Q. 40 - The Practice of Statistics for AP

Question

Do the data provide convincing evidence of a difference in the effectiveness of the four treatments? Carry out an appropriate test at the $\alpha =0.05$ significance level.

Step-by-Step Solution

Answer

Yes, there is sufficient evidence that there is an association between the variables.

Step 1: Given information

The given data is

Step 2: Explanation

Observed counts

The expected counts are the row total multiplied by the column total, divided by the sample size$n=6602$

The chi-square statistic is the sum of squared deviations (between observed and expected counts) divided by the expected count:

$$\begin{gathered} {\chi }^2=\frac{(250-205.8128{)}^2}{205.8128}+\frac{(206-205.8128{)}^2}{205.8128}+\frac{(211-206.4368{)}^2}{206.4368}+\frac{(157-205.9376{)}^2}{205.9376} \\ +\frac{(1399-1443.1872{)}^2}{1443.1872}+\frac{(1443-1443.1872{)}^2}{1443.1872}+\frac{(1443-1447.5632{)}^2}{1447.5632}+\frac{(1493-1444.0624{)}^2}{1444.0624} \\ =24.243 \end{gathered}$$

The interval for the P-value can then be found in table C in the column title which have the $X^2$-value in the corresponding interval in the row with

$$\begin{gathered} d f=(c-1)(r-1) \\ =(4-1)(2-1) \\ =3 \end{gathered}$$

The  two tailed $p$ value is $\approx 0.000044$

$P<0.0005$

If the P-value is smaller than the significance level, then the null hypothesis is rejected.

$P<0.05 \Rightarrow Reject H_0$

There is sufficient evidence that there is an association between the variables.