n a hospital 24 operating rooms were sampled. The mean noise level was 41.6 decibels and the standard deviation was 7.5. What is the 95% confidence interval of the true mean of the noise levels in the operating rooms? If possible, can you list the steps you used. I'm having a hard time calculating these results. Thanks.
Formula:
CI95 = mean + or - 1.96(sd/√n)
Substituting your data:
CI95 = 41.6 + or - 1.96(7.5/√24)
Finish the calculations for your confidence interval.
Thank you very much for the clarification.
Certainly! To calculate the 95% confidence interval for the true mean of the noise levels in the operating rooms, you can use the following steps:
Step 1: Note down the given information:
- Mean noise level (sample mean) = 41.6 decibels
- Standard deviation (sample standard deviation) = 7.5
Step 2: Determine the critical value for a 95% confidence level:
- Since we want a 95% confidence interval, we need to find the critical value corresponding to a 95% confidence level. This critical value can be found using a standard normal distribution table or a statistical calculator.
- For a 95% confidence level, the critical value (z-value) is approximately 1.96.
Step 3: Calculate the margin of error:
- The margin of error represents the range within which the true population mean is likely to fall. It is calculated by multiplying the critical value by the standard deviation of the sample and dividing by the square root of the sample size.
- Margin of error = (Critical value) * (Standard deviation / √(Sample size))
- Margin of error = 1.96 * (7.5 / √24)
Step 4: Calculate the confidence interval:
- The confidence interval is calculated by subtracting the margin of error from the sample mean and adding the margin of error to the sample mean.
- Lower limit of the confidence interval = Sample mean - Margin of error
- Upper limit of the confidence interval = Sample mean + Margin of error
Now, let's calculate the confidence interval:
Step 2 (revisited): Critical value = 1.96 (for a 95% confidence level)
Step 3 (revisited): Margin of error = 1.96 * (7.5 / √24)
Margin of error ≈ 1.96 * (7.5 / 4.899) ≈ 2.998
Step 4 (revisited):
Lower limit of the confidence interval = 41.6 - 2.998 ≈ 38.602
Upper limit of the confidence interval = 41.6 + 2.998 ≈ 44.598
Therefore, the 95% confidence interval for the true mean of the noise levels in the operating rooms is approximately 38.602 decibels to 44.598 decibels.
Please note that this calculation assumes that the sample is a simple random sample and that the noise levels in the operating rooms follow a normal distribution.