A soft drink filling machine, when in perfect adjustment, fills bottles with 12 ounces of soft drink. A random sample of 25 bottles is selected, and the contents are measured. The sample yielded a mean content of 11.88 ounces, with a sample standard deviation of 0.16 ounce.
Set up the hypotheses, and with a .05 level of significance, test to see if the machine is in perfect adjustment. Show your work.
x̅ = 11.88, s = 0.16, n = 25
Null and Alternative hypothesis:
Ho : µ = 12
H1 : µ ≠ 12
Test statistic:
t = (x̅ - µ)/(s/√n) = (11.88 - 12)/(0.16/√25) = -3.75
df = n-1 = 24
Critical value :
Critical value, t-crit = T.INV.2T(0.05, 24) = 2.064
Reject Ho if t < -2.064 or if t > 2.064
p-value = T.DIST.2T(ABS(-3.75), 24) = 0.0010
Decision:
p-value < α, Reject the null hypothesis.
Conclusion:
There is enough evidence to conclude that population mean is different than 12 at 0.05 significance level.
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