Confidence intervals come into play when we want to create a better approximation for what the true value of a parameter is. In this module, we will discuss the confidence interval for the sample mean. Up to this point, we have created point estimates, which is what we get when we compute the sample mean. This approximation is almost surely incorrect, so we can be better suited using an interval estimate, in this case the confidence interval.

The concept here is we buffer our prediction of the mean using a margin of error, which uses the Z distribution, as well as a level of confidence, c. Common confidence intervals we create are 80%, 90%, 95%, and 99% confidence intervals. The approach of a confidence interval is this: If we collect sample data and run this approach repeatedly, then approximately 100*(1-c) % of the confidence intervals will contain the true value of the parameter. So, if we construct 95% confidence intervals, we expect that approximately 95% of the intervals we create will contain the true value of the parameter of interest.

The common formula we use when construction confidence intervals for the mean is this: where E is our margin of error. This is the value that will change depending on which distribution we are using.

If we are using the Z-distribution, then = where is our critical Z value. Now we have to figure out what our critical Z values are.

Critical Z values will never change and are as follows:

80% confidence interval: = 1.28

90% confidence interval: = 1.645

95% confidence interval: = 1.96

99% confidence interval: = 2.576

So, let's walk through a confidence interval calculation using the Z distribution: Suppose we have a sample of data with a mean of 50, a population standard deviation of 10, and a sample size of 64. We want to create a 95% confidence interval for this sample: = = 1.961064 = 2.45.

Lower bound: =502.45=47.55

Upper bound: +=50+2.45=52.45

Then we write our final answer as such: (47.55, 52.45). We can then say that we are 95% confident the true value of the population mean falls between 47.55 and 52.45.

If we have a scenario in which we are computing a confidence interval for the population proportion, we need to ensure the following conditions have been met: Each trial is independent of one another, and we have seen at least 5 successes and 5 failures(5and (1)5 ). If we meet these conditions, then the distribution of the sample proportion can be approximated using the Normal distribution, and we can use the critical Z values discussed above.

The process of constructing the confidence interval for the population proportion will be similar to that for the mean, and constructed using ^, where the margin of error, E, is found as: =.

So, let's walk through an example. A survey of 500 nurses was done to see if they were satisfied with their current employer. Of these 500 nurses, 415 claimed they were satisfied. Construct a 90% confidence interval for the population proportion.

First, we want to compute the value of ^ We do that by taking the number of successes, in this case a nurse being satisfied, over the total number of nurses surveyed. This gives us ^ ==415500=0.83.

Once we have this, we can identify that ^=1^=10.83=0.17. Next, we want to verify we can use the Normal distribution by seeing at least 5 successes and 5 failures. We have definitely met this requirement as we have 415 successes and 85 failures. So now we can compute the margin of error, E, using our appropriate critical Z value. Here is the calculation: =^^=1.6450.830.17500=0.0276.

Lower bound: ^=0.830.0276=0.8024

Upper bound: ^+=0.83+0.0276=0.8576

Then we write our final answer as such: (0.8024, 0.8576). We can then say that we are 90% confident the true value of the population proportion falls between 0.8024 and 0.8576.

Instructions

For this discussion post, we are going to construct a confidence interval for the sample mean using the Z-distribution:

We would like to create an interval to estimate the average recovery time for patients undergoing a new ACL tear recovery program. We sampled 45 patients who underwent this new recovery program and saw the average recovery time to be 285 days. If the population standard deviation can be assumed to be 100 days, compute the 90% confidence interval for the mean recovery time.

Discussion Prompts

Answer the following questions in your initial post:

1. The Z-distribution will be used to create the confidence interval for the mean. Why are we able to use this distribution for this problem?
2. Create your confidence interval and report what it is.
3. The current ACL recovery program averages 320 days to fully recover. Based on the confidence interval you constructed, where does this value fall?
4. Based on the confidence interval, do you think we have enough evidence from a statistical standpoint to say the new procedure is significantly better than the current procedure? Why or why not?