Chapter 13 Chi-square tests

There are three different tests in this chapter, but only two distinct methods.

The first method is what you used in class to determine whether your sample was reasonably consistent with the hypothesized proportions by color of Goldfish, Froot Loops, or Smarties. You determine the expected counts by multiplying the hypothesized proportion by the total of objects. The number of degrees of freedom is the number of categories minus one. This was a Chi-square goodness of fit test.

The next method you will use is the Chi-square test of homogeneity. This is used when you have two populations that you are comparing to see if they have a common distribution by the categorical variable. You base this decision on your sample comparison. Using the same methods, you can perform a Chi-square test of independence. This is used to determine whether a sample described in a two-way table by two different characteristics demonstrates independence between the two variables or if there appears to be a connection. In these cases, you have to multiply the row total by the column total and divide by the table total to get the expected count for each cell. You will use (r-1)*(c-1) for the number of degrees of freedom. This is the number of cells you would have to fill in (if you knew all of the totals) before the rest of the cells' values are determined.

You have now seen every topic on the Barron's guide and on the AP exam syllabus. We're almost there!

Chapter 12 Inferences about proportions

BINOMIAL CONNECTION:
The methods of this chapter are based on the binomial distribution.

Let x be the number of successes in n trials.
If the conditions of a binomial setting hold,
then mu of x = np and
sigma sub x =sqrt(n*p*(1-p)).

Now, because p-hat, the estimator for the population proportion equals x/n,
mu sub p-hat = (mu sub x)/n Which means that p-hat is an unbiased estimator of p

and

sigma sub p-hat = (sigma sub x) / n.

Well, if you take that last part, (sigma sub x) / n, and substitute for sigma sub x,
you get sigma sub p-hat = sqrt(np(1-p)) / n

which can be rewritten as sqrt(p(1-p)/n).


WHICH P DO I USE?

If you have a hypothesized p, you use that. For instance, if your previous study or some expert indicated that p = .35, then you use .35 in your hypothesis, the standard deviation for your hypothesis test, and calculations to find the minimum sample size for a margin of error.

You also use this value when checking assumptions np>10 and n(1-p)>10.

If you have only your sample proportion, then you use p-hat to estimate the standard deviation for confidence intervals and for checking conditions for CI: n*p-hat> 10 and n*(1 - p-hat) > 10.

If you have neither, then you must be finding the minimum sample size, so use the most conservative estimate: .5.


2 proportion methods:

It helps A LOT to make a table of values as they showed in the book.

For confidence intervals, the methods just as you imagined. You are developing a confidence interval for the difference between two proportions,
so use p-hat1 - p-hat2.

For the standard deviation,look to the variances. Add the variances of the two samples and take the sqrt. Among the conditions, compare the products n1*p-hat1, n1*(1 - p-hat1), n2*p-hat2, and n2*(1 - p-hat2) to 5. Each product must exceed 5.

For hypothesis tests, there is a nifty twist. Your null hypothesis probably stated that the two proportions were the same. Therefore, their standard deviations should be combined. Take (x1+x2)/(n1+n2) to calculate a new, stronger p-hat which you use for standard deviation calculations and checking conditions.

The standard deviation would be sqrt( p-hat(1-p-hat)/n1 + p-hat(1-p-hat)/n2), but that requires that you enter p-hat too many times. Rewritten, that formula is sqrt( p-hat * (1-p-hat) * (1/n1+1/n2)). It looks nicer in the book. Go there to read all about it.

There are super examples in this chapter.

Chapter 11

The test is Tuesday, March 7th. I will not be available before school to help. Prepare early!

Sunday. 2-4. You know where.