Determine the sample size needed in order to be 95 percent confident the p, the sample proportion of HD television sets that last at least five years without a single repair, is within 0.03 of , the true proportion of sets that last at least five years without a single repair.
I get that:
standard deviation: .03
z = 1.960
so I'm thinking to use:
n = (zo/E)^2
but I'm not getting an answer that makes sense.
You are on the right track with using the formula n = (z₀/E)² to calculate the sample size needed. Let's break it down step by step:
1. Firstly, you correctly identified the standard deviation (σ) as 0.03, which is the maximum margin of error you can tolerate.
2. Secondly, you correctly identified the z-value as 1.960 for a 95% confidence level. This corresponds to a standard normal distribution.
Now let's plug these values into the formula:
n = (z₀/E)²
where:
n is the desired sample size
z₀ is the z-value corresponding to the desired confidence level (1.960 for a 95% confidence level)
E is the margin of error (0.03)
Substituting the values:
n = (1.960 / 0.03)²
n ≈ 1067.11
The sample size required is approximately 1068.
It is worth noting that the formula for calculating the sample size assumes that the population is large or infinite. If the population size N is small (e.g., less than 10,000), you should apply a finite population correction factor to adjust the formula:
n = (N * z₀² * p̂ * (1 - p̂)) / [(N - 1) * E² + z₀² * p̂ * (1 - p̂)]
where p̂ is an estimate of the proportion based on prior knowledge or pilot study (if such data is available). However, if you do not have any prior information or estimate, it is generally conservative to assume p̂ = 0.5.
I hope this helps! Let me know if you have any further questions.