6. Basic Computation: Testing p A random sample of 60 binomials trials resulted

Question

6. Basic Computation: Testing p A random sample of 60 binomials trials resulted in 18 successes. Test the claim that the population proportion of suc- cesses exceeds 18%. Use a level of significance of 0.01. Can a normal distribution be used for the p distribu- tion? Explain. (b) State the hypotheses. (c) Compute p and the corresponding standardized sample test statistic. (d) Find the P-value of the test statistic. (e) Do you reject or fail to reject Ho? Explain. (f) Interpretation What do the results tell you? For Problems 7-21, please provide the following information (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? Do you think the sample size is sufficiently large? Explain. Compute the value of (c) Find the P-value of the test statistic. Sketch the sampling distribution and (d) Based on your answers in parts (a) to (c), will you reject or fail to reject (e) Interpret your conclusion in the context of the application. the sample test statistic. show the area corresponding to the P-value. the null hypothesis? Are the data statistically significant at level a?

Explanation / Answer

Points to pass for normal approximation:
1) experiment consistes of a sequence of n identical trials
2) only 2 outcomes are possible on each trail, success or failure
3) trials are independent & below conditions should satisfy
n*p>5, 60*0.18> 5 => 10.8>5

Given that,
possibile chances (x)=18
sample size(n)=60
success rate ( p )= x/n = 0.3
success probability,( po )=0.18
failure probability,( qo) = 0.82
null, Ho:p=0.18  
alternate, H1: p>0.18
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.3-0.18/(sqrt(0.1476)/60)
zo =2.4194
| zo | =2.4194
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =2.419 & | z | =2.33
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 2.41943 ) = 0.00777
hence value of p0.01 > 0.00777,here we reject Ho
ANSWERS
---------------
a.
we use normal approximation
b.
null, Ho:p=0.18
alternate, H1: p>0.18
c.
p= 0.30, test statistic: 2.4194
critical value: 2.33
decision: reject Ho
p-value: 0.00777
we have evidence that it exceeds 18%

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