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According to the M&M candy company, the expected proportion can be found in Table #11.2.3. In...

According to the M&M candy company, the expected proportion can be found in Table #11.2.3. In addition, the table contains the number of M&M’s of each color that were found in a case of candy (Madison, 2013). At the 5% level, do the observed frequencies support the claim of M&M? Table #11.2.3: M&M Observed and Proportions

Blue Brown Green Orange Red Yellow Total
Observed Frequencies 481 371 483 544 372 369 2620
Expected Proportion 0.24 0.13 0.16 0.20 0.13 0.14

Using this link to answer all four steps: http://www.quantpsy.org/chisq/chisq.htm

1. Claim (null hypothesis, alternate hypothesis)

2. Calculate the test statistic (p-value)

3. Reject or Fail to Reject

4. Conclusion

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Answei Date: 16/04/2019 To test the hypothesis is that the color distribution for the number of M&Ms is different from thatNp = 2620 × 24 % 628.80 The expected frequency for brown color is s.- Npi = 2620 × 13 % - 340.60 The expected frequency for g2620 2090 - 524 The expected frequency for red color is, JNpi 2620 13 % 340.60 The expected frequency for yellow color is 2622)Select Goodness-of-Fit Test from Chi-Square in MegaStat option from Add Ins menu 3) Select Observed values and Expected valFrom the Excel output, the chi-square testp-value is 0.0000. The p-value is 0.0000 3) Decision The conclusion is that the sig

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