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Transcript of Skittles Statistics
We conducted a simple random sample
Population ≥ 10n
All skittles ≥ 300 = 10(30) Hypotheses Ho: µ = 55 Skittles per 2.17 oz. bag Ha: µ <55 Skittles per 2.17 oz. bag 2.17 oz bags Power Alternative mean: μ = 53
Previous study: s = 2.41
df = 29
reject at α = 0.05
t < -1.699
x < 54.25 z = 2.84
power = .9977 Formulas (not quite accurate because our bag was 2.17 oz) TEST RESULTS DATA Bag Number # of skittles Alpha Level:
α = 0.05 99.77% of the time we reject our null hypothesis, we will be correct! μ = 55
s = 2.527
n = 30
df = n-1 df = 29
α = 0.05
p=1 Conclusion If the true mean of Skittles per 2.17 ounce bag is 55, then the chance of getting a sample like ours is 1. Therefore, there is insufficient evidence to say that the mean of Skittles per 2.17 ounce bag is less than 55. The p-value of our one sample t- test is 1. We fail to reject the null hypothesis because our p-value is greater than α = .05. Sample of Data