Use the data we collected on our class to do a hypothesis test. In our class 10
ID: 3220911 • Letter: U
Question
Use the data we collected on our class to do a hypothesis test. In our class 10 out of 21 people said they enjoyed horror movies. In the general population, about 36% of people say they like horror movies. We'd like to test the hypotheses that the proportion of college students who like horror movies is the same as the general population vs the hypothesis that the proportion of college students who like horror movies is higher than the general population. Assume the assumptions and conditions for using the Normal model are met.
1. Write the hypotheses being tested.
2. Find the p-value for your hypothesis test. Your answer should include all your calculations or a description of how you used your calculator to find the answer.
3. What is your conclusion at the 0.05 significance level?
Explanation / Answer
Given that,
possibile chances (x)=10
sample size(n)=21
success rate ( p )= x/n = 0.48
success probability,( po )=0.36
failure probability,( qo) = 0.64
null, Ho:p=0.36
alternate, H1: p!=0.36
level of significance, = 0.05
from standard normal table, two tailed z /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=0.47619-0.36/(sqrt(0.2304)/21)
zo =1.11
| zo | =1.11
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.109 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.10927 ) = 0.26731
hence value of p0.05 < 0.2673,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.36
alternate, H1: p!=0.36
test statistic: 1.11
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.26731
it is as same as general population rate
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