Question

The vast majority of people have hand dominance--they are either left-handed or right-handed.
In this question we will ignore the very small percentage of the population who are
ambidextrous (i.e., those who use both hands equally well or equally badly!).
A researcher believes that men are more likely to be left-handed than women. To check this she
checked the dominant hand of random samples of 200 men and 200 women, and found 24 men

and 18 women to be left-handed. Let pM and pW be the proportions of left-handed men and left-
handed women, respectively, in the population.

Construct an appropriate 95% 1-sided confidence interval to estimate the difference pM –
pW.
[2]

b) Does your answer to part a) support the researcher’s claim that men are more likely to be
left-handed? Please explain your reasoning briefly.
[1]

c) Confirm your conclusion in part b) by formulating and testing the hypothesis (at the 5%
significance level) that men are more likely to be left-handed than women in the population.
Make sure to compute a p-value.
[5]

d) In a different study, a 95% confidence interval for the proportion of right-handed people in
the general population was constructed. The interval was 0.89 ± 0.04. Using the same
sample, would a 95% confidence interval for the proportion of left-handed people in the
general population be 0.11 ± 0.04? Please circle your answer choice.
[1]
True False Can’t tell
e) Explain if the calculations in parts (a) and (c) require any assumptions on the distribution of
data on left-handedness in the general population.

Answer 1

Let xM and xW denote the proportion of left-handed women in sample

Here xM = 24/200 = 0.12

xW = 18/200 = 0.09

a) 95% 1-sided confidence interval is  

(xM-xW) z* sqrt[( xM(1-xM) + xW(1-xW))/n]

where z is 95% one sided z value and n is sample size of men/women

(0.12-0.09) 1.645* sqrt( (0.12*0.88 + 0.09*0.91) / 200)

0.03 1.645*0.031

0.03 0.05

(-0.02, 0.08)

b) No, answer in part a) does not support the researcher's claim as difference in proportion interval contains negative value too.

c) Let Null hypothesis be

Ho : pM = pW

against Ha : pM > pW

Test statisitic is

z = (xM- xW) / sqrt [ (xM(1-xM) + xW(1-xW))/n

= (0.12-0.09) / sqrt( (0.12*0.88 + 0.09*0.91) / 200)

=0.03/0.031

= 0.96

p value of one sided z test corresponding to z=0.96 is 0.168 > 0.05. Hence it is not significant to reject Ho at 5% level

d) Cant tell, we need to more information to calculate the same.