PLEASE ANSWER ALL THE QUESTIONS In a presidential election 308 out of 611 voters

Question

PLEASE ANSWER ALL THE QUESTIONS

In a presidential election 308 out of 611 voters surveyed said that they voted for the candidate who won (based on data from ICR Survey Research Group). Use a 0.01 significance level to test the claim that among all voters, the percentage who believe that they voted for the winning candidate is equal to 43%, which is the actual percentage of votes for the winning candidate.

a) What is the format of H0?

b) What is the format of H1?

c) What number goes in the blank for the hypotheses?

d) Which distribution should be used for the test statistic and critical value?

Normal distribution (z) or

Chi-square distribution (X) or

F distribution (F) or

Student distribution (t)

e) What is the value of the test statistic?

Round your answer to three decimal places

f) Will the critical value(s) be positive, negative, or one of each?

Negative or

One positive and one negative (±) or

Positive?

g) What is the critical value(s)?

Enter the answer rounded to three decimal places. If the critical value is plus/minus, just enter the positive value. You indicated in the last question that it should be plus/minus. If the critical value is a negative number, please type the negative sign.

h) What is your conclusion?

Reject the null hypothesis.? or

Fail to reject the null hypothesis.

i) Select the appropriate final conclusion.

There is NOT sufficient evidence to support the claim that the percentage who believe that they voted for the winning candidate is equal to 43%.

There is sufficient evidence to support the claim that the percentage who believe that they voted for the winning candidate is equal to 43%.

There is sufficient evidence to warrant rejection of the claim that the percentage who believe that they voted for the winning candidate is equal to 43%.

There is NOT sufficient evidence to warrant rejection of the claim that the percentage who believe that they voted for the winning candidate is equal to 43%.

Normal distribution (z) or

Chi-square distribution (X) or

F distribution (F) or

Student distribution (t)

Explanation / Answer

Given that,
possibile chances (x)=308
sample size(n)=611
success rate ( p )= x/n = 0.5041
success probability,( po )=0.43
failure probability,( qo) = 0.57
null, Ho:p=0.43  
alternate, H1: p!=0.43
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.50409-0.43/(sqrt(0.2451)/611)
zo =3.6993
| zo | =3.6993
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =3.699 & | z | =2.58
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.69929 ) = 0.00022
hence value of p0.01 > 0.0002,here we reject Ho
ANSWERS
---------------
a.
null,percentage voted for the winning candidate is equal to 43%, Ho:p=0.43
b.
alternate,is not equal to 43% H1: p!=0.43
d.
Normal distribution (z) or
e.
test statistic: 3.6993
f.
critical value: -2.58 , 2.58
h.
decision: reject Ho
i.
There is NOT sufficient evidence to support the claim that the percentage who believe that they voted for the winning candidate is equal to 43%.

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