# Examples of Confidence Intervals for Means

One of the major parts of inferential statistics is the development of ways to calculate confidence intervals. Confidence intervals provide us with a way to estimate a population parameter. Rather than say that the parameter is equal to an exact value, we say that the parameter falls within a range of values.  This range of values is typically an estimate, along with a margin of error that we add and subtract from the estimate.

Attached to every interval is a level of confidence. The level of confidence gives a measurement of how often, in the long run, the method used to obtain our confidence interval captures the true population parameter.

It is helpful when learning about statistics to see some examples worked out. Below we will look at several examples of confidence intervals about a population mean. We will see that the method we use to construct a confidence interval about a mean depends on further information about our population. Specifically, the approach that we take depends on whether or not we know the population standard deviation or not.

## Statement of Problems

We start with a simple random sample of 25 a particular species of newts and measure their tails. The mean tail length of our sample is 5 cm.

1. If we know that 0.2 cm is the standard deviation of the tail lengths of all newts in the population, then what is a 90% confidence interval for the mean tail length of all newts in the population?
2. If we know that 0.2 cm is the standard deviation of the tail lengths of all newts in the population, then what is a 95% confidence interval for the mean tail length of all newts in the population?
3. If we find that that 0.2 cm is the standard deviation of the tail lengths of the newts in our sample the population, then what is a 90% confidence interval for the mean tail length of all newts in the population?
4. If we find that that 0.2 cm is the standard deviation of the tail lengths of the newts in our sample the population, then what is a 95% confidence interval for the mean tail length of all newts in the population?

## Discussion of the Problems

We begin by analyzing each of these problems. In the first two problems we know the value of the population standard deviation. The difference between these two problems is that the level of confidence is greater in #2 than what it is for #1.

In the second two problems the population standard deviation is unknown. For these two problems we will estimate this parameter with the sample standard deviation. As we saw in the first two problems, here we also have different levels of confidence.

## Solutions

We will calculate solutions for each of the above problems.

1. Since we know the population standard deviation, we will use a table of z-scores. The value of z that corresponds to a 90% confidence interval is 1.645. By using the formula for the margin of error we have a confidence interval of 5 – 1.645(0.2/5) to 5 + 1.645(0.2/5). (The 5 in the denominator here is because we have taken the square root of 25). After carrying out the arithmetic we have 4.934 cm to 5.066 cm as a confidence interval for the population mean.
2. Since we know the population standard deviation, we will use a table of z-scores. The value of z that corresponds to a 95% confidence interval is 1.96. By using the formula for the margin of error we have a confidence interval of 5 – 1.96(0.2/5) to 5 + 1.96(0.2/5). After carrying out the arithmetic we have 4.922 cm to 5.078 cm as a confidence interval for the population mean.
3. Here we do not know the population standard deviation, only the sample standard deviation. Thus we will use a table of t-scores. When we use a table of t scores we need to know how many degrees of freedom we have. In this case there are 24 degrees of freedom, which is one less than sample size of 25. The value of t that corresponds to a 90% confidence interval is 1.71. By using the formula for the margin of error we have a confidence interval of 5 – 1.71(0.2/5) to 5 + 1.71(0.2/5). After carrying out the arithmetic we have 4.932 cm to 5.068 cm as a confidence interval for the population mean.
4. Here we do not know the population standard deviation, only the sample standard deviation. Thus we will again use a table of t-scores. There are 24 degrees of freedom, which is one less than sample size of 25. The value of t that corresponds to a 95% confidence interval is 2.06. By using the formula for the margin of error we have a confidence interval of 5 – 2.06(0.2/5) to 5 + 2.06(0.2/5). After carrying out the arithmetic we have 4.912 cm to 5.082 cm as a confidence interval for the population mean.

## Discussion of the Solutions

There are a few things to note in comparing these solutions. The first is that in each case as our level of confidence increased, the greater the value of z or t that we ended up with. The reason for this is that in order to be more confident that we did indeed capture the population mean in our confidence interval, we need a wider interval.

The other feature to note is that for a particular confidence interval, those that use t are wider than those with z. The reason for this is that a t distribution has greater variability in its tails than a standard normal distribution.

The key to correct solutions of these types of problems is that if we know the population standard deviation we use a table of z-scores. If we do not know the population standard deviation then we use a table of t scores.