To test the dietitian's claim about the proportion of people trying to avoid trans fats, we can set up the hypothesis test as follows:
Step 1: Set up the hypotheses
- Null Hypothesis (\(H_0\)): \(p = 0.60\) (The proportion of people trying to avoid trans fats is 60%)
- Alternative Hypothesis (\(H_1\)): \(p \neq 0.60\) (The proportion of people trying to avoid trans fats is not 60%)
Step 2: Define the significance level
- Significance level (\(\alpha\)): 0.05
Step 3: Collect sample data
- Sample size (\(n\)) = 200
- Number of successes (\(X\)) = 128
- Sample proportion (\(\hat{p}\)) = \(\frac{X}{n} = \frac{128}{200} = 0.64\)
Step 4: Calculate the test statistic
We'll use the formula for the z-test statistic for proportions: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Where:
- \(\hat{p}\) is the sample proportion
- \(p_0\) is the hypothesized population proportion
- \(n\) is the sample size
Plugging in the values: \[ z = \frac{0.64 - 0.60}{\sqrt{\frac{0.60 \cdot (1 - 0.60)}{200}}} \] Calculating the denominator: \[ \sqrt{\frac{0.60 \cdot 0.40}{200}} = \sqrt{\frac{0.24}{200}} = \sqrt{0.0012} \approx 0.03464 \] Now calculate the z statistic: \[ z = \frac{0.04}{0.03464} \approx 1.1547 \]
Step 5: Determine the critical z value
For a two-tailed test at \(\alpha = 0.05\):
- The critical z values are approximately \(-1.96\) and \(1.96\).
Step 6: Make a decision
Compare the calculated z value to the critical z values:
- If \(z < -1.96\) or \(z > 1.96\), we reject \(H_0\).
- In our case, \(z \approx 1.1547\), which does not exceed the critical values.
Step 7: Conclusion
Since the calculated z value is within the range \(-1.96 < z < 1.96\), we fail to reject the null hypothesis.
Conclusion: At the 0.05 significance level, there is not enough evidence to reject the dietitian's claim that 60% of people are trying to avoid trans fats in their diets.