(1)Research has demonstrated that people tend to be attracted to others who are similar to
themselves. One study demonstrated that individuals are disproportionately more likely to
marry those with surnames that begin with the same last letter as their own (Jones, Pelham,
Carvallo, & Mirenberg, 2004). The researchers began by looking at marriage records and
recording the surname for each groom and the maiden name of each bride. From these
records it is possible to calculate the probability of randomly matching a bride and a groom
whose last names begin with the same letter. Suppose that this probability is only 6.5%. Next,
a sample of n = 200 married couples is selected and the number who shared the same last
initial at the time they were married is counted. The resulting observed frequencies are as
follows: Same Initial=19, Different Initials =181, Total=200. Do these data indicate that the
number of couples with the same last initial is significantly different that would be expected
if couples were matched randomly? Test with a=.05
𝑛 = 200
Population proportion 𝜋 = 6.50% = 0.065
19
Sample proportion 𝑝 = 200 = 0.095
Null hypothesis H0: 𝜋 = 𝜋0 = 0.065
Alternative hypothesis: 𝜋 ≠ 𝜋0 = 0.065
Test statistic
𝑧=
𝑝−𝜋
√𝜋(1 − 𝜋)/𝑛
=
0.095 − 0.065
√0.065(1 − 0.065)/200
= 1.7210
Critical region
|𝑧| > 1.96
The value of z is not in the critical region.
So we do not have sufficient evidence to reject the null hypothesis.
Conclusion: these data do not indicate that the number of couples with the same last initial is
significantly different that would be expected if couples were matched randomly.
(2)The color red is often associated with anger and male dominance. Based on this
observation, Hill and Barton (2005) monitored the outcome of four combat sports (boxing,
tae Kwan do, Greco-Roman wrestling, and freestyle wrestling) during the 2004 Olympic
games and found that participants wearing red outfits won significantly more often than those
wearing blue. a. In 50 wrestling matches involving red versus blue, suppose that the red outfit
won 31 times and lost 19 times. Is this result sufficient to conclude that red wins significantly
more than would be expected by chance? Test at the .05 level of significance. b. In 100
matches, suppose red won 62 times and lost 38. Is this sufficient to conclude that red wins
significantly more than would be expected by chance? Again, use a=.05. c. Note that the
winning percentage for red uniforms in part A is identical to the percentage in part B (31 out
of 50 is 62%, and 62 out of 100 is also 62%). Explain why the two samples lead to different
conclusions.
Part (a)
Let 𝜋 denote the proportion of times red outfit wins.
Ho: 𝜋 = 0.5
H1: 𝜋 > 0.5
𝑛 = 50
31
𝑝 = 50 = 0.62
𝑧=
𝑝−𝜋
√𝜋(1−𝜋)/𝑛
=
0.62−0.50
√0.50(1−0.50)/50
= 1.6971
The critical region is 𝑧 > 1.65
Z falls in the critical region.
So we reject the null hypothesis
Conclusion: This is sufficient to conclude that red wins significantly more than would be
expected by chance.
.
Part (b)
Let 𝜋 denote the proportion of times red outfit wins.
Ho: 𝜋 = 0.5
H1: 𝜋 > 0.5
𝑛 = 100
62
𝑝 = 100 = 0.62
𝑧=
𝑝−𝜋
√𝜋(1−𝜋)/𝑛
=
0.62−0.50
√0.50(1−0.50)/100
The critical region is 𝑧 > 1.65
= 2.4
Conclusion: This is sufficient to conclude that red wins significantly more than would be
expected by chance.
Z falls in the critical region.
So we reject the null hypothesis
Conclusion: This is sufficient to conclude that red wins significantly more than would be
expected by chance.
Note: The two samples do not lead to different conclusions (here). But the two z-values are
different. This is because when sample size increases, the standard error of the sample proportion
decreases. Consequently, the z –score increases. The two conclusions can differ in general. Suppose
the alternative hypothesis is 𝜋 ≠ 0.5, Then we will have the following analysis.
Part (a)
Let 𝜋 denote the proportion of times red outfit wins.
Ho: 𝜋 = 0.5
H1: 𝜋 ≠ 0.5
𝑛 = 50
31
𝑝 = 50 = 0.62
𝑧=
𝑝−𝜋
√𝜋(1−𝜋)/𝑛
=
0.62−0.50
√0.50(1−0.50)/50
= 1.6971
The critical region is |𝑧| > 1.96
Z does not fall in the critical region.
So we cannot reject the null hypothesis
Conclusion: This is not sufficient to conclude that red wins significantly more than would be
expected by chance.
.
Part (b)
Let 𝜋 denote the proportion of times red outfit wins.
Ho: 𝜋 = 0.5
H1: 𝜋 ≠ 0.5
𝑛 = 100
62
𝑝 = 100 = 0.62
𝑧=
𝑝−𝜋
√𝜋(1−𝜋)/𝑛
=
0.62−0.50
√0.50(1−0.50)/100
= 2.4
The critical region is 𝑧 > 1.96
Z falls in the critical region.
So we reject the null hypothesis
Conclusion: This is sufficient to conclude that red wins significantly more than would be
expected by chance.
Part(c)
The conclusions in parts a and b are different. This is because the z-scores are different. This
because, the denominator of the z-formula, which is the standard error of the sample proportion
decreases when n increases. Then z increases and leads to the rejection of the null hypothesis with
more probability.
In other words, when the sample size increases, the power of a statistical test increases.