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M&M Project - Essay Example

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M&M Project Introduction: The purpose of the project is to compare different proportion of candies in each colour between the two sample proportions and give an analysis report and construct confidence interval for proportion of each color of candies in the bag…
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M&M Project Essay
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Download file to see previous pages There were 1587 orange candies in the bag out of 7356 candies and the proportion of orange candies is 0.2157. The 95% Confidence Interval is (0.2063,0.2251). There were 1320 green candies in the bag out of 7356 candies and the proportion of green candies is 0.1794. The 95% Confidence Interval is (0.1707, 0.1882). There were 925 yellow candies in the bag out of 7356 candies and the proportion of yellow candies is 0.1257. The 95% Confidence Interval is (0.1182, 0.1333). There were 990 red candies in the bag out of 7356 candies and the proportion of red candies is 0.1346. The 95% Confidence Interval is (0.1268, 0.1424). There were 999 brown candies in the bag out of 7356 candies and the proportion of brown candies is 0.1358. The 95% Confidence Interval is (0.128, 0.1436). Part 3: The sample size in estimating the proportion is 438. In testing whether the given sample proportion of blue candies (0.2087) matches with the population proportion 0.24 of blue candies. The hypothesis is rejected (Z=-6.291) since observed value is less than the expected value -1.96. Hence there is a significant difference between the sample proportion and population proportion. In testing whether the given sample proportion of orange candies (0.2157) matches with the population proportion 0.2 of orange candies, the hypothesis is rejected (Z=3.3754) since observed value is greater than the expected value 1.96. ...
Hence there is a significant difference between the sample proportion of green candies and population proportion of green candies. In testing whether the given sample proportion of yellow candies (0.1257) matches with the population proportion 0.14 of yellow candies, the hypothesis is rejected (Z=-3.523) since observed value is less than the expected value of -1.96. Hence there is a significant difference between the sample proportion of yellow candies and population proportion of yellow candies. In testing whether the given sample proportion of red candies (0.1346) matches with the population proportion 0.13 of red candies, the hypothesis is rejected (Z=1.1691) since observed value is less than the expected value of 1.96. Hence there is no significant difference between the sample proportion of red candies and population proportion of red candies. In testing whether the given sample proportion of brown candies (0.1358) matches with the population proportion 0.13 of brown candies, the hypothesis is rejected (Z=1.4811) since observed value is less than the expected value of 1.96. Hence there is no significant difference between the sample proportion of brown candies and population proportion of brown candies. Part 4 In testing the claim that a 1.69 oz bag contains more than 54 candies at 1% level of significance, the sample mean is 55.7273 and the population mean is 54 and n=123. Since the test statistic is positive (2.9372) and is greater than 1.645, we REJECT. We have sufficient evidence to suggest the true SD is not 1.5. To compare the proportion of red and brown candies in the bag: Since the test statistic is greater than -1.96, we have sufficient evidence to NOT TO REJECT null hypothesis H0. We accept the null hypothesis and conclude that ...Download file to see next pagesRead More
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