In: Statistics and Probability
•Write the claim as a math statement, its opposite, and then determine H0and H1and the tail type.
•Test the claim using the traditional (critical value) method:
•Sketch the appropriate distribution showing the tail(s) and their area, and give the critical value(s) determined from the appropriate distribution table.
•Show the test statistic formula with the relevant numbers in place, and determine the test statistic value. Show where this value is located on the sketch of the distribution.
•State the proper conclusion of the test with regard to H0and the conclusion with regard to the original claim.
•Determine the p-value using the distribution tables and confirm the conclusion about H0.Note: the p-value will be a single number if the test requires the use of a standard normal distribution, otherwise you will obtain a range of values (for example p>0.01).
4)An exit poll is done at a polling station. 105 people are randomly sampled and 62 said they voted for Jane Doe for congress. Use a significance level of 5% to test the claim that a majority of people voted for Jane.
4.
Given that,
possibile chances (x)=62
sample size(n)=105
success rate ( p )= x/n = 0.5905
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p!=0.5
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.59048-0.5/(sqrt(0.25)/105)
zo =1.8542
| zo | =1.8542
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.854 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject
Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.85421 )
= 0.06371
hence value of p0.05 < 0.0637,here we do not reject Ho
ANSWERS
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null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: 1.8542
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.06371
we do not have enough evidence to support the claim that a majority
of people voted for Jane.