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Mathematics & Economics
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Math Problem
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BOTTLING COMPANY CASE STUDY (Math Problem Sample)

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Perform calculations. source..
Content:
BOTTLING COMPANY CASE STUDY Student’s Name: Professor’s Name: Course title: Date: Question 1: Mean, median, and standard deviation for ounces in the bottles Solution: The sampled values are given as: 14.23, 14.32, 14.98, 15.00, 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, 16.91, 16.96 * Mean Sample Mean=1n i=1i=nxi =130i=1i=30xi =130{14.23+14.32+…+16.91+16.96} =130×475.62 =15.854 * Median Since size of the sample = 30, an even number, then; The median will be given by average of the two meddle observations, that is, [n2]th and [n2+1]th Therefore, median = [n2]th + [n2+1]th 2 [n2]th=302 15thobservation =15.98 And [n2+1]th=16thobservation=16.00 Median=15.98+16.002 =15.99 * Standard deviation standard deviation s= i=1i=30(xi-x)2n-1 But x̅=15.854 and n=30 s= (14.23-15.854)2 +…+(16.96-15.854)230-1 s=(-1.624)2+…+(1.106)2 29 s=12.68629 s=0.6614 Question 2: A 95% Confidence Interval for the ounces in the bottles Solution: Confidence interval will be given by the summation of mean and margin error confidence Interval=X±Zsn whereby Zsn is the margin error, But X̅=15.854, s=0.6614 and n=30 At 95% confidence interval Z=1.96 =15.854±1.960.661430 =15.854±0.237 =(15.617, 16.091) Therefore, we are 95% confident that population mean falls between 15.617 and 16.091 Question 3: Hypothesis test Solution: Set null hypothesis (H0) and alternative hypothesis (H1) Note: The null hypothesis is usually a negative statement to the claim. In this case, the null hypothesis should suggest that bottles of the product do not contain less than 16 ounces. On the other hand, the alternative hypothesis is the claim we wish to test, that is, is mean volume less than 16? H0:u0=16 ounces H1:u1<16 ounces Taking 95% as the acceptance level, then significance level would be 5% (0.05) For one tailed test (lower-tail in this case), δ=0.052=0.025 Using normal distribution table to identify rejection region, we find Zδ=Z0.025=1.645 Calculating the test statistic; Zc=x-u0s/√n x=15.854, u0=16, s=0.6614 and n=30 Zc=15.854-160.6614/√30 =-0.1460.12075 =-1.2091 Comparing tabulated and calculated values Zδ=-1.645<ZC=-1.2092 28289261155710390525113447 Rejection 66675041275region 2000250110490 11239501600202000250160020010096492667001114425-19051114425-1905005143508572500428625150495428625112395409575150495 390525-635 -1.645 Zc= -1.2091 0 00 -1.645 Zc= -1.2091 0 4743456273808553451351280 Acceptance region From the sketch, the calculated value Zc falls within the accep...
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