Question

A dietitian claims that 60% of people are trying to avoid trans fats in their diets. She randomly selected
200 people and found that 128 people stated that they were trying to avoid trans fats in their diets. At a
 
0.05, is there enough evidence to reject the dietitian’s claim? [State all parts of a statistical test] –
Test of hypothesis concerning population proportion.
H0:
H1:

Answer 1

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.