Question

In the 2015 federal election, 39.47% of the electorate voted for the Liberal party, 31.89% for the Conservative party, 19.71% for the NDP, 4.66% for the Bloc Quebecois, and 3.45% for the Green party.

(a) A recent poll of 1500 respondents found that 29.3% support the Conservative party. Test whether this is sufficient evidence to conclude that the level of support has dropped since the election. Use the 5% level of significance and show your manual calculations.

(b) Suppose you want to estimate the national level of support for the Conservatives using a 95% 2-sided confidence interval with a margin of error of 1%. What sample size would be required?

(c) Suppose that, in a random sample of 23 University of Ottawa (UOttawa) students, only 3 indicated a preference for the Conservatives. Test whether this is sufficient evidence to indicate that the level of support for the Conservatives among UOttawa students is lower than the 31.89% share of the popular vote in 2015. Use the .05 level of significance and show how you would calculate the p-value for this test.

Answer 1

a.
Given that,
possibile chances (x)=439.5
sample size(n)=1500
success rate ( p )= x/n = 0.293
success probability,( po )=0.3189
failure probability,( qo) = 0.6811
null, Ho:p=0.3189
alternate, H1: p<0.3189
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.293-0.3189/(sqrt(0.21720279)/1500)
zo =-2.152
| zo | =2.152
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =2.152 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -2.15235 ) = 0.01568
hence value of p0.05 > 0.01568,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.3189
alternate, H1: p<0.3189
test statistic: -2.152
critical value: -1.645
decision: reject Ho
p-value: 0.01568
we have enough evidence to support the claim that the level of support has dropped since the election.
b.
national support for conservatives is 31.89%
margin of error is 1%
confidence level is 95%
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.3189
ME = 0.01
n = ( 1.96 / 0.01 )^2 * 0.3189*0.6811
= 8344.0624 ~ 8345          
c.
Given that,
possibile chances (x)=3
sample size(n)=23
success rate ( p )= x/n = 0.13
success probability,( po )=0.3189
failure probability,( qo) = 0.6811
null, Ho:p=0.3189
alternate, H1: p<0.3189
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.13043-0.3189/(sqrt(0.21720279)/23)
zo =-1.939
| zo | =1.939
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.939 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -1.93938 ) = 0.02623
hence value of p0.05 > 0.02623,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.3189
alternate, H1: p<0.3189
test statistic: -1.939
critical value: -1.645
decision: reject Ho
p-value: 0.02623
we have enough evidence to support the claim that the level of support for the Conservatives among UOttawa students is lower than the 31.89% share of the popular vote in 2015.