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Bottling.e

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RUNNING HEAD: BOTTLING COMPANY CASE ANALYSIS . BOTTLING COMPANY CASE ANALYSIS Student Name: Institution affiliation: Date 1 BOTTLING COMPANY CASE ANALYSIS 2 1) Mean (14.23+14.32+14.98+15+15.11+15.21+15.42+15.47+15.65+15.74+15.77+15.80+15.82 +15.87+15.98+16.00+16.02+16.05+16.21+16.21+16.23+16.25+16.31+16.32+16.34+16 .46+16.47+16.51+16.96)/30=15.81 ounces Median= (15.93+15.95)/2=15.94 mean 15.81 Median 15.94 standard deviation 0.66538 2) 95% Confidence Interval x̅ = 15.81, standard deviation=0.66538, significance level= 0.05, n=30 𝑡0.025 = 2.045 Confidence interval (15.81 − 2.045 ( 0.66538 √30 ) , 15.81+2.045 ( 0.66538 √30 )) = (15.56157092, 16.05842908) Therefore, as per the results above, we can be 95 percent confident that the mean of the ounces in the bottle is between 15.561 ounces & 16.05842908 ounces 3) Hypothesis test Hypotheses The Null hypothesis: BOTTLING COMPANY CASE ANALYSIS 3 The mean of the ounces of soda in the bottle is equal to 16 ounces H0: µ =16 ounces Alternative hypothesis: The mean of the ounces of soda in the bottle is less than 16 ounces H1 < 16 ounces. x̅ = 15.81, standard deviation=0.66538, significance level= 0.05, n=30 DF = 30-1= 29 𝑡= 𝑥̅ − 16 15.81 − 16 = = −1.56403 𝑠 0.66538 ( ) ( ) 30^0.5 √𝑛 The p-value with the degrees of freedom of 29= 0.06433 Since the p-value arrived at (0.06433) > the alpha value 0.05. The null hypothesis that ounces of soda in the bottle is equal to 16 is accepted. The ...
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