Please help me solve these problems!!.. Please give me right answer and save my
ID: 3306921 • Letter: P
Question
Please help me solve these problems!!.. Please give me right answer and save my life..
2. [ 10 marks ]
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) An August poll of 1150 respondents found that 35% would support the Conservative party. Test whether this is sufficient evidence to conclude that the level of support has changed since the election. Use the 1% level of significance and show your manual calculations (you do not need to calculate the summary statistics manually).
(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 ±1.5%. What sample size would be required?
(c) To test whether the level of support for the Conservatives among U of O students is lower than the 31.89% share of the popular vote in 2015, you found 3 out of 23 randomly selected University of Ottawa students supported the Conservatives. Use the 5% level of significance and show how you would calculate the p-value for this test.
Explanation / Answer
Solution:
(a)
Giben that sample Size = 1150
P(Conservative Party) = 0.3189
p^ = 0.35
Standard error of the proportion se0 = sqrt (0.3189 * 0.6811/ 1150) = 0.0137
Z = (p^ - ph ) / se0 = (0.35 - 0.3189)/ 0.0137 = 2.27
Level of significane = 1 %
Zcritical = 2.33
so here, Z < Zcritical so we can not reject the null hypothesis and can conclude that there is no change in support has changed the elcetion.
(b)
Margin of error = 0.015 = Test statistic * standard error of the proportion = Z95% * sqrt {pc * (1-pc)/ n]
0.015 = 1.96 * sqrt [0.3189 * 0.6811/ n] (where n = sample size)
so required sample size n = 3708
(c)
Proportion of conservatives among Ottawa students = 3/23 = 0.1304
Standard error of the proportion se0 = sqrt (0.3189 * 0.6811/ 23) = 0.0972
Z = (p^ - ph ) / se0 = (0.1304 - 0.3189)/ 0.0972 = -1.94
for 5% level of significance of one sided interval Zcritical = 1.65
Pr (Z < -1.94) = 0.0262 < 0.05
Therefore, null hypothesis is rejected and hence we conclude that support for conservatives among Ottawa students has lowered.