Lecture 42 Section Tue, Apr 8, 2008


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1 the Lecture 42 Section 14.3 HampdenSydney College Tue, Apr 8, 2008
2 Outline the 1 2 the 3 4 5
3 the The will compute χ 2 areas, but not χ 2 percentiles. (That s ok.) After performing the χ 2 test by hand, we will see how to do it on the. It turns out that there is no special function for a goodnessoffit test. We will also see that the χ 2 test with one degree of freedom is equivalent to the singleproportion Z test.
4  ChiSquare Probabilities the To find a chisquare probability (pvalue) on the, Press DISTR. Select χ 2 cdf. Press ENTER. Enter the lower endpoint, the upper endpoint, and the degrees of freedom. Press ENTER. The probability appears.
5 Computing the pvalue the In this example, df = 5. To find the pvalue, use the to calculate the probability that χ 2 5 would be at least as large as 3.4. pvalue = χ 2 cdf(3.4,e99,5) = Accept H 0. We conclude that the die is fair.
6 the Be careful! There is a function called χ 2 Test, but it does not perform this test. Some TI84s may have a GOFTest. To perform a goodnessoffit test on the, do the following. Put the observed counts in list L 1. Put the hypothetical proportions in list L 1. Multiply L 1 by the sample size and store as L 1. These are the expected counts. Calculate (L 1 L 2 ) 2 /L 2. Go to LIST > MATH and select sum (item #5). Enter Ans and press ENTER. The value of χ 2 appears. Then use χ 2 cdf to find the pvalue.
7 Example  the Suppose we observe 1000 births and find that 520 are male and 480 are female. Does this indicate that male births and female births are not equally likely? Let p 1 = proportion of male births. Let p 2 = proportion of female births. H 0 : p 1 = 0.50, p 2 = 0.50 H 1 : H 0 is not true. α = 0.05.
8 Example  the We have the table Male Female Observed (Expected) (500) (500)
9 Example  the The test statistic is Calculate χ 2 = all cells χ 2 = ( ) = = 1.6 (O E) 2. E ( )2 500
10 Example  the The pvalue is pvalue = χ 2 cdf(1.6,e99,1) = Accept H 0. The proportion of male births is 50%.
11 Example  the Perform the above test as a twotailed oneproportion Z test. That is, let the alternative hypothesis be What is the pvalue? H 1 : p 1 p 2. What is the value of the test statistic Z? Square that number. What do you get?
12 the The will compute χ 2 areas, but not χ 2 percentiles. (That s ok.) The will not perform the goodnessoffit test, although we can do it using lists. The goodnessoffit test with with only two cells is equivalent to the oneproportion Z test.
The GoodnessofFit Test
on the Lecture 49 Section 14.3 HampdenSydney College Tue, Apr 21, 2009 Outline 1 on the 2 3 on the 4 5 Hypotheses on the (Steps 1 and 2) (1) H 0 : H 1 : H 0 is false. (2) α = 0.05. p 1 = 0.24 p 2 = 0.20
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