The impact of three different advertising techniques is being studied by a marketing firm. Sales, in thousand dollars, categorized in four groups are shown for each advertising technique for 200 randomly selected customers exposed to each technique. Can we conclude that the advertising technique has an impact on sales? Use $\alpha=0.10$. What is p value?
\begin{array}{|c|c|c|c|} \hline Sales& Technique 1 & Technique 2 & Technique 3 \\ \hline 0-99&60 &40 &75\\ \hline 100-199&85 &30 &70\\ \hline 200-299&20 &70 &30\\ \hline 300-399& 35 &60 &25\\ \hline \end{array}
So $H_o$= All techniques are equal
and $H_\alpha$= At least one technique has impact on sales
Now I know we need to be using this formula:
$\sum_{i=1}^k\frac{(n_i-E(n_i))^2}{E(n_i)}$, where $E(n_i)=np_i$
We can see that n=200, but i'm having trouble getting the $p_i$, do we just focus on one row of sales, where we can get all 3 techniques and use that as our sample size? For example
Sales 0-99, n= 175, so Tech 1= $\frac{60}{175}$, Tech 2= $\frac{40}{175}$ or am I supposed to apprach this question differently? Any hints are appreciated.
The $4\times 4$ contingency table: $$\begin{array}{|c|c|c|c|c|} \hline Sales& Technique 1 & Technique 2 & Technique 3 & \text{Total}\\ \hline 0-99&60 &40 &75 & 175\\ \hline 100-199&85 &30 &70 & 185\\ \hline 200-299&20 &70 &30 & 120\\ \hline 300-399& 35 &60 &25 & 120\\ \hline \text{Total}& 200 & 200 & 200 & 600\\ \hline \end{array}$$ Test statistic: $$\chi^2=\sum \frac{(n_{ij}-E_{ij})^2}{E_{ij}}= \frac{(60-\frac{175\cdot 200}{600})^2}{\frac{175\cdot 200}{600}}+ \frac{(40-\frac{175\cdot 200}{600})^2}{\frac{175\cdot 200}{600}}+\cdots + \frac{(25-\frac{120\cdot 200}{600})^2}{\frac{120\cdot 200}{600}}=88.04$$ Critical value: $$\chi^2_{0.1, df=(3-1)(4-1)=6}= 10.64.$$ Since $88.04>10.64$, we reject the null hypothesis, that is, the technique and the sales are dependent.
Note: $p$-$value=P(\chi^2>88.04)=7.7\cdot 10^{-17}<\alpha=0.1 \Rightarrow$ Reject $H_0$.