2. We want to test the alternative hypothesis that p 0.25 using a sample ofn-500

Question

2. We want to test the alternative hypothesis that p 0.25 using a sample ofn-500. In that sample of 500, there were 135 "successes." [35 points] Hint: Your answers for parts A, D, and F should be very similar. A. Use StatKey to construct a randomization distribution to find thep-value. (Use the randomization methods from Lesson 5) B. From your randomization distribution, what is the standard error? C. Model the randomization distribution with a normal distribution. To do this you will construct a normal distribution with a mean of 0.25 (null parameter) and standard deviation equal to this standard error that from part B. Find the p-value by finding the area to the right of 0.270 (observed sample statistic) in this normal distribution D. Compute the standardized test statistic using the formula for the general form of a test statistic E. Construct a z distribution to find the p-value given your test statistic is part D. F. Explain why your answers in parts A, C, and E were similar

Explanation / Answer

PART A.
Given that,
possibile chances (x)=135
sample size(n)=500
success rate ( p )= x/n = 0.27
success probability,( po )=0.25
failure probability,( qo) = 0.75
null, Ho:p=0.25  
alternate, H1: p>0.25
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=0.27-0.25/(sqrt(0.1875)/500)
zo =1.0328
| zo | =1.0328
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.033 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.0328 ) = 0.15085
hence value of p0.05 < 0.15085,here we do not reject Ho

ANSWERS
---------------
success rate ( p )= x/n = 0.27

PART B.
The PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.27
standard error ( sd )= sqrt(PQ/n) = sqrt(0.27*0.73/500)
=0.0199

PART C.
The PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 0.25
standard Deviation ( sd )= 0.019
P(X > 0.27) = (0.27-0.25)/0.019
= 0.02/0.019 = 1.0526
= P ( Z >1.0526) From Standard Normal Table
= 0.1463

PART D.
from PART A solved, zo =1.0328

PART E.
p-value: 0.15085

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