Thank you 6.3 Edwards and Fraccaro [1960] present Swedish data about the gender

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Thank you

6.3 Edwards and Fraccaro [1960] present Swedish data about the gender of a child and the parity. These data are: Order of Birth Gender Males 2846 25542162 1667 1341 987 666 12,223 Females 2631 2361 1996 1676 1230 914 668 11,476 Total 5477 4915 4158 3343 25711901 1334 23,699 4 7 Total (a) Find the p-value for testing the hypothesis that a birth is equally likely to be of either gender using the combined data and binomial assumptions. 196 COUNTING DATA Construct a 90% confidence interval for the probability that a birth is a female child Repeat parts (a) and (b) using only the data for birth order 6. (b) (c)

Explanation / Answer

a. For combined data we have proportion of males = p = 12223/23699 = 0.5158

We need to check if p =0.5 , then gender is equally likely else not

SE(Male proportion) = sqrt(p*(1-p)/n) = sqrt(0.5158*0.4842/23699) = 0.00325

Ho : p=0.5

Ha : p not equal to 0.5

t stat = (p-0.5)/SE(p) = (0.5158-0.5)/0.00325 = 4.86

p value at t=2.33 is approx 0.01 .Hence p value <0.01 here

b. Proportion of female child = f = 11476/23699 = 0.4842

SE(female child) = sqrt(f*(1-f)/n) = sqrt(0.4842*0.5158/23699) = 0.00325

So, 90% CI is (f - t0.05,23698 * SE(f), f + t0.05,23698 * SE(f)) = (0.4842 - 1.65*0.00325, 0.4842 + 1.65*0.00325) = (0.4788, 0.4896)

c. For birth order 6 we have :

proportion of males = p = 987/1901 = 0.5192

We need to check if p =0.5 , then gender is equally likely else not

SE(Male proportion) = sqrt(p*(1-p)/n) = sqrt(0.5192*0.4808/1901) = 0.0114

Ho : p=0.5

Ha : p not equal to 0.5

t stat = (p-0.5)/SE(p) = (0.5192-0.5)/0.0114 = 1.684

p value from t distribution table is found as 0.092

Proportion of female child = f = 914/1901 = 0.4808

SE(female child) = sqrt(f*(1-f)/n) = sqrt(0.4808*0.5192/1901)= 0.0114

So, 90% CI is (f - t0.05,1900 * SE(f), f + t0.05,23698 * SE(f)) = (0.4808 - 1.65*0.0114, 0.4808 + 1.65*0.0114) = (0.462, 0.500)

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