Past studies have indicated that 85.6% of all enrolled college students in the U
ID: 3066835 • Letter: P
Question
Past studies have indicated that 85.6% of all enrolled college students in the US. are undergraduates. A random sample of 500 enrolled college students in a particular stats revealed that 420 of the students were undergraduates. Is there enough evidence to conclude that the percentage of undergraduates from this particular state differs from the national percentage? What type of test would be appropriate for this situation? (name the test and indicate if it should be a one tailed test or two tailed test?) a) b) State the null and the alternate hypotheses. c) Perform the appropriate test and report the P-value accepting or rjecting Use your P-value to make a conclusion about the problem. Are you accepting or rejecting the null hypothesis? d) e) Are your resuts satisticay sgniin? Practicl significanr?Explanation / Answer
Given that,
possibile chances (x)=420
sample size(n)=500
success rate ( p )= x/n = 0.84
success probability,( po )=0.856
failure probability,( qo) = 0.144
null, Ho:p=0.856
alternate, H1: p!=0.856
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=0.84-0.856/(sqrt(0.123264)/500)
zo =-1.019
| zo | =1.019
critical value
the value of |z ?| at los 0.05% is 1.96
we got |zo| =1.019 & | 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 != -1.01903 ) = 0.30819
hence value of p0.05 < 0.3082,here we do not reject Ho
ANSWERS
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z test for proportion, two tailed test
null, Ho:p=0.856
alternate, H1: p!=0.856
test statistic: -1.019
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.30819
enough evidence to conclude that the percentage of undergraduates from this particular state is similar to
the national percentage