The events e the person selected has in excess of The events e the person select

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The events e the person selected has in excess of
The events e the person selected has in excess of

mers who gane oRsole manufacturer his assertion against the default that the difference still has the historic value of 30 percentage poi asserts that the proportion of play on consoles exceeds that of adult gamers by more than 30 percentage points. t the 5% level of significance, using the data Teen gamers Adult gamers n1=1100 n2 = 1100 =0.89 = 0.54 The sample are suficiently large for the procedure to be valid.) (a) State the null and alternative hypotheses for the test. 12 points (b) State the formula for the test statistic and compute its value. Justify your answer. (4 pointsl 4 points! (c) Construct the rejection region and make a decision (but note part (d) below first) (d) State a conclusion in the context of the problem, based on the test you performed. 12 points) (e) Compute the p-value (the observed significance) of the test and state what it means in the co of this problem. [2 points

Explanation / Answer

Given that,
sample one, x1 =0.89, n1 =1100, p1= x1/n1=0.001
sample two, x2 =0.54, n2 =1100, p2= x2/n2=0
null, Ho: p1 = p2
alternate, H1: p1 != p2
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) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.001-0)/sqrt((0.001*0.999(1/1100+1/1100))
zo =0.293
| zo | =0.293
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =0.293 & | 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 != 0.2928 ) = 0.7697
hence value of p0.05 < 0.7697,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: p1 = p2
alternate, H1: p1 != p2
b.
test statistic: 0.293
critical value: -1.96 , 1.96
d.
decision: do not reject Ho
e.
p-value: 0.7697

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