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Mathematics 13 Online
OpenStudy (amorfide):

I just need someone to tell me what they get for the test statistic, with their working out, so I know where I am going wrong please.

OpenStudy (amorfide):

OpenStudy (amorfide):

@agent0smith

OpenStudy (amorfide):

this is what I did, it is wrong clearly lol

OpenStudy (amorfide):

@Luigi0210 @e.mccormick

OpenStudy (amorfide):

@Notamathgenius

OpenStudy (notamathgenius):

I have no clue m8 sorry ;-;

OpenStudy (amorfide):

@thomaster

OpenStudy (amorfide):

@ybarrap

OpenStudy (ybarrap):

a) Null Hypotheses: \(\mu = \mu_0\) Alternative: \(\mu > \mu_0\) b) Compute the \(t\) value: $$ \large{ t = \frac{\overline{x} - \mu_0}{s/\sqrt{n}} } $$ Where \(\bar{x}=1065\), \(\mu_0=1050\), \(s=19\) and \(n=50\). c) Compare \(t\) to \(\large T_{critical}\): \(\large T_{critical}\) at the \(5\text{%}\) level is 1.6766: see - http://easycalculation.com/statistics/critical-t-test.php If \(t < T_{critical}\) then the manufacturer's claims are justified because you can not reject the null hypothesis and the difference is likely just due to random variation.

OpenStudy (ybarrap):

Can you see that? If not, I'll screenshot it for you.

OpenStudy (ybarrap):

(just in case)

OpenStudy (amorfide):

ybarrap that gives me 5.58 on part b, but my lecturer has 1.537, and I don't know how... and considering the critical value is 1.645, 1.537 seems like the more logical answer. Any idea how I can get to that?

OpenStudy (amorfide):

@ybarrap

OpenStudy (amorfide):

this is my lecturers answer

OpenStudy (ybarrap):

I would have to say that the lecturer made an error -- you and I came up with the same result independently. The formula is pretty straight-forward and we all agree on all the other parameters.

OpenStudy (amorfide):

Thank you, I have been getting mad at this alot, but I have never had a answer that high on a hypothesis test, I am inclined to believe there is something missing but we did get the same answer so thank you for helping

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