Question

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

Answer 1

Answei Date: 16/04/2019 To test the hypothesis is that the color distribution for the number of M&M's is different from that expected at 5% level of significance. 1) The null and alternative hypothesis is, The null hypothesis is that the color distribution for the number of M&M's is same from that expected The alternative hypothesis is that the color distribution for the number of M&M's is different from that expected. 2) The chi-square test statistic for goodness-of-fit test is, First, compute frequency expected then find chi-square test statistics The expected frequency for blue color is,

Np = 2620 × 24 % 628.80 The expected frequency for brown color is s.- Npi = 2620 × 13 % - 340.60 The expected frequency for green color is, 2620 x1690 419.20 The expected frequency for orange color is,

2620 2090 - 524 The expected frequency for red color is, JNpi 2620 13 % 340.60 The expected frequency for yellow color is 2620 x14% 366.80 The chi-square test statistic for goodness-of-fit test is, By using Excel software, find chi-square test statistics with the help of following steps 1) Import the data.

2)Select Goodness-of-Fit Test from Chi-Square in MegaStat option from Add Ins menu 3) Select Observed values and Expected values 4)Click Ok Goodness of Fit Test O-E 628.800147.800 30.400 63.800 20.000 31.400 2.200 0.000 (0-E)2 / E 34.741 2.713 9.710 0.763 2.895 0.013 50.835 observed expected %ofchis 340.600 419.200 524.000 340.600 366.800 2620 2620.000 481 371 483 544 372 369 68.4 5.34 19.10 1.50 5.69 0.03 100.00 50.835 chi-square 5 df 9.35E-10 p-value From the Excel output, the chi-square test statistic for goodness-of-fit test is 50.835 The p-value is,

From the Excel output, the chi-square testp-value is 0.0000. The p-value is 0.0000 3) Decision The conclusion is that the significant p-value is less than 0.05 which is 0.0000, so the null hypothesis is rejected at 5% level of significance 4) There is sufficient evidence to indicate that the color distribution for the number of M&M's is different from that expected. The result is statistically significant.