Statistics 7.1: Confidence Intervals

Method 2 of finding Zα/2

Look up Zα/2 in Table F using confidence level and the last row, which is labeled Z^∞ (look in your notes if you need a visual of how to do this).

Confidence interval for the mean μ formula

*x̄ - Zα/2 (σ/√ n) < μ < x̄ + Zα/2 (σ/√ n)*

Method 1 of finding Zα/2

1. Compute α = 1 - confidence level. 2. Compute α/2. 3. Set α/2 equal the tail area on the left and right side of the normal distribution. 4. Use skills learned in Ch. 6 to find Zα/2.

Interval estimate

A range of values that is used to estimate a parameter. The range may or may not contain the parameter.

Statistics

A value that's computed using all of the data in a sample of size n. Example: x̄, s, p̂

Parameter

A value that's computed using all the data in a population. These values are typically difficult to compute. Example: μ, σ, and p

What is alpha (α) ?

Alpha is the probability that the confidence interval does not contain μ. Confidence level: 95% α = 5% α is always 1 - confidence level.

Example using method 1

Cl = 90% α = 1 - 0.9 = 0.1 α/2 = 0.1/2 = 0.05 Then you would draw one of those curves and shade in 0.5 on each tail. If you look for the z-value that is closest to 0.05, you will find that z = -1.65 Thus, Zα/2 = 1.65

Point estimate

Found by setting a parameter equal to the corresponding statistic.

Minimum sample size needed to construct a confidence interval for the mean.

Here, we're looking for n. n = [Zα/2(σ)/E]^2 E is the maximum allowed error. Round n upwards to the next whole number!

Confidence interval

Is an interval estimate that is computed using statistics and a given confidence level.

Note

It is important to solve the Zα/2 (σ/√ n) first and then add and subtract from x̄

Confidence level

The probability that an interval estimate contains the parameter. Why not 100% confidence? You're not 100% sure because you didn't use the whole population. Only a smaller sample.

What is the problem with method 2?

The problem with Table F is that it only has 5 main confidence levels.

Example 14b on p. 366

n = 50 x̄ = 7.2 σ = 2.1 Cl = 95% Zα/2 = 1.96 Use: x̄ - Zα/2 (σ/√ n) < μ < x̄ + Zα/2 (σ/√ n) 7.2 - 1.96(2.1/√50) < μ < 7.2 + 1.96(2.1/√50) 7.2 - 0.6 < μ < 7.2 + 0.6 *6.6 < μ < 7.8*

Example 25

variance = 900 *MAKE SURE TO CONVERT VARIANCE TO STANDARD DEVIATION!!!!* σ = √900 = 30 E = 5 Cl = 99% Zα/2 = 2.58 n = [2.58(30)/5)^2 n = 239.6304 n ≈ *240*