A medical researcher wants to construct an 80% confidence interval for the proportion of knee replacement surgeries that result in complications.

Question

(a) An article in a medical journal suggested that approximately 9% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.03?

use a ti-84 calculator

Answer 1

To determine the sample size needed for an 80% confidence interval for the proportion of knee replacement surgeries that result in complications with a specified margin of error, you can use the formula for the sample size for a proportion.

The formula to calculate the required sample size $n$ is:

$$n=\left(\frac{Z^2\cdot p\cdot (1-p)}{E^2}\right)$$

Where:

• $Z$ is the z-score corresponding to the desired confidence level,
• $p$ is the estimated proportion,
• $E$ is the margin of error.

Step 1: Determine the z-score for an 80% confidence level.

For an 80% confidence level, we can find the z-score corresponding to the cumulative probability of $(1-\text{confidence level})/2=(1-0.80)/2=0.10$.

Using a z-table or calculator, the z-score for 80% confidence is approximately 1.28.

Step 2: Identify the estimated proportion and margin of error.

From the article:

• The estimated proportion $p$ is 0.09 (or 9%).
• The desired margin of error $E$ is 0.03.

Step 3: Plug the values into the sample size formula.

Now, substituting into the formula:

$$n=\left(\frac{(1.28{)}^2\cdot (0.09)\cdot (1-0.09)}{(0.03{)}^2}\right)$$

Calculating this step by step:

1. Calculate $(1.28{)}^2=1.6384$.
2. Calculate $(1-0.09)=0.91$.
3. Now multiply:

$$1.6384\cdot 0.09\cdot 0.91\approx 0.13448496$$

4. Now divide by $(0.03{)}^2=0.0009$:

$$n=\frac{0.13448496}{0.0009}\approx 149.43884$$

Step 4: Round up to the nearest whole number.

Since the sample size must be a whole number, we round up:

$$n\approx 150$$

Conclusion:

The researcher needs a sample size of 150 to ensure the 80% confidence interval for the proportion of knee replacement surgeries that result in complications has a margin of error of 0.03.

Using a TI-84 Calculator:

To perform these calculations on a TI-84 calculator, you can perform the following sequence:

1. Use the 1/x button to enter the formula for sample size $n$.
2. Use the x^2 for the square of the z-score.
3. Enter the estimate for proportion and subtract from 1.
4. Calculate the margin of error square.
5. Perform the multiplication and division as demonstrated above.

Make sure to round the final value appropriately to find the sample size as shown.

Answer 2

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

Answer 3

When you don't have a prior estimate of the proportion $p$, you can use the most conservative estimate, which is $p=0.5$. This maximizes the variability and ensures that the sample size calculated will be adequate regardless of the actual population proportion.

Step 1: Keep the confidence level and margin of error.

• Z-score for an 80% confidence level: $Z\approx 1.28$
• Desired margin of error $E=0.03$

Step 2: Apply the sample size formula.

Using $p=0.5$:

$$n=\left(\frac{Z^2\cdot p\cdot (1-p)}{E^2}\right)$$

Substituting in the known values:

$$n=\left(\frac{(1.28{)}^2\cdot (0.5)\cdot (1-0.5)}{(0.03{)}^2}\right)$$

Step 3: Perform the calculations.

1. Calculate $(1.28{)}^2$:

   $$(1.28{)}^2=1.6384$$

2. Calculate $(0.5)\cdot (1-0.5)$:

   $$0.5\cdot 0.5=0.25$$

3. Substitute these values back into the formula:

   $$n=\left(\frac{1.6384\cdot 0.25}{(0.03{)}^2}\right)$$

4. Now, calculate $(0.03{)}^2$:

   $$(0.03{)}^2=0.0009$$

5. Now multiply $1.6384\cdot 0.25$:

   $$1.6384\cdot 0.25=0.4096$$

6. Now divide by $0.0009$:

   $$n=\frac{0.4096}{0.0009}\approx 454.6667$$

Step 4: Round up to the nearest whole number.

Since the sample size must be a whole number:

$$n\approx 455$$

Conclusion:

If no prior estimate of the proportion $p$ is available, the researcher should plan for a sample size of approximately 455 to ensure the 80% confidence interval for the proportion of knee replacement surgeries that result in complications has a margin of error of 0.03.