Question

Female primates visibly display their fertile window, often with red or pink coloration. Do humans also do this? A study looked at whether human females are more likely to wear red or pink during their fertile window (days 6-14 of their cycle). They collected data on 24 female undergraduates at the University of British Columbia, and asked each how many days it had been since her last period, and observed the color of her shirt. Of the 10 females in their fertile window, 4 were wearing red or pink shirts. Of the 14 females not in their fertile window, only 1 was wearing a red or pink shirt.

(a) State the null and alternative hypotheses. (2 points)

(b) Calculate the relevant sample statistic, ?̂? −?̂??, for the difference in proportion wearing a pink or red shirt between the fertile and not fertile groups. (2 points)

(c) For the 1000 statistics obtained from the simulated randomization samples, only 6 different values of the statistic, ?̂? − ?̂?? are possible. The table below shows the number of times each difference occurred among the 1000 randomizations. Calculate the p-value. (2 points) Randomization distribution for difference in 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.

Answer 1

Given data

Sample data n=24

observed the color of her shirt. Of the 10 females in their fertile window, 4 were wearing red or pink shirts. Of the 14 females not in their fertile window, only 1 was wearing a red or pink shirt.

Let group 1 be females wearing a red or pink shirt in the fertile group; and group 2 be females wearing a red or pink shirt in the not fertile group.

(a)

null hypotheses :H₀ : p_{1 =} p₂

null and alternative hypotheses : H_a : p₁ > p₂

(b) We have to find

p1^-p2^, for the difference in proportion wearing a pink or red shirt between the fertile and not fertile groups.

p^₁ - p^₂ = 4/10 - 1/14 = 0.3286

(c) Given data

1000 statistics obtained from the simulated randomization samples

p1^-p2^	-0.357	-0.186	-0.014	0.157	0.329	0.500
Count	39	223	401	264	68	5

so here as we know that p^₁ - p^₂ = 0.329

so p - value = Pr( p^₁ - p^₂ >= 0.329 ; p^₁ - p^₂ = 0) = (68+5) /1000 = 0.073