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Background

In my university class, we've been discussing the following experiment:

Consider an experiment on artificially raised salmon, with two treat-ments (one a control) and $20$ fish per treatment. Average weight gains(g) over the experimental period were $1210$ and $1320$ grams. The estimate of variation between fish within a group was $s = 135\mathrm{g}$. Did treatment improve growth rate?

My professor established the

  • observed difference between group means $1320 - 1210 = 110\mathrm{g}$
  • variation between two means expected solely from chance $135×\left(\frac{2}{20}\right)^{0.5}= 42.7$
  • test statistic = $\frac{110}{42.7} = 2.58$

My work so far

I've wanted to work through the formula, to get better intuition on the process

$$\begin{align*} t&=\frac{\overline{x}-\overline{y}}{\sqrt{S^2_p\left(\frac{1}{n_1}\frac{1}{n_2}\right)}}\\[5pt] t&=\frac{1320-1210}{\sqrt{S^2_p\left(\frac{1}{20}+\frac{1}{20}\right)}}\\[5pt] &=\frac{110}{^{135\sqrt{\frac{2}{20}}}}\\[5pt] &\approx2.57667\dots \\[5pt] \end{align*}$$

I understand that $2.58$ is statistically signifigant. My professor also stated the $t$-table shows this being $38$, and the chance of a value as large as $2.58$ is about $1$ in $100$. I've had no success getting this, using the following table, to derive the $1$ in $100$ chance.

How would this be derived from the $t$-table?

Pitouille
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Jessie
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  • What is your question? Is it about the value for the DF (as stated in the title), or about using the table, or about getting better intuition, or perhaps something else? – whuber Sep 18 '21 at 19:09
  • @whuber, I've updated my title and question accordingy. Thank you for raising this. – Jessie Sep 18 '21 at 20:11
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    I think your misunderstanding starts after your calculations: you got a t-value of 2.58. Then, you use the t-table (for a degree of freedom of 38) to determine the p-value… which gives you 0.01 for a one-tailed test. – Pitouille Sep 18 '21 at 20:25
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    1. Your professor appears to be using the z-table rather than the t-table (2.58 is the two-tailed critical value at the 1% level for the normal table -- as you can see in the $\infty$-row at the bottom of of the table you link, while you have 38 df) ... and arguably might seek a one-tailed value instead. $:$ 2. See "How do I find values not given in statistical tables?" for how to get an approximate value for 38 d.f. when the table doesn't have a row for it and you don't have a program that will do it for any d.f. available. – Glen_b Sep 19 '21 at 02:36

1 Answers1

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You are doing a one-sided, two-sample pooled t test. You are correct that under the null hypothesis (no difference) the test statistic $T$ has Student's t distribution with DF $= 40-2=38.$

The critical value $c$ for a test at significance level $0.01 = 1\%$ cuts probability $0.01$ from the upper tail of Student's t distribution with 38 degrees of freedom. You would reject the null hypothesis (no change) against the alternative that the treatment improves growth at the 1% level of significance if $T \ge c.$

The table in your link does not show values for DF = 38, skipping from DF = 35 to DF = 40, on account of the small change in values as DF increases in that part of the table. According to the table, the $c$ is between 2.44 and 2.42. Using software, you could get the exact value. For example, R statistical software gives $c = 2.429.$ Because your observed value of the test statistic is $T = 2.577 > 2.492,$ you would reject the null hypothesis at the 1% level of significance.

qt(.99, 38)
[1] 2.428568

The P-value of your test is the probability of getting a $T$-value greater than or equal to $2.577$ if the null hypothesis is true. Printed tables of t distributions are seldom adequate to give exact P-values, but from your table you can guess that the P-value must be between $0.01$ and 0.005.$

The exact P-value from R (where pt is the CDF of a t distribution) is $P(T \ge 2.577 | H_0) = 0.00598.$ Similar precision for P-values is provided as output from t tests in most statistical software, but this degree of precision is not necessary to know whether to reject at the 1% level of significance.

1 - pt(2.577, 38)
[1] 0.006986493

Here is a plot of the density function of Student's t distribution with DF = 38. The solid vertical line is the observed value of the $T$ statistic and the P-value is the area (probability) under the density curve to the right of that line.

enter image description here

R code for figure:

curve(dt(x,38), -3.5, 3.5, ylab="PDF", xlab="t", 
      main = "T(38) Density")
 abline(h = 0, col="green2")
 abline(v = 0, col="green2")
 abline(v = 2.577, lwd=2)
BruceET
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