A survey of 8 university student revealed a mean income of $180.98 with a standard deviation of $143.042. Calculate a 95% confidence interval for the mean income of all students:
Question
A survey of 8 university student revealed a mean income of 143.042. Calculate a 95% confidence interval for the mean income of all students:
Solution
To calculate a 95% confidence interval for the mean income of all students, we will use the formula for the confidence interval which is:
CI = X ± Z * (s/√n)
where:
- X is the sample mean
- Z is the Z-value from the standard normal distribution for the desired confidence level (for 95% confidence, Z = 1.96)
- s is the standard deviation of the sample
- n is the sample size
Step 1: Identify a sample statistic. Since we are trying to estimate a population mean, we choose the sample mean (X) as the sample statistic. X = $180.98
Step 2: Select a confidence level. In this analysis, the confidence level is defined for us in the problem. We are working with a 95% confidence level.
Step 3: Find the standard deviation or standard error. Since we do not know the standard deviation of the population, we cannot compute the standard deviation of the sample mean. Instead, we compute the standard error (SE). SE = s/√n = 50.58
Step 4: Substitute the values into the formula and solve:
CI = X ± Z * SE = 50.58 = 99.14
So, the 95% confidence interval for the mean income of all students is (280.12). This means we are 95% confident that the true population mean falls within this range.
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