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Can someone show me how to use the table? I know that T is 0.745 but how do I find P and use the table.

[img]https://media.cheggcdn.com/media%2F449%2F449a694b-eeab-4b17-b8f9-339b1b7f263c%2FphpFR2xCP.png[/img]

Notice two things in the table.

• For a given value of $nu$, an increase of the $t$-value corresponds to a decrease of the $alpha$-level (or p-value). That means high $t$-values are rarer when the null-hypothesis is true (if you observe a high $t$-value then this is 'special').




• For a given $alpha$-level the t-values to obtain this level are lower when $nu$ increases.

(sidenote: the table is for positive t-values, but the same can be done for negative t-values)

Intuitively: you find the t-value by dividing the mean by the estimate of the variance. This estimate of the variance is a variable whose variance depends on the size of the sample. Every time you perform an experiment it will be different, and the smaller the sample the larger this difference.

So a smaller sample size will cause the $t$-score to differ to a larger extent from experiment to experiment. When your sample is smaller, then the variance in the $t$-score will be larger and therefore larger $t$-score values will be less 'special'.

You should look at the row for $nu = 29$

 alpha 

 0.40 0.25 0.10 0.05 

nu 29 0.256 0.683 1.311 1.6999

You are not gonna find the value exactly but, what kind of $alpha$ or $p$ does the t-value $0.745$ correspond to? Between which two $p$ values should it be?

I put your summarized data into Minitab's 'one-sample t` procedure.
Here are results.

One-Sample T 

Test of μ = 98.2 vs ≠ 98.2

 N Mean StDev SE Mean 95% CI T P
30 98.285 0.625 0.114 (98.052, 98.518) 0.74 0.462

Here $T = frac98.285 - 98.20.625/sqrt30 = 0.7449027,$ which
Minitab rounds to 0.74.

If $T sim mathsfT(29)$ then one can use software to find that
$P(T < .7449) approx 0.2312.$ For a two-sided t test the P-value
is $P(|T| > .7449) approx 2(0.2312) approx 0.4624,$ which Minitab
rounds to 0.46.

P-values are 'creatures' of the computer age. Computations beyond elementary-school arithmetic are required to find exact P-values. Once you know that the
P-value of a test is $0.46 > .05,$ you know you can't reject $H_0$
at the 5% level (or the 10% level or at any other reasonable level).

As @MartijnWeterings (+1) has shown, you can use a sufficiently detailed
t table (row for 29 DF) to see that $0.256 < T < 0.683$ implies
$0.80 > textP-value > 0.50$ for 2-sided P-values. But you usually won't be able to find
exact P-values from printed tables.

The figure below shows the density function of $mathsfT(29).$
Our observed $T$-statistic is shown by the vertical heavy blue line, which cuts area 0.2312 from the upper tail. Vertical dotted red lines show areas
0.40, 0.25, and 0.10 cut from the upper tail by tabled values
0.256, 0.683, and 1.311, respectively.

Finally, you can't use hypothesis testing to "determine' mean human body
temperature. You can say your data are 'consistent with' 98.2. Or (from Minitab's confidence interval) with
lots of other values between 98.05 and 98.52 degrees Fahrenheit.

Note: For the reverse procedure, getting $T$ from a P-value, see
this Q&A. Another look at the connection between
$T$ and P-value.