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

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To sell the pipes for specification.Sample mean for strength is 50 pipes to be turned to =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!

Mar 14th, 2015

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)

= 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 = (x̄-μ)/SEM

= (2460-2400)/28.28


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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