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 p-value?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 two-tailed (a difference in either direction causes you to reject the null hypothesis).
I'll work through the latter (two-tailed) 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 z-value for a two-tailed hypothesis at alpha = 0.05 is
norminv(p = 1-alpha/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)
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)
= 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 = (x̄-μ)/SEM
Since this is greater than the critical z value of 1.96, we can reject the null hypothesis. We can also find the p-value:
P(z>2.121) = 1 - P(z<2.121)
= 1 - 0.983052573
Mar 14th, 2015
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