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

Question Description

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!

Final Answer

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


iainharlow (1041)
New York University

I was on a very tight deadline but thanks to Studypool I was able to deliver my assignment on time.

The tutor was pretty knowledgeable, efficient and polite. Great service!

I did not know how to approach this question, Studypool helped me a lot.

Similar Questions
Related Tags

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors