtained from the simulated randomization samples
sohere as we know that p^1 – p^2 = 0.329
sop – value = Pr( p^1 – p^2>= 0.329 ; p^1 – p^2 = 0) = (68+5) /1000 =0.073