Individual)Mini)Project) Donye’)
Dickens)
Now solving this problem wasn’t easy for me. I was a little confused about the wording
that had set up. The first question tells us to conduct a hypothesis test for each sample at the .01
level of significance and determine what action, if any, should be taken. Provide the test statistic
and p–value for each test. With sample one, you have to do the z–test and find the p–value for each
one. So when I computed the equation, I got z=1.08 and the p–value= 2*p (x<–1.08). Then you
have to do 2*0.1401, which equals 0.2802.After that you do 0.28>0.01. After you do all of that
you decide if you accept it or reject it. For sample one we accept it because 0.28 is greater than
the level of significance. Moving on to sample two, your z–value= 0.75. You do the same thing
that you did for sample one. Next you do 2*P (x<0.75). Next you do 2*0.7734, which you’ll get
1.5468. Since 1.55 is greater than 0.01, then we accept it. Sample three is different from samples
one and two because we reject it! The z–value equals –2.90.
The you multiply 2 * the p–value, opening parenthesis, x<–2.90, and then closing the
parenthesis (2*P (x<–2.90)). After that you times 2 by 0.0019, which equals 0.0038. 0.01 is
greater than .004 so we reject it because it’s less than the level of significance. Sample four, is
exactly like sample one and two. The z–value =2.12. You multiply 2*P (x< 2.12) and then do
2*0.9830, which gives you 1.9660. 1.97 is greater than the level of significance. For question
number two, we have to compute the standard deviation for each of the four samples. Does the
assumption of .21 for the population standard deviation appear reasonable? If the margin or error
is small enough to adjust difference in value, then the average is standard deviation. For problem
number three, we have to compute limits for the sample mean x around m=12 such that, as long