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A pollster wants to construct an 80% confidence interval for the proportion of adults who believe that economic conditions are...Question
A pollster wants to construct an 98% confidence interval for the proportion of adults who believe that economic conditions are getting better. A poll taken in July 2010 estimates this proportion to be 0.28. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.02?
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(b) Estimate the sample size needed if no estimate of p is available.
Answered by
GPT-4o mini
To determine the required sample size for constructing a confidence interval with a specified margin of error, we can use the formula for the margin of error (ME) in the context of proportion estimates:
\[
ME = z \cdot \sqrt{\frac{p(1 - p)}{n}}
\]
Where:
- \( ME \) is the margin of error
- \( z \) is the z-score corresponding to the desired confidence level
- \( p \) is the estimated proportion
- \( n \) is the sample size
Given:
- Confidence level = 98%, which corresponds to a z-score of approximately 2.33 (this can be found in a z-table or calculated using statistical software).
- Estimated proportion \( p = 0.28 \)
- Desired margin of error \( ME = 0.02 \)
We need to rearrange the margin of error formula to solve for \( n \):
\[
n = \left(\frac{z \cdot \sqrt{p(1 - p)}}{ME}\right)^2
\]
Now, substitute the known values into the equation.
1. Calculate \( p(1 - p) \):
\[
p(1 - p) = 0.28 \times (1 - 0.28) = 0.28 \times 0.72 = 0.2016
\]
2. Substitute into the sample size formula:
\[
n = \left(\frac{2.33 \cdot \sqrt{0.2016}}{0.02}\right)^2
\]
3. Calculate \( \sqrt{0.2016} \):
\[
\sqrt{0.2016} \approx 0.4485
\]
4. Now plug this back into the formula:
\[
n = \left(\frac{2.33 \cdot 0.4485}{0.02}\right)^2
\]
5. Calculate:
\[
\frac{2.33 \cdot 0.4485}{0.02} \approx \frac{1.0444}{0.02} \approx 52.22
\]
6. Now square that result to find \( n \):
\[
n \approx (52.22)^2 \approx 2731.04
\]
Since the sample size must be a whole number, we round up to ensure the margin of error will be within the specified level.
Thus, the required sample size is:
\[
\boxed{2732}
\]
\[
ME = z \cdot \sqrt{\frac{p(1 - p)}{n}}
\]
Where:
- \( ME \) is the margin of error
- \( z \) is the z-score corresponding to the desired confidence level
- \( p \) is the estimated proportion
- \( n \) is the sample size
Given:
- Confidence level = 98%, which corresponds to a z-score of approximately 2.33 (this can be found in a z-table or calculated using statistical software).
- Estimated proportion \( p = 0.28 \)
- Desired margin of error \( ME = 0.02 \)
We need to rearrange the margin of error formula to solve for \( n \):
\[
n = \left(\frac{z \cdot \sqrt{p(1 - p)}}{ME}\right)^2
\]
Now, substitute the known values into the equation.
1. Calculate \( p(1 - p) \):
\[
p(1 - p) = 0.28 \times (1 - 0.28) = 0.28 \times 0.72 = 0.2016
\]
2. Substitute into the sample size formula:
\[
n = \left(\frac{2.33 \cdot \sqrt{0.2016}}{0.02}\right)^2
\]
3. Calculate \( \sqrt{0.2016} \):
\[
\sqrt{0.2016} \approx 0.4485
\]
4. Now plug this back into the formula:
\[
n = \left(\frac{2.33 \cdot 0.4485}{0.02}\right)^2
\]
5. Calculate:
\[
\frac{2.33 \cdot 0.4485}{0.02} \approx \frac{1.0444}{0.02} \approx 52.22
\]
6. Now square that result to find \( n \):
\[
n \approx (52.22)^2 \approx 2731.04
\]
Since the sample size must be a whole number, we round up to ensure the margin of error will be within the specified level.
Thus, the required sample size is:
\[
\boxed{2732}
\]
Answered by
GPT-4o mini
Sure! Let’s go through the calculation step by step to ensure we get the correct sample size for the required margin of error in the confidence interval for the proportion.
### Given Data
- Confidence level = 98%, which corresponds to a z-score of approximately \( z = 2.33 \) (you can find this value in a z-table).
- Estimated proportion \( p = 0.28 \).
- Desired margin of error \( ME = 0.02 \).
### Formula for Sample Size
We’ll use the formula to calculate the required sample size \( n \):
\[
n = \left(\frac{z \cdot \sqrt{p(1 - p)}}{ME}\right)^2
\]
### Step 1: Calculate \( p(1 - p) \)
\[
p(1 - p) = 0.28 \times (1 - 0.28) = 0.28 \times 0.72 = 0.2016
\]
### Step 2: Calculate \( \sqrt{p(1 - p)} \)
\[
\sqrt{p(1 - p)} = \sqrt{0.2016} \approx 0.4485
\]
### Step 3: Substitute into the Sample Size Formula
Now substitute the known values into the sample size formula:
\[
n = \left(\frac{2.33 \cdot 0.4485}{0.02}\right)^2
\]
### Step 4: Calculate the Numerator
Calculating the numerator:
\[
2.33 \cdot 0.4485 \approx 1.044305
\]
### Step 5: Divide by the Margin of Error
Now divide by \( ME \):
\[
\frac{1.044305}{0.02} = 52.21525
\]
### Step 6: Square the Result
Now, we need to square \( 52.21525 \):
\[
n \approx (52.21525)^2 \approx 2732.487
\]
### Step 7: Round Up to Nearest Whole Number
Since we cannot have a fraction of a sample, we round up:
\[
n = 2733
\]
Thus, the required sample size is:
\[
\boxed{2733}
\]
Thank you for your patience, and I hope this calculation is now clear and correct!
### Given Data
- Confidence level = 98%, which corresponds to a z-score of approximately \( z = 2.33 \) (you can find this value in a z-table).
- Estimated proportion \( p = 0.28 \).
- Desired margin of error \( ME = 0.02 \).
### Formula for Sample Size
We’ll use the formula to calculate the required sample size \( n \):
\[
n = \left(\frac{z \cdot \sqrt{p(1 - p)}}{ME}\right)^2
\]
### Step 1: Calculate \( p(1 - p) \)
\[
p(1 - p) = 0.28 \times (1 - 0.28) = 0.28 \times 0.72 = 0.2016
\]
### Step 2: Calculate \( \sqrt{p(1 - p)} \)
\[
\sqrt{p(1 - p)} = \sqrt{0.2016} \approx 0.4485
\]
### Step 3: Substitute into the Sample Size Formula
Now substitute the known values into the sample size formula:
\[
n = \left(\frac{2.33 \cdot 0.4485}{0.02}\right)^2
\]
### Step 4: Calculate the Numerator
Calculating the numerator:
\[
2.33 \cdot 0.4485 \approx 1.044305
\]
### Step 5: Divide by the Margin of Error
Now divide by \( ME \):
\[
\frac{1.044305}{0.02} = 52.21525
\]
### Step 6: Square the Result
Now, we need to square \( 52.21525 \):
\[
n \approx (52.21525)^2 \approx 2732.487
\]
### Step 7: Round Up to Nearest Whole Number
Since we cannot have a fraction of a sample, we round up:
\[
n = 2733
\]
Thus, the required sample size is:
\[
\boxed{2733}
\]
Thank you for your patience, and I hope this calculation is now clear and correct!
Answered by
GPT-4o mini
When no estimate of \( p \) is available, the conservative approach is to use \( p = 0.5 \). This is because \( p(1 - p) \) is maximized when \( p = 0.5 \), leading to the largest possible required sample size for a given margin of error.
### Given Data
- Confidence level = 98%, which corresponds to a z-score of approximately \( z = 2.33 \).
- No prior estimate of \( p \), so we use \( p = 0.5 \).
- Desired margin of error \( ME = 0.02 \).
### Formula for Sample Size
The formula for sample size remains the same:
\[
n = \left(\frac{z \cdot \sqrt{p(1 - p)}}{ME}\right)^2
\]
### Step 1: Calculate \( p(1 - p) \)
Since \( p = 0.5 \):
\[
p(1 - p) = 0.5 \times (1 - 0.5) = 0.5 \times 0.5 = 0.25
\]
### Step 2: Calculate \( \sqrt{p(1 - p)} \)
\[
\sqrt{p(1 - p)} = \sqrt{0.25} = 0.5
\]
### Step 3: Substitute into the Sample Size Formula
Now substitute the known values into the sample size formula:
\[
n = \left(\frac{2.33 \cdot 0.5}{0.02}\right)^2
\]
### Step 4: Calculate the Numerator
Calculating the numerator:
\[
2.33 \cdot 0.5 = 1.165
\]
### Step 5: Divide by the Margin of Error
Now divide by \( ME \):
\[
\frac{1.165}{0.02} = 58.25
\]
### Step 6: Square the Result
Now, we need to square \( 58.25 \):
\[
n \approx (58.25)^2 \approx 3396.5625
\]
### Step 7: Round Up to Nearest Whole Number
Since we cannot have a fraction of a sample, we round up:
\[
n = 3397
\]
Thus, the required sample size, when no prior estimate of \( p \) is available, is:
\[
\boxed{3397}
\]
This provides a conservative estimate for the sample size needed to achieve the desired margin of error at the specified confidence level.
### Given Data
- Confidence level = 98%, which corresponds to a z-score of approximately \( z = 2.33 \).
- No prior estimate of \( p \), so we use \( p = 0.5 \).
- Desired margin of error \( ME = 0.02 \).
### Formula for Sample Size
The formula for sample size remains the same:
\[
n = \left(\frac{z \cdot \sqrt{p(1 - p)}}{ME}\right)^2
\]
### Step 1: Calculate \( p(1 - p) \)
Since \( p = 0.5 \):
\[
p(1 - p) = 0.5 \times (1 - 0.5) = 0.5 \times 0.5 = 0.25
\]
### Step 2: Calculate \( \sqrt{p(1 - p)} \)
\[
\sqrt{p(1 - p)} = \sqrt{0.25} = 0.5
\]
### Step 3: Substitute into the Sample Size Formula
Now substitute the known values into the sample size formula:
\[
n = \left(\frac{2.33 \cdot 0.5}{0.02}\right)^2
\]
### Step 4: Calculate the Numerator
Calculating the numerator:
\[
2.33 \cdot 0.5 = 1.165
\]
### Step 5: Divide by the Margin of Error
Now divide by \( ME \):
\[
\frac{1.165}{0.02} = 58.25
\]
### Step 6: Square the Result
Now, we need to square \( 58.25 \):
\[
n \approx (58.25)^2 \approx 3396.5625
\]
### Step 7: Round Up to Nearest Whole Number
Since we cannot have a fraction of a sample, we round up:
\[
n = 3397
\]
Thus, the required sample size, when no prior estimate of \( p \) is available, is:
\[
\boxed{3397}
\]
This provides a conservative estimate for the sample size needed to achieve the desired margin of error at the specified confidence level.