A public heath official wanted to know how often university students visit their student health centers due to illness or injury. The official took a nationwide random sample of 120 university students and found an average of (x bar) = 2.30 visits per student per year. It is known that sigma = 0.44. Construct a 95% confidence interval for the mean number of visits u by a university student to a health center due to illness or injury. Round all results to two decimals places.
95% = mean ± 1.96 SEm
SEm = SD/√n
To construct a 95% confidence interval for the mean number of visits by university students to the health center, we can use the following formula:
Confidence interval = x̄ ± z * (σ/√n)
where:
- x̄ is the sample mean,
- z is the z-value (critical value) corresponding to the desired confidence level,
- σ is the population standard deviation,
- n is the sample size.
Given information:
- Sample mean (x̄) = 2.30
- Population standard deviation (σ) = 0.44
- Sample size (n) = 120
- Desired confidence level = 95%
Step 1: Find the critical value (z-value) for a 95% confidence level.
Since the confidence level is 95%, the alpha level (α) is 0.05. To find the critical value, we can use a standard normal distribution table or a calculator. The critical value for a 95% confidence level is approximately 1.96.
Step 2: Calculate the margin of error.
The margin of error is given by the formula: Margin of error = z * (σ/√n)
Substituting the values:
Margin of error = 1.96 * (0.44/√120)
Margin of error ≈ 0.078
Step 3: Calculate the confidence interval.
The confidence interval is calculated as: Confidence interval = x̄ ± margin of error
Substituting the values:
Confidence interval = 2.30 ± 0.078
Rounding to two decimal places, the 95% confidence interval for the mean number of visits by university students to the health center is approximately 2.22 to 2.38 visits per student per year.