## Presentation on theme: "How plenty of Skittles space In a 2.17 ounce Bag? By: Ryan Riling & Tom Dougherty."— Presentation transcript:

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2 How plenty of Skittles space In a 2.17 ounce Bag? By: Ryan Riling & Tom Dougherty

3 HistoryHistory -Skittles manufacturing originated in England -First introduced to United claims in 1974 -Owned through Mars Inc. -Skittles factory are situated in U.S, Victoria, Australia, and brand-new Zealand -Advertising campaigns are connected with rainbows -“Taste the Rainbow”

4 PurposePurpose -We wanted to determine whether or not Mars Inc. (producer that Skittles) was reasonably filling their bags v the declared amount. -We decided to acquisition 35 standard sized bags of skittles (2.17 ounce) and test to determine if Skittles consumers are acquiring their money’s worth.

5 sleeve Stores -Acme  five 2.17 oz. Bags -Genuardi’s  5 2.17 oz. Bags -Giant  five 2.17 oz. Bags -Redner’s  five 2.17 oz. Bags -CVS  five 2.17 oz. Bags -Wawa  five 2.17 oz. Bags -7-11  5 2.17 oz. Bags total = 35 BAGS

7 GraphsGraphs

8 Graphs (Cont.) 5254565860626466687072 5 Number review Minimum = 53 Quartile 3 = 63 Quartile 1 = 56 preferably = 68 average = 59 number of Skittles

9 Stem Plot 5 53 54 5 5 5 56 6 6 6 7 7 7 7 58 8 8 9 9 9 9 9 9 60 62 2 2 3 3 3 64 4 5 66 6 68 shape = about Symmetric center = 59 spread out = Minimum – 53 maximum – 68

10 1 Var Stats x = 59.4286 Σx = 2080 Σx² = 124110 Sx = 3.8293 n = 35 Minimum = 53 Quartile 1 = 56 mean = 59 Quartile 3 = 63 preferably = 68

11 assumptions 1). SRS 1). 2). Normal populace 2). 35 ≥ 30 OR n ≥ 30

12 HypothesisHypothesis -Ho:  = 60 skittles per 2.17 oz. Bag -Ha:  ≠ 60 cones per 2.17 oz. Bag

) =.3835 levels Freedom: Df = n-1 =" > 13 TestsTests One Sample T-Test check Statistic: t* = x - µ s/ √n = P-Value: 2 * P(µ > -.8828) =.3835 degrees Freedom: Df = n-1 = 34 -.8828

) =.3835 levels Freedom: Df = n-1 =" title="TestsTests One Sample T-Test check Statistic: t* = x - µ s/ √n = P-Value: 2 * P(µ > ) =.3835 levels Freedom: Df = n-1 =">

14 tests (Cont.) Conclusion: us fail to refuse the null hypothesis because our p-value is better than  =.05. Us have sufficient evidence the the mean number of Skittles every 2.17 oz. Bag is 60 Skittles.

15 confidence Level (95%) to trust Level = x ± t*(s/ √ n) = (58.113, 60.744) We space 95% Confident the the mean number of Skittles every 2.17 oz. Bag is between 58.113 and 60.744 Skittles.

16 an individual Opinions -We felt as though it was very tedious to counting the lot of cones in each of the 35 bags -It to be time-consuming to travel to every of the 7 stores to acquire the compelled amount that samples -We agree with our T-Test results and also feel together though where ever you choose to buy her Skittles from, you are gaining a same amount per bag for the price.

17 ApplicationApplication -Although huge had the best average variety of Skittles every bag, us feel as though that is unnecessary come go out of your method just to buy skittles at Giant. -We feel as though Mars Inc. Relatively manufactures and also packages your Skittles bags.

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- 7-11 skittles bags room packaged most fairly and have actually an typical of 60.2 skittles per bag.

18 Bias/ErrorBias/Error -Incorrect Skittles count -Mistake entering data right into lists -Obtaining cones at miscellaneous stores  determined the very first available bags -Counting broken or deformed cones