In a recent study a SRS of 525 employed Americans, 53 individuals indicated that they would fire their boss if they could. Calculate a 90% confidence interval for the population proportion who would fire their boss if they could.
To calculate a confidence interval for a population proportion, we can use the formula:
CI = p̂ ± Z * √((p̂(1-p̂))/n)
where:
- CI represents the confidence interval,
- p̂ is the sample proportion,
- Z is the z-score corresponding to the desired level of confidence, and
- n is the sample size.
In this case, we have a sample size (n) of 525, and the sample proportion (p̂) is calculated by dividing the number of individuals who indicated they would fire their boss (53) by the total sample size (525):
p̂ = 53/525 = 0.101
To determine the z-score for a 90% confidence level, we need to find the critical value associated with a 5% significance level. Since we have a two-tailed test, we divide the significance level by 2, resulting in a 2.5% value. From a standard normal distribution table, we find the z-score that corresponds to a cumulative probability of 0.975 (1 - 0.025).
Looking up the value in a standard normal distribution table or using a calculator, we find that the z-score is approximately 1.645.
Now, we can substitute the values into the formula:
CI = 0.101 ± 1.645 * √((0.101 * (1-0.101))/525)
Simplifying the calculation:
CI = 0.101 ± 1.645 * √(0.0917/525)
CI = 0.101 ± 1.645 * 0.0136
CI = 0.101 ± 0.0224
Therefore, the 90% confidence interval for the population proportion who would fire their boss if they could is approximately:
0.0786 to 0.1234