Confidence Intervals for Normally Distributed Population: An Example - Prof. N. Phillips, Study notes of Probability and Statistics

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Math 243: Lecture File 8

N. Christopher Phillips

23 April 2009

Confidence intervals

“Statistics is never having to say you are certain.”

(Outside a UO statistician’s office door.)

Confidence intervals

“Statistics is never having to say you are certain.”

(Outside a UO statistician’s office door.)

A 95% confidence interval means:

I got this result using a method that gives a correct statement 95% of the time.

Confidence intervals: an example

Crumple-horned snorkacks have horn lengths (in centimeters) which are normally distributed with standard deviation σ = 6, but we don’t know the mean μ.

Confidence intervals: an example

Crumple-horned snorkacks have horn lengths (in centimeters) which are normally distributed with standard deviation σ = 6, but we don’t know the mean μ.

We choose a simple random sample of 9 crumple-horned snorkacks, and find that the mean x of the horn lengths of the snorkacks in our sample is

Confidence intervals: an example

Crumple-horned snorkacks have horn lengths (in centimeters) which are normally distributed with standard deviation σ = 6, but we don’t know the mean μ.

We choose a simple random sample of 9 crumple-horned snorkacks, and find that the mean x of the horn lengths of the snorkacks in our sample is

The sampling distribution of x is N(μ, 6 /

9. = N(μ, 2). (Remember, we don’t know μ.)

We want a 95% confidence interval for the mean horn length μ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗^ such that 95% of all data is within z∗^ standard deviations of the mean.

We want a 95% confidence interval for the mean horn length μ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗^ such that 95% of all data is within z∗^ standard deviations of the mean.

That is, 95% of all data has z-scores between −z∗^ and z∗.

We want a 95% confidence interval for the mean horn length μ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗^ such that 95% of all data is within z∗^ standard deviations of the mean.

That is, 95% of all data has z-scores between −z∗^ and z∗.

That is, for N(μ, σ), 95% of all data lies in (μ − z∗σ, μ + z∗σ).

These numbers are in the 3rd last line of Table C—see the next page. We get z∗^ ≈ 1. 960. (The Rule of Thumb gives z∗^ ≈ 2 .)

We want a 95% confidence interval for the mean horn length μ of all crumple-horned snorkacks. For a normal distribution, we find the number z∗^ such that 95% of all data is within z∗^ standard deviations of the mean.

That is, 95% of all data has z-scores between −z∗^ and z∗.

That is, for N(μ, σ), 95% of all data lies in (μ − z∗σ, μ + z∗σ).

These numbers are in the 3rd last line of Table C—see the next page. We get z∗^ ≈ 1. 960. (The Rule of Thumb gives z∗^ ≈ 2 .)

We apply this to the sampling distribution, which is N(μ, 2).

Interruption: Table C

The first row of Table C, and the third row from the bottom, sideways (so that it fits on my page):

Confidence level C z∗ 50% 0. 674 60% 0. 841 70% 1. 036 80% 1. 282 90% 1. 645 95% 1. 960 96% 2. 054 98% 2. 326 99% 2. 576 99 .5% 2. 807 99 .8% 3. 091 99 .9% 3. 291

What the entries in Table C mean

Table entries: 90% gives 1. 645.

• 4 - 3 - 2 - 1 0 1 2 3 4

This shows the standard normal distribution. The shaded region extends from − 1 .645 to 1. 645 , and has area very close to 0. 9.

95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x such that

μ − (1.960)(2) ≤ x ≤ μ + (1.960)(2)).

95% of all simple random samples of 9 crumple-horned snorkacks have mean horn length x such that

μ − (1.960)(2) ≤ x ≤ μ + (1.960)(2)).

(x is within distance (1.960)(2) of μ.)