Confidential Interval

Assignment: M&M Project Part 3 M&Ms Project Part 3 Strayer University maths 300 Statistics Professor Ahmed Rawish March 14, 2012 make over a 95% Confidence separation for the similarity of distressing M&Ms candies. 95% Confidence insularity 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% trustingness we can claim that the proportion of blue M&Ms® candies is within (0.190492031, 0.214868103). Details |Confidence time interval Estimate for keeping | | | | |Data |  | |Sample Size |4179 | | descend of Successes |847 | |Confidence level |95% | |  |  | |Intermediate Calculations | |Sample Proportion |0.202680067 | |Z comfort |-1.

959963985 | |Standard Error of the Proportion |0.0062185 | |Interva! l 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 condition 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, youth hostel it on our website: BestEssayCheap.com

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