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?

Answer 1

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:

\[ \boxed{2732} \]

Answer 2

try again

Answer 3

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!

Answer 4

(b) Estimate the sample size needed if no estimate of p is available.

Answer 5

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.