Question 11.4: Test of Normality Use chi-square to determine if the variabl......

$H_{0}$ : The variable is normally distributed.

$H_{1}$ : The variable is not normally distributed.

First find the mean and standard deviation of the variable. (Note: $s$ is used to approximate $\sigma$.)

$\begin{array}{|r|c|r|r|r|} \hline \textbf{ Boundaries } & \boldsymbol{f} & \boldsymbol{X}_{\boldsymbol{m}} & \boldsymbol{f} \cdot \boldsymbol{X}_{\boldsymbol{m}} & \boldsymbol{f} \cdot \boldsymbol{X}_{\boldsymbol{m}}^{\mathbf{2}} \\ \hline 89.5-104.5 & 24 & 97 & 2,328 & 225,816 \\ 104.5-119.5 & 62 & 112 & 6,944 & 777,728 \\ 119.5-134.5 & 72 & 127 & 9,144 & 1,161,288 \\ 134.5-149.5 & 26 & 142 & 3,692 & 524,264 \\ 149.5-164.5 & 12 & 157 & 1,884 & 295,788 \\ 164.5-179.5 & 4 & 172 & 688 & 118,336 \\ &\overline{200}&&\overline{24,680} &\overline{3,103,220} \\ \hline \end{array}$

$\begin{aligned}\bar{X} & =\frac{24,680}{200}=123.4 \\s & =\sqrt{\frac{200(3,103,220)-24,680^{2}}{200(199)}}=\sqrt{290}=17.03\end{aligned}$

Next find the area under the standard normal distribution, using $z$ values and Table E for each class.
The $z$ score for $x=104.5$ is found as

$z=\frac{104.5-123.4}{17.03}=-1.11$

The area for $z<-1.11$ is $0.1335$.

The $z$ score for $119.5$ is found as

$z=\frac{119.5-123.4}{17.03}=-0.23$

The area for $-1.11<z<-0.23$ is $0.4090-0.1335=0.2755$.

The $z$ score for $134.5$ is found as

$z=\frac{134.5-123.4}{17.03}=0.65$

The area for $-0.23<z<0.65$ is $0.7422-0.4090=0.3332$.

The $z$ score for $149.5$ is found as

$z=\frac{149.5-123.4}{17.03}=1.53$

The area for $0.65<z<1.53$ is $0.9370-0.7422=0.1948$.

The $z$ score for $164.5$ is found as

$z=\frac{164.5-123.4}{17.03}=2.41$

The area for $1.53<z<2.41$ is $0.9920-0.9370=0.0550$.

The area for $z>2.41$ is $1.0000-0.9920=0.0080$.

Find the expected frequencies for each class by multiplying the area by 200 . The expected frequencies are found by

$\begin{aligned}& 0.1335 \cdot 200=26.7 \\& 0.2755 \cdot 200=55.1 \\& 0.3332 \cdot 200=66.64 \\& 0.1948 \cdot 200=38.96 \\& 0.0550 \cdot 200=11.0 \\& 0.0080 \cdot 200=1.6\end{aligned}$

Note: Since the expected frequency for the last category is less than 5, it can be combined with the previous category.

The table looks like this.

$\begin{array}{|l|l|l|l|l|l|} \hline \mathbf{O} & 24 & 62 & 72 & 26 & 16 \\ \hline \boldsymbol{E} & 26.7& 55.1 & 66.64& 38.96 &12.6 \\ \hline \end{array}$

Finally, find the chi-square test value using the formula $\chi^{2}=\sum \frac{(O-E)^{2}}{E}$.

$\begin{aligned}\chi^{2}= & \frac{(24-26.7)^{2}}{26.7}+\frac{(62-55.1)^{2}}{55.1}+\frac{(72-66.64)^{2}}{66.64}+\frac{(26-38.96)^{2}}{38.96} \\& +\frac{(16-12.6)^{2}}{12.6} \\= & 6.797\end{aligned}$

The critical value in this test has the degrees of freedom equal to the number of categories minus 3 since 1 degree of freedom is lost for each parameter that is estimated. In this case, the mean and standard deviation have been estimated, so 2 additional degrees of freedom are needed.

The C.V. with d.f. $=5-3=2$ and $\alpha=0.05$ is $5.991$, so the null hypothesis is rejected. Hence, the distribution can be considered not normally distributed. Note: At $\alpha=0.01$, the C.V. $=9.210$ and the null hypothesis would not be rejected. Hence, we could consider that the variable is normally distributed at $\alpha=0.01$. So it is important to decide which level of significance you want to use prior to conducting the test.