how to find p value from chi square on ti 84

If one thinks dorsum to the problems that considered the difference of two proportions, the method considered a binomial variables, with probability of success p; and compared this with another proportion in order to notice a statistically significant difference. The x ²-exam is an extension of this concept to a multinomial trials where there are grand outcomes, each with probability of success p ₁, p ₂, … , p _m. The x ² compares the observed outcomes with the outcomes expected past assuming a aught hypothesis H ₀ is true. The utilise of the chi-squared test statistic is only advisable when all of the expected counts are greater than or equal to 5!

The test statistic for the chi-squared test is a measure out of how far apart the observed values are from the expected in all cells of the 2-way tables.

Returning to our example, our exam statistic is:

This tells us that large values of the test statistic will signal that the values are far apart, or rather that the distributions are different. This will give us evidence to suggest that our goose egg hypothesis is not true. Exist careful, every bit we will see later the x ⁱⁱdistribution is not symmetric, and the alternative hypothesis has many options for sides and directions only chi is 1-sided. As such, any violation of the null hypothesis volition produce a big exam statistic. However, pocket-size values of x ² are non evidence against the nil hypothesis.

The chi-Squared Distribution

The chi-squared distribution is different depending on the degrees of freedom present in our data, just like was truthful for the Student'due south t-distribution. For this reason computing p-values on this particular distribution is all-time done using engineering science or tables. Since there are ii categorical variables to consider our degrees of freedom is thousand = (r – one)(c – 1), where r and c are the number of rows and columns, respectively. The graph beneath is how the distribution of chi-squared looks for three different values of chiliad.

Example 2

The commodity "Decision of Carboxyhemoglobin Levels and Health Effect on Officers Working at the Istanbul Bosphorus Bridge"(G. Kocasoy and H. Yalin, Journal of Ecology Science and Health, 2004: 1129-1139) presents assessments of wellness outcomes of people working in an environment with high levels of carbon monoxide (CO). Following are the numbers of workers reporting various symptoms, categorized by work shift. Can yous conclude that the proportions of workers with the various symptoms differ amidst shifts?

Ailment	Shift			Totals
	Morning	Evening	Night
Flu	16	xiii	18	47
Headache	24	33	6	63
Weakness	11	16	5	32
Shortness of Jiff	7	nine	ix	25
Total	58	71	38	167

Solution: Showtime the null hypothesis is stated, "H ₀ is there is no divergence in the proportion of workers with the various symptoms between the shifts." This is used to generate the following expected table.

Expected Ailment	Shift
	Morning	Evening	Nighttime	Totals
Influenza	sixteen.232	19.982	ten.695	47
Headache	21880	26.784	14.335	63
Weakness	xi.114	xiii.605	7.281	32
Shortness of Breath	eight.683	ten.629	5.689	25
Total	58	71	38	167

Adjacent we calculate the x ²-examination statistic.

If you are familiar with Excel, you can create a table of the actual counts and a table of the expected counts, and then use the command CHISQ.Examination(actual_range,expected_range), to calculate x ⁱⁱ-examination statistic. Now we calculate the degrees of freedom k = (four – i) (iii – i) = vi. Then we can use the command CHIDIST(x,degrees_freedom, to calculate the p-value in Excel. For this instance we do CHIDIST(17.570, 6) =0.007402. This p-value is less than 5%, so one may conclude that there is enough bear witness to reject the cypher hypothesis. So we conclude that there is evidence to suggest that the proportion of workers with the various symptoms differs amid the shifts.

Alternatively, nosotros could calculate a disquisitional value from the distribution of x ² with half-dozen degrees of freedom and compare it with our test statistic 17.570. This can be done from a table or on Excel using the CHIINV(probability,degrees_freedom) command. Delight note that this gives the critical value for the upper one-tailed exam. In our example, the critical value is CHIINV(0.05,vi) = 12.592. When we compare the critical value is less than our exam statistic, as 12.592 < 17.570. Hence our test statistic falls in the rejection region, which follows our earlier decision.

Additional Uses of the chi-Squared Examination

The chi-squared distribution can as well be used to test of significance virtually variance or the standard deviation of the normal distribution. Information technology likewise can exist used to test the goodness of fit for a theoretical model against sample data.

Chi-Squared Testing with the TI-83/84

All of these exam tin can be found by hitting the [STAT] button and arrowing over to the TESTS menu.

Reckoner Example: The Chi-Squared Goodness of fit exam.

If births were uniformly distributed across the calendar week, nosotros would await that about 1/7 of all births occur during each twenty-four hour period of the week. How closely do the observed number of births fit this expected distribution? The chi-square goodness-of-fit examination is used to decide whether an observed frequency distribution is significantly dissimilar from the expected distribution, or how "good" (sic) the two distributions fit each other. If we were only interested in one 24-hour interval of the week, we could deport a i-proportion z exam. Yet, because we take seven hypothesized proportions, we need to conduct a exam that considers all of them together and gives an overall indication of whether the observed distribution differs from the expected i. The chi-square goodness-of-fit test is only what we demand. Let's consider the frequency distribution of all 2008 Wisconsin births by day of the week.

NOTE: This is not an option on all calculators, yours must accept the GOF test on it..

Solution for the TI-84:

1. Enter the observed data into L1.

ii. Here nosotros are hypothesizing that the births all occur in equal proportions for every day of the week. Now compute the expected frequencies as Expected=northward/thou, due north is the full number of trials (births) and yard is the number of different categories(days of the week). For this case E=116823/7=16689 for all the days since they are all the same. Enter 16689 into all the rows for L2.

3. At present hits [Stat] arrow over to the TESTS menu, arrow down to D: χ^two  GOF-Test striking ENTER. Then enter in the post-obit for the screen:

4. The degrees of liberty is k-1 highlight and hit enter to get:

5. It gives y'all both the value of the χ2 test statistic and its associated P-value. CNTRB provides a list of the CoNTRiButions of each category to the overall χ2 value. Employ the arrow key to coil through these numbers. Round chi-square values to iii decimal places and P-values to 3 significant figures. Yous could study these results every bit P( χ2> 3679.867) ≈0.

What does this mean?

If births were in fact distributed uniformly across the seven days of the week, an observed χ2value of 3679.867 would occur about 0% of the fourth dimension. This result is certainly unusual, so we reject H0 and conclude that the sample data are consistent with births being non-uniformly distributed across the seven days of the week.

Source: https://mat117.wisconsin.edu/the-chi-squared-test/

Posted by: paulinoalonese.blogspot.com

Share this post