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My try:

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Is it correct? The reason I am asking this because , when I tried to calculate the value of t by std . dev and mean , I got a different value from 1.9527887

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2 Answers 2

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If "mean" is taken to be the mean of $15$ differences between the $15$ measured cut-on voltages and the hypothetical average of $0.6$, then my software agrees with the output in the posted question above.

However, in that case I would say the problem is not expressed clearly.

One should have $$ T = \frac{\overline X - 0.6}{s/\sqrt{15}}. $$ If $\overline X-0.6 = 0.0453333$ then this is $$ T = \frac{0.0453333}{0.08991/\sqrt{15}} = 1.952787 $$ which agrees with what is reported. Then one finds that $\Pr(|T| > 1.962787) = 0.07113305$ where $T$ has a t-distribution with $14=15-1$ degrees of freedom, and that is greater than $0.01$, so at the $0.01$ level one does not reject the null hypothesis.

My software reports $\Pr(T>1.952782) = 0.03556652$, and that is what must be multiplied by $2$ to get $0.07113305$. You shouldn't be multipllying $0.07113305$ by $2$.

But I would would have thought that the output meant $\overline X = 0.0453333$ rather than $\overline X - 0.6 = 0.0453333$. That is what is confusingly stated.

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  • $\begingroup$ I agree that the output is unclear. It would be better if it said “Mean difference” or even “Mean difference from 0.6.” The hypothesized population mean was probably required as input for the test somewhere but doesn’t show up in this part of the output. $\endgroup$
    – Steve Kass
    Apr 8, 2016 at 1:19
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    $\begingroup$ Thanks for clearing the confusion $\endgroup$
    – max
    Apr 8, 2016 at 1:24
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The output shown in the question indicates under “Mean” appears to be the mean difference in cut-on voltage from $0.60$ volts. The observed mean cut-on voltage was therefore $0.0453333$ more than the hypothesized population mean, or $0.6453333$. The standard deviation in the sample is given as $0.08991$, so the standard error is $\frac{0.08991}{\sqrt{15}}\approx0.023215$, and the observed mean was $\frac{0.0453333}{0.023215}\approx1.95276$ standard errors from the hypothesized population mean. The difference between $1.95276$ and the $t$-statistic provided is probably due to rounding or keeping a different number of decimal places somewhere.

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