MAT 300 Assignment 1

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The Calculation of the Mean, Median, Standard Deviation and Sum • Mean: The element of mean is calculated by taking the total of all the ounce values and dividing the sum by the number of values. ... For the values included for this assignment, the mean is: ̅=15.854SUM = 475.62/N=30 =15.854 • Median MD: This is determined by identifying the midpoint of the values. For the data provided for this assignment, the median is: MD =15.99 • Standard Deviation: Determining the standard deviation is a value that is utilized to determine how broadly the dataset differs. Standard deviation calculations based on the data provided for this assignment is: SD = 0.661 • Sum: The sum is for this case study is the total number of ounces in Σ = 475.62 N √ i N Confidence Intervals Constructed at 95% The mean sample of bottled sodasis: ̅=15.854 1 n ´x= ∑xi i=1 There are 30 samples to make up the sample size that is a part of the data, so n=30 The standard deviation sample observed is: s = 0.661 s= 1 N−1 n ∑( x −´x) 2 i=1 The population standard deviation: σ = 0.650 √ Population Variance: σ 2 = 0.423 2 1 n 2 σ = ∑( xi−μ) i =1 As a part of this case study, I will work to the 95% confidence level; the corresponding confidence interval is set at +0.24 The lower limit utilizing the 95% confidence levels is: = 15.617 x−1.96 σ √n 15.854−1.96 0.661=15.6 17 √30 The upper limit of the 95% confidence level is: = 16.090 x+1.96 σ √n 15.854+1.96 0.661=16.090 √30 σ= i 1 N n ∑( x −μ) 2 i=1 By utilizing a 95% confidence level interval, I can determine with 95% confidence that the level interval of 15.617 to 16.090 will make up the mean population. This outcome is based on a 30 bottle sample of sodas. Testing of Hypothesis Because I understand that every bottle of soda should contain at least 16 ounces o [Show More]

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