Question
In a poll of 1000 likely voters, 560 say that the United States spends too little on fighting hunger at home. Find a 95% confidence interval for the true proportion of voters who fell this way.
Please round to the nearest thousandth and write your answer in the form ( , ); for example (31,34) or (25,46). Do not type in any extra spaces.
Please round to the nearest thousandth and write your answer in the form ( , ); for example (31,34) or (25,46). Do not type in any extra spaces.
Answers
GPT-4o mini
To find a 95% confidence interval for the true proportion of voters who believe that the United States spends too little on fighting hunger at home, we can use the formula for the confidence interval for a proportion:
\[
\text{CI} = \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Where:
- \(\hat{p}\) is the sample proportion
- \(n\) is the sample size
- \(z\) is the z-score corresponding to the desired confidence level
### Step 1: Calculate the sample proportion (\(\hat{p}\)):
\[
\hat{p} = \frac{560}{1000} = 0.56
\]
### Step 2: Determine the sample size (\(n\)):
\[
n = 1000
\]
### Step 3: Find the z-score for a 95% confidence level:
The z-score for a 95% confidence level is approximately \(1.96\).
### Step 4: Calculate the standard error (SE):
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.56 \times (1 - 0.56)}{1000}} = \sqrt{\frac{0.56 \times 0.44}{1000}} = \sqrt{\frac{0.2464}{1000}} \approx \sqrt{0.0002464} \approx 0.0167
\]
### Step 5: Calculate the margin of error (ME):
\[
ME = z \cdot SE = 1.96 \cdot 0.0167 \approx 0.0327
\]
### Step 6: Calculate the confidence interval:
\[
\text{CI} = \hat{p} \pm ME = 0.56 \pm 0.0327
\]
Lower limit:
\[
0.56 - 0.0327 = 0.5273
\]
Upper limit:
\[
0.56 + 0.0327 = 0.5927
\]
### Step 7: Round to the nearest thousandth:
Lower limit: \(0.527 \)
Upper limit: \(0.593 \)
### Final Answer:
The 95% confidence interval for the true proportion of voters who feel this way is:
\[
(527,593)
\]
\[
\text{CI} = \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Where:
- \(\hat{p}\) is the sample proportion
- \(n\) is the sample size
- \(z\) is the z-score corresponding to the desired confidence level
### Step 1: Calculate the sample proportion (\(\hat{p}\)):
\[
\hat{p} = \frac{560}{1000} = 0.56
\]
### Step 2: Determine the sample size (\(n\)):
\[
n = 1000
\]
### Step 3: Find the z-score for a 95% confidence level:
The z-score for a 95% confidence level is approximately \(1.96\).
### Step 4: Calculate the standard error (SE):
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.56 \times (1 - 0.56)}{1000}} = \sqrt{\frac{0.56 \times 0.44}{1000}} = \sqrt{\frac{0.2464}{1000}} \approx \sqrt{0.0002464} \approx 0.0167
\]
### Step 5: Calculate the margin of error (ME):
\[
ME = z \cdot SE = 1.96 \cdot 0.0167 \approx 0.0327
\]
### Step 6: Calculate the confidence interval:
\[
\text{CI} = \hat{p} \pm ME = 0.56 \pm 0.0327
\]
Lower limit:
\[
0.56 - 0.0327 = 0.5273
\]
Upper limit:
\[
0.56 + 0.0327 = 0.5927
\]
### Step 7: Round to the nearest thousandth:
Lower limit: \(0.527 \)
Upper limit: \(0.593 \)
### Final Answer:
The 95% confidence interval for the true proportion of voters who feel this way is:
\[
(527,593)
\]