SOLUTION: You are the incoming inspector for potato chips – you are to ensure that each bag has 16 ounces or more in it. You want your testing to be at the level of significance of 0.05. Y
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Question 133946: You are the incoming inspector for potato chips – you are to ensure that each bag has 16 ounces or more in it. You want your testing to be at the level of significance of 0.05. You pull a sample of 49 bags of chips from a recent truckload. Your sample statistics are:
x-bar (the sample mean) = 15.9 ounces
s ( the sample standard deviation) = 0.35 ounces
(a) what is the null and alternative hypothesis
(b) one or two tailed test ??
(c) what is the critical z value for your test at the 0.05 level of significance??
(d) what is the calculated z value ??
(e) what is your decision about the load of potato chips ??
-- reject ?? -- not-reject ?? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! You are the incoming inspector for potato chips – you are to ensure that each bag has 16 ounces or more in it. You want your testing to be at the level of significance of 0.05. You pull a sample of 49 bags of chips from a recent truckload. Your sample statistics are:
x-bar (the sample mean) = 15.9 ounces
s ( the sample standard deviation) = 0.35 ounces
(a) what is the null and alternative hypothesis
Ho: mu = 16 oz (Claim)
Ha: mu < 16 oz
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(b) one or two tailed test ??
Ans: one
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(c) what is the critical z value for your test at the 0.05 level of significance??
z= -1.645
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(d) what is the calculated z value ??
z(15.9) = (15.9-16)/[0.35/sqrt(49)]= -0.1*7/0.35= -2
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(e) what is your decision about the load of potato chips ??
-- reject ?? -- not-reject ??
since -2 is less that -1.645, reject Ho
The mean weight of the bags is statistically less than 16 oz.
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Cheers,
Stan H.
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