The strength of residential pipes equal 2400lbs per feet. Each manufacture wants
Statistics

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To sell the pipes for specification.Sample mean for strength is 50 pipes to be turned to x̅ =2460lbs per linear foot.Assume the population Standard deviation =200lbs per linear foot. can you reject the null hypothesis at 5%? what is the pvalue?Help!
You're testing, presumably, against the null hypothesis that strength = 2400. But it's not specified in your question whether this is one tailed (null hypothesis is strength >= 2400, i.e. greater strength is OK) or twotailed (a difference in either direction causes you to reject the null hypothesis).
I'll work through the latter (twotailed) case since you obviously wouldn't reject the hypothesis that the strength is >=2400 (your sample mean is greater, therefore your p will be >0.5). That question would be trivial.
OK, since you know the population stdev and the sample size is reasonable (n>30) we can use the normal distribution for testing (z values instead of t). The critical zvalue for a twotailed hypothesis at alpha = 0.05 is
norminv(p = 1alpha/2, mean = 0, sd = 1) = 1.96
To get the sample z and compare, we first find the standard error SEM:
SEM = stdev/sqrt(n)
= 200/sqrt(50)
The critical z value for the sample mean is found by adjusting the standard deviation by the root of the sample size n to give the standard error:
SEM = sd/sqrt(n)
= 4.5/sqrt(5)
= 28.28
(think of this as the standard deviation around the mean you get from your sample; the larger your sample the lower the error around the mean).
Then z = (x̄μ)/SEM
= (24602400)/28.28
= 2.121320344
Since this is greater than the critical z value of 1.96, we can reject the null hypothesis. We can also find the pvalue:
P(z>2.121) = 1  P(z<2.121)
= 1  0.983052573
= 0.016947427
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