Calculate the sample proportion and the margin of error for the given data.

In a poll of 1,000 randomly selected voters in a local election, 706 voted in favor of school bond measures.

1. What is the sample proportion ([tex]\(\hat{p}\)[/tex])?

2. What is the margin of error (m) for the 90% confidence level?



Answer :

Certainly, let's walk through the solution step by step for the given problem:

### Given Data:
- Sample Size ([tex]\( n \)[/tex]): 1000
- Number of Voters in Favor ([tex]\( x \)[/tex]): 706
- Confidence Level: 90%

### Step 1: Calculate the Sample Proportion ([tex]\( \hat{p} \)[/tex])
The sample proportion ([tex]\( \hat{p} \)[/tex]) is calculated by dividing the number of favorable responses by the total sample size.

[tex]\[ \hat{p} = \frac{x}{n} = \frac{706}{1000} = 0.706 \][/tex]

So, the sample proportion ([tex]\( \hat{p} \)[/tex]) is 0.706.

### Step 2: Determine the Z-Score for the 90% Confidence Level
From the standard Z-table, for a 90% confidence level, the critical value (Z-score, [tex]\( z^* \)[/tex]) is approximately 1.645.

### Step 3: Calculate the Margin of Error ([tex]\( m \)[/tex])
The margin of error is given by the formula:

[tex]\[ m = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]

Substitute the given values into the formula:

[tex]\[ \hat{p} = 0.706 \][/tex]

[tex]\[ 1 - \hat{p} = 1 - 0.706 = 0.294 \][/tex]

[tex]\[ n = 1000 \][/tex]

Now, calculate the standard error:

[tex]\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.706 \cdot 0.294}{1000}} \approx 0.0144 \][/tex]

Finally, multiply by the Z-score:

[tex]\[ m = 1.645 \cdot 0.0144 \approx 0.0237 \][/tex]

### Conclusion:
- The sample proportion ([tex]\( \hat{p} \)[/tex]) is 0.706.
- The margin of error ([tex]\( m \)[/tex]) for the 90% confidence level is approximately 0.0237.

Therefore, the calculated results are:
- Sample proportion, [tex]\( \hat{p} \)[/tex]: 0.706
- Margin of error, [tex]\( m \)[/tex]: 0.0237