[quote="ChaoticSlumber"]I need help with the following; I've tried putting them into the calculator several different ways and cant get even close to the answers I'm given. Please help me.
For a "before and after" test, 16 of a sample of 25 people improved their scores on a test after receiving computer-based instruction. For H0 : p = 0.50; H1:p is not equal to 0.50; and a significance level of .05:
A. z = 1.2, fail to reject the null hypothesis.
B. z = 1.4, reject the null hypothesis.
C. z = 1.4, fail to reject the null hypothesis.
D. z = 1.64, reject the null hypothesis.
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I am absolutely no stats expert but I will try to help.
This is a binomial distribution because there are only 2 possibilities, either the score improves or it doesn't. (I am also assuming that trial outcomes are indepentant of one another)
The letters I will use are
n = number of trials = 25
p = probability of success if null hypothesis is true = 0.5
q = 1-p = probability of failure if null hypothesis is true = 0.5
X is the number of successes in the 25 trials = 16
If the null hypothesis is true then the expected mean will be n*p = 25*0.5 = 12.5
You could use a binomial probablility table to answer this question but since the np>=5 and nq>=5 , the normal curve is considered to be a good approximation.
The variance (sigma squared) = npq = 25*0.5*0.5 = 25/4
The standard error (like a standard deviation) sigma = sqrt(25/4) = 5/2 = 2.5
z = (x - np) / sigma
= (16 - 12.5) / 2.5
= 3.5 / 2.5
= 1.4
At 5% significance test and a two tail test, the critical z score is +-1.96
That is, Ho will be rejected if |z| > 1.96
|z| = 1.4 < 1.96 Therefore the null hypothesis is NOT rejected.
Here are 2 web pages that I refered to, I looked at others as well but were the ones I used.
http://www.regentsprep.org/regents/math/algtrig/ATS7/BLesson3.htm
http://www.math.armstrong.edu/statsonline/5/5.3.2.html
The second question is very similar, you can try it yourself.
+4 
+118654 
