Q. 39 - The Practice of Statistics for AP

Question

Refer to Exercise 37. 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

In the question, there are various methods for quitting smoking, as well as a table with various methods and their success. In a two-way table, the summary of this data can be represented as:

As a result, the expected counts are the total of the rows multiplied by the total of the columns, divided by the sample size.$n=893$ as:

The chi-square statistic is defined as the total of squared deviations between expected and observed counts divided by expected count, as follows:

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

    $=34.937$

The degrees of freedom are determined as follows:

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

The test statistic has three degrees of freedom and follows the $X^2$ distribution.

Therefore,

$p\text{-value }\approx 0$

Thus, the P-value interval can be discovered in table C in the column title that has $X^2$-value in the appropriate interval in the row with degree of freedom as:

$P<0.0005$

The null hypothesis is rejected if the $P$-value is less than the significance level, as we know. As a result,$P<0.05\Rightarrow \text{ Reject }{H}_0$

As a result, we find that there is sufficient evidence of a relationship between the variables.