Answer :
Sure, let's walk through this question step by step.
### Step 1: Determine the null and alternative hypotheses
To determine the null and alternative hypotheses, we need to specify what we are testing. In this case, we are assessing whether the method of helping people quit smoking (patch, medication, no intervention) has an effect on the outcome (quit or not quit).
The correct pair of hypotheses are as follows:
- [tex]\( H_o \)[/tex]: The methods for helping people quit smoking are independent (i.e., quitting rates are the same regardless of the method used).
- [tex]\( H_a \)[/tex]: The methods for helping people quit smoking are dependent (i.e., quitting rates differ depending on the method used).
### Step 2: Determine the test Statistic
Given this scenario, we use the chi-square test for independence to compare the observed frequencies in each group. The test statistic is calculated to compare the observed data with the expected data if there was no association between quitting and the method.
From the provided solution, the test statistic ([tex]\(\chi^2\)[/tex]) is:
[tex]\[ \chi^2 = 3.96 \][/tex]
### Step 3: Determine the p-value
The p-value helps in determining the significance of the test statistic. It measures the probability of observing the data, or something more extreme, if the null hypothesis is true.
From the provided solution, the p-value is:
[tex]\[ p\text{-value} = 0.1381 \][/tex]
### Step 4: Make a decision
To make a decision, we compare the p-value with the significance level ([tex]\(\alpha\)[/tex]). Here, [tex]\(\alpha = 0.05\)[/tex].
- If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis.
- If the p-value is greater than or equal to [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
Given:
[tex]\[ p\text{-value} = 0.1381 > 0.05 \][/tex]
Since the p-value is greater than 0.05, we fail to reject the null hypothesis.
### Summary of the results:
1. Hypotheses
- [tex]\( H_o \)[/tex]: The methods for helping people quit smoking are independent.
- [tex]\( H_a \)[/tex]: The methods for helping people quit smoking are dependent.
2. Test Statistic
- [tex]\(\chi^2 = 3.96\)[/tex]
3. p-value
- [tex]\(p\text{-value} = 0.1381\)[/tex]
4. Decision
- Fail to reject the null hypothesis.
Thus, based on this analysis, the test does not provide sufficient evidence to conclude that the method of intervention has a significant impact on whether individuals quit smoking or not.
### Step 1: Determine the null and alternative hypotheses
To determine the null and alternative hypotheses, we need to specify what we are testing. In this case, we are assessing whether the method of helping people quit smoking (patch, medication, no intervention) has an effect on the outcome (quit or not quit).
The correct pair of hypotheses are as follows:
- [tex]\( H_o \)[/tex]: The methods for helping people quit smoking are independent (i.e., quitting rates are the same regardless of the method used).
- [tex]\( H_a \)[/tex]: The methods for helping people quit smoking are dependent (i.e., quitting rates differ depending on the method used).
### Step 2: Determine the test Statistic
Given this scenario, we use the chi-square test for independence to compare the observed frequencies in each group. The test statistic is calculated to compare the observed data with the expected data if there was no association between quitting and the method.
From the provided solution, the test statistic ([tex]\(\chi^2\)[/tex]) is:
[tex]\[ \chi^2 = 3.96 \][/tex]
### Step 3: Determine the p-value
The p-value helps in determining the significance of the test statistic. It measures the probability of observing the data, or something more extreme, if the null hypothesis is true.
From the provided solution, the p-value is:
[tex]\[ p\text{-value} = 0.1381 \][/tex]
### Step 4: Make a decision
To make a decision, we compare the p-value with the significance level ([tex]\(\alpha\)[/tex]). Here, [tex]\(\alpha = 0.05\)[/tex].
- If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis.
- If the p-value is greater than or equal to [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.
Given:
[tex]\[ p\text{-value} = 0.1381 > 0.05 \][/tex]
Since the p-value is greater than 0.05, we fail to reject the null hypothesis.
### Summary of the results:
1. Hypotheses
- [tex]\( H_o \)[/tex]: The methods for helping people quit smoking are independent.
- [tex]\( H_a \)[/tex]: The methods for helping people quit smoking are dependent.
2. Test Statistic
- [tex]\(\chi^2 = 3.96\)[/tex]
3. p-value
- [tex]\(p\text{-value} = 0.1381\)[/tex]
4. Decision
- Fail to reject the null hypothesis.
Thus, based on this analysis, the test does not provide sufficient evidence to conclude that the method of intervention has a significant impact on whether individuals quit smoking or not.
