An article described an investigation into the coating weigh
Solution
a)
As we can see, the middle 50% of data (box part) is greater than 200.
Thus,
OPTION A: It appears that the true average weight could be significantly off from the production specification of 200 lb per pipe. [ANSWER]
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b)
Ho: u= 200
Ha: u > 200 [ANSWER]
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As we can see, this is a right tailed test.
Getting the test statistic, as
X = sample mean = 206.38
uo = hypothesized mean = 200
n = sample size = 30
s = standard deviation = 6.35
Thus, t = (X - uo) * sqrt(n) / s = 5.503102231 [ANSWER]
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Also, the p value is
p = 3.13638*10^-6 [ANSWER]
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As P < 0.05, we REJECT THE NULL HYPOTHESIS.
Thus, OPTION B: REJECT THE NULL HYPOTHESIS. There is sufficient evidence... [ANSWER, B]
