Chi Square Assignment MAT308
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Wilmington University *
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MAT-205
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Industrial Engineering
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Feb 20, 2024
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Chi Square Test (GOF and Independence) Name M&é@f\ Gf L’fifl Per. M j”)‘% Chi Square Modeling Using M & M’s Candies Introduction: Have you ever wondered why the package of M&Ms you just bought never seems to have enough of your favorite color? Or, why is it that you always seem to get the package of mostly brown M&Ms? What’s going on at the Mars Company? Is the number of the different colors of M&Ms in a package really different from one package to the next, or does the Mars Company do something to insure that each package gets the correct number of each color of M&M? You’ve probably stayed up nights pondering this! One way that we could determine if the Mars Co. is true to its word is to sample a package of M&M:s and do a type of statistical test known as a “goodness of fit” test. These type of statistical tests allow us to determine if any differences between our observed measurements (counts of colors from our M&M sample) and our expected (what the Mars Co. claims) are simply due to chance sample error or some other reason (i.e. the Mars Co.’s sorters aren’t really doing a very good job of putting the correct number of M&M’s in each package). The goodness of fit test we will be doing today is called a Chi Square Analysis. This test is generally used when we are dealing with discrete data (i.e. count data, or non continuous data). We will be calculating a statistic called a Chi square or X* We will be using a table to determine a probability of getting a particular X? value. Remember, our probability values tell us what the chances are that the differences in our data are due simply to chance alone (sample error). The Chi Square test (X?) is often used in science to test if data you observe from an experiment is the same as the data that you would predict from the experiment. This investigation will help you to use the Chi Square test by allowing you to practice it with a population of familiar objects, M&M candies. Objectives: After this investigation you should be able to: write a null hypothesis that pertains to the investigation; determine the degrees of freedom (df) for an investigation; calculate the X value for a given set of data; use the critical values table to determine if the calculated value is equal to or less than the critical value; e determine if the Chi Square value exceeds the critical value and if the null hypothesis is accepted or rejected.
Chi Square Test (GOF and Independence) M&M DATA (Individual) Here are the percentages given by M&M on their website for each color. —./- & « Green=15%*5b % Orange = 23%+54-13.34 Blue =23% , 5¢ - 13, 34 . Brown=12%a5‘5;(n.9b e Red=12% « 59 =144 * Yellow=15%.59- ¢ 1> 1) Open 2 bags of M&Ms. (If you do not have 2 bags of M&M’s email me and | will send you a set of data.) 2) Separate the M&Ms into color categories and count the number of each color. 3) Record your M&M color totals in the data table. Table 1 (4 st Gond 5 set ot Py oy bamse T 00T b 4m: r emal ) Brown Red Yellow Green Orggge Blue 1) | £ ] 4 = Total Number of M&M’s S Cfi’ 4) Calculate the expected number of M&Ms in your package by multiplying the total number of M&Ms in the package by the color percent listed on page 1 of the activity. For example, if your package contains 500 M&Ms and you want to find the expected number of red M&Ms you will need to multiply 500 by 20% (500 x 0.20). Record your calculations in the data table. 5) Calculate the difference between the observed and expected numbers for each M&M color. Record your calculations in the data table. 6) Square the difference between the observed and expected. Record your calculations in the data table. 7) Divide the square of the difference by the expected. Record your calculations in the data table. 8) Total all the answers from step 7 to determine the chi-square (A2 value. Record the chi-square (A?) in the data table. 0
Chi Square Test (GOF and Independence) Table 2 Colors Observed 4) Expected 5) o-e 6) (0-e)? 7) (0-e)? (0) (e) , £ Brown |4 .40 ZOH “74 QU 70.43 Red I8 (.40 e (1 [oM q. 29 Yellow 9 q.7 (.3 ] &.0f i %9 4.7 05 [0%6 [np Orange y 1. %4 J0.5u R % .0l Biue | [3.34 o, |39z | $.0l 8.5 = 4 le.lle Analysis Questions: 1. What are the null and alternative Hypothesis? Ho Q' = U} z /*‘:} Ha U-. g # O Now you must determine the probability that the difference between the observed and expected values (as summarized by the calculated value of chi square) occurred simply by chance. To do this you will need to compare the calculated value of chi-square with the appropriate value from the Chi Square Distribution Table on the next page. Examine the table. Note the term “degrees of freedom.” For this statistical test the degrees of freedom is equal to the number of classes (color categories) minus one. Complete the following to determine the degrees of freedom for the M&M analysis: # of color categories (fl« degrees of freedom h The reason why it is important to consider degrees of freedom is that the value of the chi-square statistic is calculated as the sum of the squared differences for all classes. The natural increase in the value of chi-square with an increase in classes must be taken into account. Scan across the row corresponding to 5 degrees of freedom. Values of the chi-square are given for several different probabilities ranging from 0.95 on the left to 0.001 on the right. Note that the chi-square increases as the probability increases. Notice that a chi-square value of 1.63 would be expected by chance in 95% (0.95) of the cases, whereas one of 12.59 would be expected in 5% (0.05) of the cases. Use the chi-square value calculated and recorded on the data table to determine the probability for the M&M analysis. If the exact chi square value is not listed in the table estimate the probability. Record your answer below.
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Chi Square Test (GOF and Independence) CHI-SQUARE DISTRIBUTION TABLE Accept Hypothesis Reject Hypothesis ‘ M BN WS AN NS Probability (p) Degrees of 0.95 1 0.90/0.80|0.70 0.50| 0.30 | 0.20 | 0.10 ] 0.05 | 0.01 | 0.001 Freedom 1 0.004 | 0.02 { 0.060.15 0.46| 1.07 164 | 2.71 } 3.84 | 6.64 | 10.83 2 0.10 10.210.45/0.71 /1,39 2.41 | 3.22 | 4.60 5,99 | 9.21 13.82 3 0.35 10.5811.01/1.42 237| 366 | 464 | 6.25] 7.82 | 11.3416.27 4 0.7t 1 1,06/1.65|2.20 3.36| 4.88 | 599 | 7.78 ] 9.49 | 13.38 18.47 5 1.14 [ 1.61]2.34|3.00/4.35| 6.06 | 7.29 | 9.24 }11.07 | 15.09 | 20.52 6 1.63 12.203.07|3.83/5,35| 7.23 | 8,56 | 10.64}12,59 | 16.81 | 22.46 7 2.17 12.83/3.82/4.676.35| 8,38 | 9.80 [12.02]114.07 | 18,48 | 24.32 8 2,73 1 3.49|4.59|553 7.34| 9,52 11.03]13.36]15.51 | 20.09 | 26.12 S 3.32 4.1715.38/6.39 8.34|10.6612.24 | 14.68]116.92 | 21.67 | 27.88 10 3,94 4.866.18[7.27 9.34|11.78 13.4415.99]18.31 | 23.21 29.59 2, Draw your Chi Square Curve and put in the critical value (p = 0.05). 0.1~ 0.1 - 040 0.65 - G-60 : ' } ] 0 5 ot § % - 1.0 3. What is the A? value for your data? _!| M . Is your null hypothesis accepted or rejected? Explain why or why not. Tf/"' Nl \n\ M’}u€s 5 e fed brawse wompantt X's 4o 105 ' 0ot F m beiare YT sl assig ;. .0/\7 e region o 049, Fhorelon