Please give me right answer and don\'t copay and paste.. please save my life.. I
ID: 2921326 • Letter: P
Question
Please give me right answer and don't copay and paste.. please save my life..
Instructions:
1. You may use Minitab or other software for any calculations. However, you must show your manual calculations when asked. You may paste your output onto your assignment to show your use of Minitab; however, this output does not replace any of the steps outlined below. This means that answers that are exclusively Minitab output may receive only one mark.
2. If you are performing a hypothesis test, make sure you state the hypotheses, the level of significance, the rejection region, the test statistic (and p-value, if requested), your decision (whether to reject or not to reject the null hypothesis), and a conclusion in managerial terms that answers the question posed. These steps must be completed in addition to any Minitab output.
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
PART A.
Given that,
possibile chances (x)=402.5
sample size(n)=1150
success rate ( p )= x/n = 0.35
success probability,( po )=0.3189
failure probability,( qo) = 0.6811
null, Ho:p<0.3189
alternate, H1: p>0.3189
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.35-0.3189/(sqrt(0.21720279)/1150)
zo =2.263
| zo | =2.263
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =2.263 & | z | =2.33
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 2.26296 ) = 0.01182
hence value of p0.01 < 0.01182,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.3189
alternate, H1: p>0.3189
test statistic: 2.263
critical value: 2.33
decision: do not reject Ho
p-value: 0.01182
PART B.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.5 is = 0.674
Sample Proportion = 0.35
ME = 0.015
n = ( 0.674 / 0.015 )^2 * 0.35*0.65
= 459.324 ~ 460
PART C.
Given that,
possibile chances (x)=0.13043
sample size(n)=23
success rate ( p )= x/n = 0.0057
success probability,( po )=0.3189
failure probability,( qo) = 0.6811
null, Ho:p>0.3189
alternate, H1: p<0.3189
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.00567-0.3189/(sqrt(0.21720279)/23)
zo =-3.2232
| zo | =3.2232
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =3.223 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -3.22325 ) = 0.00063
hence value of p0.05 > 0.00063,here we reject Ho
ANSWERS
---------------
null, Ho:p>0.3189
alternate, H1: p<0.3189
test statistic: -3.2232
critical value: -1.64
decision: reject Ho
p-value: 0.00063