Question

1. What is degrees of freedom?

2. A researcher wants to know if students who take foreign language classes have higher IQs than the general population. The average population IQ is 100. The IQ scores for 5 students in a foreign language class are as follows 120, 98, 116, 100, 100. Compute t.

Answer 1

Solution :

(1) What is degrees of freedom ?

The Degrees of Freedom (df) is defined as the difference of the number of observations present in the calculation independently and the number of linearly independent constraints (or estimated parameters) present in the calculation.

$\begin{bmatrix} Degrees\ of\\ Freedom \end{bmatrix}=\begin{bmatrix} No.of\ independent\\ observations\ in\ the\ data \end{bmatrix}-\begin{bmatrix} No.of\ linearly\ independent\\ constraints\ in\ the\ data \end{bmatrix}$

For example , for the calculation of sample variance of observations , the actual degrees of freedom used is . This happens due to the fact that the "Mean" term ($\overline{x}$) present in the expression of the Sample Variance is estimated which results in the loss of 1 degree of freedom.

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2) It is given that a researcher wants to know if students who take foreign language classes have higher IQs than the general population. The average population IQ is 100. The IQ scores for 5 students in a foreign language class are as follows 120, 98, 116, 100, 100. We have to compute " "

Thus , here we are to test the following hypothesis ,

$H_0\ :\ \mu=100\ \ vs\ \ H_1\ :\ \mu>100\ ;\ where,\ \mu \rightarrow mean\ IQ\ scores\ of\ students$

We have the following information given ,

$X=\{120,98,116,100,100\}\ ;\ Sample\ Size\ (n) = 5\ ;\ \mu_0=100$

$Sample\ Mean\ (\overline{X})=\frac{1}{n}\sum_{i=1}^{n}X_i=\frac{1}{5}(120+98+116+100+100)=106.8$

$Sample\ Standard\ Deviation\ (s)=\sqrt{\frac{1}{(n-1)}\sum_{1=1}^{n}(X_i-\overline{X})^2}$

$=\sqrt{\frac{1}{(5-1)}[(120-106.8)^2+...\ +(100-106.8)^2]}=10.35374$

Thus , the appropiate test statistic is given as ,

$t=\frac{\overline{X}-\mu_0}{s/\sqrt{n}}\sim t_{n-2}\ [under\ H_0]\ ;\ t_{n-2}\rightarrow t\ distribution\ with\ (n-2)\ dfs$

Thus, the value of the test statistic " " is given as ,

$t=\frac{\overline{X}-\mu_0}{s/\sqrt{n}}=\frac{106.8-100}{10.35374/\sqrt{5}}=1.46858............................(Ans)$

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The p - value of the test is , $p-value=0.1079344\ [considering\ \alpha =0.05]$

Since , 0.1079344 > 0.05  $\Rightarrow$ p - value > $\alpha$ , thus , we accept H₀ at 5% level of significance and conclude on the basis of the given data that there is no difference in the IQs of the students taking foreign language classes and the general popuplation.