Question

Calculate the lower limit of a 95% confidence interval for a sample (N=100) with a mean of 250mg/dL

I'm not sure if you need the other information in the problem... but this is what it asks. The last time I put the entire problem I got a response not even to the question regarding the sample and it was given as population!

Other Information:

- population mean for triglycerides is 180 mg/dL

- standard deviation is 25 mg/dL

Answer 1

The lower limit of a 95% confidence interval for a sample with a mean of 250 mg/dL and a sample size of 100 is 247.5 mg/dL.

Step-by-step explanation:

The lower limit of a 95% confidence interval for a sample with a mean of 250 mg/dL and a sample size of 100 can be calculated using the formula:

Lower Limit = Sample Mean - Margin of Error

To find the margin of error, we need to know the standard deviation of the population. The standard deviation given in the problem, which is 25 mg/dL, is the standard deviation of the population. Since the sample size is large (N=100), we can use the standard error of the mean, which is equal to the population standard deviation divided by the square root of the sample size. Therefore, the margin of error is:

Margin of Error = (Standard deviation of population) / √(Sample size)

Substituting the given values, we have:

Margin of Error = 25 / √(100) = 25 / 10 = 2.5 mg/dL

Finally, the lower limit of the 95% confidence interval is:

Lower Limit = Sample Mean - Margin of Error = 250 - 2.5 = 247.5 mg/dL