# 9.2.1 - Confidence Intervals

Given that the populations are known to be normally distributed, or if both sample sizes are at least 30, then the sampling distribution can be approximated using the $$t$$ distribution, and the formulas below may be used. Here you will be introduced to the formulas to construct a confidence interval using the $$t$$ distribution. Minitab Express will do all of these calculations for you, however, it uses a more sophisticated method to compute the degrees of freedom so answers may vary slightly, particularly with smaller sample sizes.

General Form of a Confidence Interval
$$point \;estimate \pm (multiplier) (standard \;error)$$

Here, the point estimate is the difference between the two mean, $$\overline X _1 - \overline X_2$$.

Standard Error
$$\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$
Confidence Interval for Two Independent Means
$$(\bar{x}_1-\bar{x}_2) \pm t^\ast{ \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$

The degrees of freedom can be approximated as the smallest sample size minus one.

Estimated Degrees of Freedom

$$df=smallest\;n-1$$

## Example: Exam Scores by Learner Type Section

A STAT 200 instructor wants to know how traditional students and adult learners differ in terms of their final exam scores. She collected the following data from a sample of students:

 Traditional Students Adult Learners $$\overline x$$ 41.48 40.79 $$s$$ 6.03 6.79 $$n$$ 239 138

She wants to construct a 95% confidence interval to estimate the mean difference.

The point estimate, or "best estimate," is the difference in sample means: $$\overline x _1 - \overline x_2 = 41.48-40.79=0.69$$

The standard error can be computed next: $$\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\sqrt{\frac{6.03^2}{239}+\frac{6.79^2}{138}}=0.697$$

To find the multiplier, we need to construct a t distribution with $$df=smaller\;n-1=138-1=137$$ to find the t scores that separate the middle 95% of the distribution from the outer 5% of the distribution: $$t^*=1.97743$$

Now, we can combine all of these values to construct our confidence interval:

$$point \;estimate \pm (multiplier) (standard \;error)$$

$$0.69 \pm 1.97743 (0.697)$$

$$0.69 \pm 1.379$$ The margin of error is 1.379

$$[-0.689, 2.069]$$

We are 95% confident that the mean difference in traditional students' and adult learners' final exam scores is between -0.689 points and +2.069 points.