**Introduction to Confidence Intervals and Tests of Hypotheses for Means**

*The basic ideas of confidence intervals and hypothesis testing were used to develop large sample confidence intervals and hypothesis tests for population proportions. Here, we will learn how these ideas apply when we are interested in the population mean. Because rarely, if ever, is the population standard deviation known when the population mean is unknown, we will consider only the case where both are unknown.*

**Confidence Intervals for a Mean**

Suppose we have a random sample from a distribution with unknown mean and standard deviation. Further, the distribution may or may not be normal. We would estimate the population mean using the point estimate . How would we set the confidence interval on the population mean? First, we need to determine whether the conditions for the statistical methods used here have been met.

To use the statistical methods we will present here to set a confidence interval on the population mean, two conditions must be satisfied for the methods to be valid. (1) The sample must be randomly selected from the population. (2) The population distribution must be normal, or the sample size must be large enough (at least 30) to assume the sampling distribution of is approximately normal by the Central Limit Theorem.

We learned that the general form of a confidence interval is *point estimate* ± *multiplier* × *standard error*, and this is the form we will use. The point estimate of the population mean is , and the standard error of is . Because *t** with (*n* – 1) degrees of freedom is the multiplier. Thus, the form of the confidence interval for a sample mean is:

if the population distribution is normal with known standard deviation. The choice of *t** depends on the confidence level chosen, just as *z** did when determining the multiplier for proportions. For a 100 (1 – α)% confidence interval, *t** is the *t*-value for which *P*(*t* > *t**) = , where *t* is a random selection from a *t*-distribution with (*n* –1) degrees of freedom. (By having in each tail, a total of a is in the tails, leaving 1 – α a for the confidence level. See Figure 17.1.)

**Example**

A particular species of finch exhibits a polymorphism in bill size that is unrelated to gender. There is a large-billed morph and a small-billed morph. The bill widths within a morph are normally distributed. A random sample of 20 finches is selected from a 100- hectare study region. The average bill width of the sampled finches is 15.92, and the sample standard deviation is 0.066 millimeter. Set a 95% confidence interval on the mean bill width of this species of finch within the study region.

**Solution**

Because the sample was selected randomly from a normal distribution, the conditions for inference are satisfied. The point estimate of the mean bill width is = 15.92. We are given that *s* = 0.066 and *n* = 20, so the standard error of is = = 0.0148. For a 95% confidence interval, the *t*-value corresponding to a t with *n* – 1 = 19 degrees of freedom and = 0.025 probability in each tail is found by looking in the *t*-table in Sampling Distributions and the *t*-Distribution. The column for 0.025 probability in the upper tail and the column for 19 degrees of freedom intersect to give *t** = 2.093.Thus, the confidence interval limits are 15.92 ± 2.093 × or 15.92 ± 0.03.We are 95% confident that the mean bill length of the finch species is within 0.03 mm of 15.92 mm in the study region.

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