A teacher intends to verify reliable information that the illiteracy rate at the high school is about 2%. How many randomly selected subjects should be tested if he wants 96% confi... A teacher intends to verify reliable information that the illiteracy rate at the high school is about 2%. How many randomly selected subjects should be tested if he wants 96% confidence that the sample is in error by not more than one percentage point? Read more

Steps to Solve

1. Identify the variables We need to identify the key components of the formula:

• Proportion (( p )): 2% or 0.02
• Confidence level (( z )) for 96%: Using a z-table, for 96%, ( z \approx 2.05 )
• Margin of error (( E )): 1% or 0.01

2. Use the sample size formula The sample size ( n ) can be calculated using the formula: $$ n = \left( \frac{z^2 \cdot p \cdot (1 - p)}{E^2} \right) $$

3. Insert the values into the formula Now, substituting the identified values into the formula: $$ n = \left( \frac{(2.05)^2 \cdot 0.02 \cdot (1 - 0.02)}{(0.01)^2} \right) $$

4. Calculate the components Start by calculating ( p(1 - p) ): $$ p(1 - p) = 0.02 \cdot 0.98 = 0.0196 $$

Now, square the ( z ) value: $$ (2.05)^2 = 4.2025 $$

5. Put everything together Substituting back, we get: $$ n = \left( \frac{4.2025 \cdot 0.0196}{0.0001} \right) $$

6. Complete the calculation Now, perform the final multiplication and division: $$ n = \frac{0.082333}{0.0001} = 823.33 $$

Since we need a whole number, we round up to 824.