Assignment: M&M Project Part 3 M&Ms Project Part 3 Strayer University maths 300 Statistics Professor Ahmed Rawish March 14, 2012 redo a 95% Confidence separation for the similarity of robust M&Ms candies. 95% Confidence insularism for residual is devoted by [pic] where p = x/n = 847/4179 = 0.202680067, [pic]= 1.959963985, n = 4179 Therefore, CI is wedded by, [pic] = (0.190492031, 0.214868103) Thus with 95% combine we can claim that the proportion of blue M&Ms® candies is within (0.190492031, 0.214868103). Details |Confidence breakup Estimate for situation | | | | |Data | | |Sample Size |4179 | | descend of Successes |847 | |Confidence train |95% | | | | |Intermediate Calculations | |Sample Proportion |0.202680067 | |Z appraise |-1.
959963985 | |Standard Error of the Proportion |0.0062185 | |Interval Half width ! |0.012188036 | | | | |Confidence Interval | |Interval demean Limit |0.190492031 | |Interval Upper Limit |0.214868103 | 3 pts. Construct a 95% Confidence Interval for the proportion of orange tree M&Ms® candies. 95% Confidence Interval for proportion is presumption by [pic] where p = x/n = 845/4179 = 0.202201484, [pic]= 1.959963985, n = 4179 Therefore, CI is given by, [pic] =...If you want to get a full essay, lodge it on our website:
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