Question

All persons running for public office must report the amount of money spent during their campaign. Political scientists have contended that female candidates generally find it difficult to raise money and therefore spend less in their campaign than male candidates. Suppose the accompanying data represent the campaign expenditures of a randomly selected group of male and female candidates for the state legislature. Do the data support the claim that female candidates generally spend less in their campaigns for public office than male candidates?

Female 169 206 257 294 252 283 240 207 230 183 1298 269 256 277 300 126 318 184 252 305 289 334 278 268 336 438 388 388 394 394 425 386 356 342 305 365 355 312 209 458

a) State the null and alternative hypotheses. b) Check conditions in order to proceed with your choice of procedures to analyze the data. c) Perform the testing at .05 level. d) Estimate the size of the difference in campaign expenditures for female and male candidates by a 95% confidence interval. e) Is the difference of practical significance? (why?

Answer 1

Given that,
mean(x)=245.3
standard deviation , s.d1=51.9545
number(n1)=20
y(mean)=351
standard deviation, s.d2 =61.921
number(n2)=20
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.729
since our test is left-tailed
reject Ho, if to < -1.729
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =245.3-351/sqrt((2699.27007/20)+(3834.21024/20))
to =-5.8481
| to | =5.8481
critical value
the value of |t α| with min (n1-1, n2-1) i.e 19 d.f is 1.729
we got |to| = 5.84814 & | t α | = 1.729
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -5.8481 ) = 0.00001
hence value of p0.05 > 0.00001,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 < u2
b.
test statistic: -5.8481
critical value: -1.729
decision: reject Ho
c.
p-value: 0.00001
we have enough evidence to support the claim that female candidates spend less in their
campaigns for public office than male candidates.
d.
effective size = modulus of (mean 1 - mean 2 )/S.D pooled
S.D pooled = sqrt ((s.d1^2 +s.d2 ^2)/2)
S.D pooled= sqrt ((51.9545^2 +61.921^2)/2)
S.D pooled = 57.155
effective size =modulus of (245.3 -351)/57.155
effective size = modulus of (-1.849)
effective size = 1.849
large effect.
e.
Given that,
mean(x)=245.3
standard deviation , s.d1=51.9545
number(n1)=20
y(mean)=351
standard deviation, s.d2 =61.921
number(n2)=20
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =245.3-351/sqrt((2699.27007/20)+(3834.21024/20))
to =-5.8481
| to | =5.8481
critical value
the value of |t α| with min (n1-1, n2-1) i.e 19 d.f is 2.093
we got |to| = 5.84814 & | t α | = 2.093
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -5.8481 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -5.8481
critical value: -2.093 , 2.093
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference of means of females and males.