Hard Ap Stats Multiple Choice Questions

Test Preparation
1

<p>any1 have an idea what the answer is for these questions</p>

<p>If the expected value of successes in a binomial experiment of 100 trials is 55, the standard deviation of the number of sucesses is approximately…
a 2.23
b 2.49
c 4.97
d 24.75
e None of theses</p>

<p>A residual plot of a linear regression provides a good fit if which of the following statements are ture?
I. The points are randomly scattered above and below the horizontal axis
II. The distances from the horizontal axis are about the same
III. The values of the deviations from the horizonal are small
a. I only
b. I and II only
c I,II,III
d II only
e II and III only</p>

<p>A CEO of a small corporation wishes to hire your consulting firm to conduct a simple random sample of its customers to determine the proportion of its customers that consider her company as their primary source of her product. She tells you staff that she requires the margin of error to be no more then 3% with a 95% confidence interval. Earlier studies have indicated that the approximate proportion is 37%. Using the CEO’s research, your staff calculates that a sample size of 995 or more will metter her requirements. By how many MORE people would you increase the size if you use the most conservative estimate of the sample proportion?
a. 12
b. 52
c. 73
d. 112
e. 1068</p>

2

<p>bumperoooo</p>

3

<h1>1 - the standard deviation of a binomial distribution is given by the formula sqrt[(n)(p)(1-p)]. Expected value is just (n)(p), so n is 100 (given) and p is .55 (55/100). So, using the formula: sqrt[(100)(.55)(.45)] = 4.97 (C).</h1>

<h1>2 - I. - definitely. random scattering is a requirement.</h1>

<p>II. - i don’t think that this is true…if they were the same then it wouldn’t be random scattering…more like equidistant inconsistencies
III. - well, a linear regression is almost always a LEAST squares regression line, so small deviations are a big plus.</p>

<p>but this doesn’t fit any of the choices, so i dunno</p>

<h1>3 - use this formula:</h1>

<p>“margin of error” < z*(sqrt[(p)(1-p)/n)])</p>

<p>.03 = 1.95996(sqrt[(.37)(.63)/n]</p>

<p>then we get a number (n = 994.3) that corresponds to their 995, so we know we’re going in the right direction. </p>

<p>and, by common sense, we should know that a p of .5 would give us the largest n and, thus, the most conservative estimate of the sample size we need.</p>

<p>therefore:
.03 = 1.95996(sqrt[(.5)(.5)/n])
n = 1067.06, round up to be safe, so 1068 (E).</p>

4

<p>thanks a lot</p>

5

<p>please tell me if you find out the answer to number 2</p>

6

<p>number two is c. I know I is definitely right, 80% on the others.</p>

<p>Anyone know how to graph a residual line?</p>