Chapter 11.2, Problem 39E

To determine

To explain do the data provide convincing evidence of a difference in the effectiveness of the four treatments and carry out an appropriate test at the $\alpha =0.05$ significance level.

Answer to Problem 39E

Yes, the data provide convincing evidence of a difference in the effectiveness of the four treatments.

Explanation of Solution

In the question, it is given different ways so that a person can quit smoking and the table with different ways and with their success. Thus, the summary of these data in a two-way table can be represented as:

  

Thus, the expected counts are the row total multiplied by the column total, divided by the sample size $n=893$ as:

  

Thus, the chi-square statistic is the sum of squared deviations between observed and expected counts divided by the expected count as:

  $\begin{array}{l}{\chi }^2=\frac{{(40-61.7514)}^2}{61.7514}+\frac{{(74-61.7514)}^2}{61.7514}+\frac{{(87-62.0045)}^2}{62.0045}+\frac{{(25-40.4927)}^2}{40.4927}\\ +\frac{{(204-182.2486)}^2}{182.2486}+\frac{{(170-182.2486)}^2}{182.2486}+\frac{{(158-182.9955)}^2}{182.9955}+\frac{{(135-119.5073)}^2}{119.5073}\\ =34.937\end{array}$

The degrees of freedom is calculated as:

  $\begin{array}{l}df=(c-1)(r-1)\\ =(4-1)(2-1)\\ =3\end{array}$

Thus, the interval for the P-value can then be found in table C in the column title which have ${\chi }^2$ -value in the corresponding interval in the row with degree of freedom as:

  $P<0.0005$

As we know that if the P-value is smaller than the significance level, then the null hypothesis is rejected. Thus, we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is sufficient evidence that there is an association between the variables. Yes, the data provide convincing evidence of a difference in the effectiveness of the four treatments.