t intervals for a mean

Three quick questions from earlier lessons.

To estimate a mean we usually do not know σ, so we use the sample's $s$ . Does that add uncertainty?

The one-sample t interval.

\bar{x} \pm t^*\dfrac{s}{\sqrt{n}}

, with

t^*

from the t distribution with

n - 1

degrees of freedom. Conditions: random sample; population roughly normal or

n

large (about 30 or more), with no strong outliers for small

n

$n = 16$ , $\bar{x} = 50$ , $s = 8$ , $t^* = 2.131$ (15 df, 95%).

SE = 2

8/4.

ME = 4.262

2.131 × 2.

(45.738, 54.262)

50 ± 4.262.

Your turn: $n = 9$ , $s = 6$ . The standard error.

2

6/√9.

A random sample of 25 has standard deviation s = 16. With t* = 2.064 (24 df, 95%), find the margin of error for the mean.
A random sample of 25 has standard deviation s = 13. With t* = 2.064 (24 df, 95%), find the margin of error for the mean.

Why is a t critical value used for a confidence interval for a mean from a sample of 40?
1. Because σ is estimated by s, which adds variability
2. Because n is less than 1,000
3. Because the data are skewed

Three questions to check you have got it.

A random sample of 25 has standard deviation s = 16. With t* = 2.064 (24 df, 95%), find the margin of error for the mean.
A random sample of 25 has standard deviation s = 13. With t* = 2.064 (24 df, 95%), find the margin of error for the mean.
A random sample of 25 has standard deviation s = 19. With t* = 2.064 (24 df, 95%), find the margin of error for the mean.

t interval for μ

$\bar{x} \pm t^* s/\sqrt{n}$ , df $= n - 1$ .

Watch out: t*, not z*, when σ is estimated.

End of the lesson