Question

A recent opinion poll of funding for the arts asked, “Would you favor spending more federal tax money on the arts?” In a random sample of 220 women from the U.S., 59 responded ‘yes.’ In a random sample of 175 men from the U.S., 56 responded ‘yes.’ Use these data as follows, to test whether there is any difference in the proportion of men vs. women who favor spending more for the arts:

c. From the output, report the test statistic (i.e., the standardized score), and the p-value.

d. Make a formal conclusion at the 5% significance level (in terms of whether you reject the null hypothesis or not and how you are deciding), and then also say what your conclusion means in terms of whether the study give sufficient evidence that there is any difference in the proportion of men versus women who favor spending more for the arts

e.The output also shows a 95% confidence interval (“95% CI”). Briefly say how the interval supports the formal conclusion.

(your answer should involve whether a certain number does or does not fall in the interval.)

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Test and Cl for Two Proportions X Sample 1 2 59 56 n 220 175 Difference = P1 p2 Estimate for difference: -0.0518182 95% CI for difference: (-0.142391, 0.0387551) Test for difference = 0 (vs * 0): z = -1.13 P-Value = 0.26

Answer 1

a.
Given that,
sample one, x1 =59, n1 =220, p1= x1/n1=0.268
sample two, x2 =56, n2 =175, p2= x2/n2=0.32
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.268-0.32)/sqrt((0.291*0.709(1/220+1/175))
zo =-1.126
| zo | =1.126
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.126 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.1261 ) = 0.2601
hence value of p0.05 < 0.2601,here we do not reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 != p2
c.
test statistic: -1.126
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.2601
d.
we do not have enough evidence to support the claim that there is any difference in the proportion of men vs. women who favor spending more for the arts:
e.
TRADITIONAL METHOD
given that,
sample one, x1 =59, n1 =220, p1= x1/n1=0.268
sample two, x2 =56, n2 =175, p2= x2/n2=0.32
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.268*0.732/220) +(0.32 * 0.68/175))
=0.046
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.046
=0.091
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.268-0.32) ±0.091]
= [ -0.142 , 0.039]
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DIRECT METHOD
given that,
sample one, x1 =59, n1 =220, p1= x1/n1=0.268
sample two, x2 =56, n2 =175, p2= x2/n2=0.32
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.268-0.32) ± 1.96 * 0.046]
= [ -0.142 , 0.039 ]
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interpretations:
1) we are 95% sure that the interval [ -0.142 , 0.039] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2