12.2: Hypothesis Test of Proportion A medical school claims that more than 28% o

Question

12.2: Hypothesis Test of Proportion A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 students, 42 of them plan to go into general practice. Use the P-value method to test the hypothesis with a significance level of a- 0.5. 16) State the null and alternate hypotheses. A) H0: p = 28% B)Ho: p 28% D) Ho: p 32% 17) State the null and alternate hypotheses A)HA: p 28% C)HA: p#28% D) HA: p=28% 18) What type of test will be used? A) Left-tail B) Right-tail C) No-tail D) Two-tail 19) What is the point estimate for the proportion, p? Round to four decimal places A) .3231 B) .2800 C).0500 D) 3.0952 20) Compute the test statistic, z. Round to three decimal places. A) 1.96 B).137 C) 1.094 D).323

Explanation / Answer

Given that,
possibile chances (x)=42
sample size(n)=130
success rate ( p )= x/n = 0.3231
success probability,( po )=0.28
failure probability,( qo) = 0.72
null, Ho:p=0.28  
alternate, H1: p>0.28
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.32308-0.28/(sqrt(0.2016)/130)
zo =1.0939
| zo | =1.0939
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.094 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.09388 ) = 0.137
hence value of p0.05 < 0.137,here we do not reject Ho
ANSWERS
---------------
Q16. null, Ho:p=0.28
Q17. alternate, H1: p>0.28
Q18. one tailed
Q19. 0.3231
Q20. test statistic: 1.094
Q21. p-value: 0.137
Q22. decision: failed to reject null hypothesis

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