Questions: A researcher wanted to determine which method would help smokers quit smoking. He divided a group of 183 smokers into three groups: Group 1 is given a patch, Group 2 is given oral medication, and Group 3 has no intervention. The results of the study are in the below table. Test the researcher's claim using a significance level of α=0.05. Group 1 Group 2 Group 3 --------- Did Not Quit 52 54 51 Quit 13 9 4 "Source: Harris Poll a. Determine the null and alternative hypotheses. H0: The methods for helping people quit smoking are independent. Ha: The methods for helping people quit smoking are dependent. H0: The methods for helping people quit smoking are the same. Ha: The methods for helping people quit smoking follow a different distribution. b. Determine the test statistic. Round your answer to two decimal places. x^2= c. Determine the p-value. Round your answer to four decimal places. p-value= d. Make a decision. - Reject the null hypothesis. - Fail to reject the null hypothesis. e. Make a conclusion. - There is sufficient evidence to support the claim that the methods for helping people quit smoking are different. - There is not sufficient evidence to support the claim that the methods for helping people quit smoking are different.

Solution Steps

The null and alternative hypotheses for this study are defined as follows:

• \( H_0 \): The methods for helping people quit smoking are independent.
• \( H_a \): The methods for helping people quit smoking are dependent.

The expected frequencies for each cell in the contingency table are calculated using the formula:

\[ E = \frac{R_i \times C_j}{N} \]

where \( R_i \) is the total for row \( i \), \( C_j \) is the total for column \( j \), and \( N \) is the total number of observations.

The expected frequencies are:

• For cell (1, 1): \( E = \frac{157 \times 65}{183} = 55.765 \)
• For cell (1, 2): \( E = \frac{157 \times 63}{183} = 54.0492 \)
• For cell (1, 3): \( E = \frac{157 \times 55}{183} = 47.1858 \)
• For cell (2, 1): \( E = \frac{26 \times 65}{183} = 9.235 \)
• For cell (2, 2): \( E = \frac{26 \times 63}{183} = 8.9508 \)
• For cell (2, 3): \( E = \frac{26 \times 55}{183} = 7.8142 \)

Thus, the expected frequencies are:

\[ \begin{bmatrix} 55.765 & 54.0492 & 47.1858 \\ 9.235 & 8.9508 & 7.8142 \end{bmatrix} \]

The Chi-Square test statistic \( \chi^2 \) is calculated using the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where \( O \) is the observed frequency and \( E \) is the expected frequency. The calculations for each cell are as follows:

• For cell (1, 1): \( O = 52, E = 55.765 \) \[ \frac{(52 - 55.765)^2}{55.765} = 0.2542 \]

• For cell (1, 2): \( O = 54, E = 54.0492 \) \[ \frac{(54 - 54.0492)^2}{54.0492} = 0.0 \]

• For cell (1, 3): \( O = 51, E = 47.1858 \) \[ \frac{(51 - 47.1858)^2}{47.1858} = 0.3083 \]

• For cell (2, 1): \( O = 13, E = 9.235 \) \[ \frac{(13 - 9.235)^2}{9.235} = 1.535 \]

• For cell (2, 2): \( O = 9, E = 8.9508 \) \[ \frac{(9 - 8.9508)^2}{8.9508} = 0.0003 \]

• For cell (2, 3): \( O = 4, E = 7.8142 \) \[ \frac{(4 - 7.8142)^2}{7.8142} = 1.8618 \]

Summing these values gives:

\[ \chi^2 = 0.2542 + 0.0 + 0.3083 + 1.535 + 0.0003 + 1.8618 = 3.9596 \]

The critical value for \( \chi^2 \) at \( \alpha = 0.05 \) with 2 degrees of freedom is:

\[ \chi^2_{\alpha, df} = 5.9915 \]

The p-value associated with the calculated Chi-Square statistic is:

\[ P = P(\chi^2 > 3.9596) = 0.1381 \]

Since the p-value \( 0.1381 \) is greater than the significance level \( \alpha = 0.05 \), we fail to reject the null hypothesis.

Final Answer