(g) Interpret the results. 14. | Basic Computation: Test p-P2 For one binomial e
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(g) Interpret the results. 14. | Basic Computation: Test p-P2 For one binomial experiment, 200 binomial trials produced 60 successes. For a second independent binomial experimen 400 binomial trials produced 156 successes. At the 5% level of significance. test the claim that the probability of success for the second binomial experi- ment is greater than that for the first. (a) Compute the pooled probability of success for the two experiments (b) Check Requirements What distribution does the sample test statistic follow? Explain. (c) State the hypotheses (d) Compute pi p2 and the corresponding sample test statistic. (e) Find the P-value of the sample test statistic. (f) Conclude the test. (g) Interpret the results.Explanation / Answer
14.
a.
probability of success of two experiments x1 =60, n1 =200, p1= x1/n1=0.3
x2 =156, n2 =400, p2= x2/n2=0.39
b.
Z test for difference of proportion is required
c.
null, Ho: p1 = p2
alternate, H1: p1 < p2
d.
Given that,
sample one, x1 =60, n1 =200, p1= x1/n1=0.3
sample two, x2 =156, n2 =400, p2= x2/n2=0.39
null, Ho: p1 = p2
alternate, H1: p1 < p2
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.3-0.39)/sqrt((0.36*0.64(1/200+1/400))
zo =-2.1651
| zo | =2.1651
critical value
the value of |z | at los 0.05% is 1.645
we got |zo| =2.165 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -2.1651 ) = 0.01519
hence value of p0.05 > 0.01519,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -2.1651
critical value: -1.645
e.
p-value: 0.01519
f.
decision: reject Ho
g.
probability of success second experiment is greater than the first experiment
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