It is claimed that 60 percent of households now subscribe to cable TV. You would like to verify this statement for your class in mass communications. If you want your estimate to be within 5 percent, with a 95 percent level of confidence, how large of a sample is required?
To find out the sample size required to estimate the percentage of households subscribing to cable TV within a particular confidence level and margin of error, we can use the formula for sample size calculation for proportions.
The formula is:
n = (Z² * p * (1 - p)) / E²
Where:
- n is the required sample size
- Z is the z-score corresponding to the desired confidence level
- p is the estimated proportion of the population
- E is the desired margin of error
In this case, our desired confidence level is 95% (which corresponds to a z-score of 1.96) and the desired margin of error is 5%. We also have an estimated proportion of 60% subscribing to cable TV.
Plugging in these values into the formula, we get:
n = (1.96² * 0.6 * (1 - 0.6)) / (0.05²)
Simplifying the equation, we have:
n = (3.8416 * 0.6 * 0.4) / 0.0025
n = 9.2616 / 0.0025
n ≈ 3704.64
Therefore, to estimate the percentage of households subscribing to cable TV within a 5% margin of error and 95% confidence level, you would need a sample size of approximately 3705 households.