Mean Diameter of Ball Bearings in a Hypothesis Test - Prof. N. Phillips, Study notes of Probability and Statistics

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Math 243: Lecture File 9

N. Christopher Phillips

28 April 2009

How big does the sample have to be?

Example: As before, spiral-horned snorkacks have horn lengths (in cm)

distributed N(μ, 12) with μ unknown. A simple random sample of 100

spiral-horned snorkacks has sample mean horn length 121.502 (again, in

cm). We found a 95% confidence interval for the true mean horn length of

spiral-horned snorkacks.

How big does the sample have to be?

Example: As before, spiral-horned snorkacks have horn lengths (in cm)

distributed N(μ, 12) with μ unknown. A simple random sample of 100

spiral-horned snorkacks has sample mean horn length 121.502 (again, in

cm). We found a 95% confidence interval for the true mean horn length of

spiral-horned snorkacks.

The margin of error for a confidence interval is

m = z

∗

σ √ n

For this case, we had σ = 12, x = 121. 502 , n = 100, and z ∗ ≈ 1. 960. We

got

m = (1.960)

The resulting confidence interval was

Your boss isn’t satisfied. He wants a margin of error of at most 2.000 for

the 95% confidence interval. What to do?

Your boss isn’t satisfied. He wants a margin of error of at most 2.000 for

the 95% confidence interval. What to do?

First, the sample size n must be larger. This reduces the sampling

standard deviation σ/

n.

Caution: In situations with unknown population standard deviation σ, we

will have to use the standard deviation s of the sample instead of σ (and

make other changes—we will do this soon). Since the standard deviation

of the sample depends on the sample, you are not guaranteed that larger

sample sizes reduce s/

n. But it usually happens.

The margin of error for a confidence interval is

m = z ∗

σ √ n

We had σ = 12, x = 121. 502 , n = 100, and z ∗ ≈ 1. 960. We want a

margin of error of at most 2.000 for the 95% confidence interval.

The margin of error for a confidence interval is

m = z ∗

σ √ n

We had σ = 12, x = 121. 502 , n = 100, and z ∗ ≈ 1. 960. We want a

margin of error of at most 2.000 for the 95% confidence interval.

Substitute 2.000 for m, and the known values for σ and z ∗ :

n

Now solve for n:

(2.000)

n ≈ (1.960)(12)

The margin of error for a confidence interval is

m = z ∗

σ √ n

We had σ = 12, x = 121. 502 , n = 100, and z ∗ ≈ 1. 960. We want a

margin of error of at most 2.000 for the 95% confidence interval.

Substitute 2.000 for m, and the known values for σ and z ∗ :

n

Now solve for n:

(2.000)

n ≈ (1.960)(12)

n ≈

Substitute 2.000 for m, and the known values for σ and z ∗ :

n

Substitute 2.000 for m, and the known values for σ and z ∗ :

n

Now solve for n:

(2.000)

n ≈ (1.960)(12)

Substitute 2.000 for m, and the known values for σ and z ∗ :

n

Now solve for n:

(2.000)

n ≈ (1.960)(12)

n ≈

n ≈

Substitute 2.000 for m, and the known values for σ and z ∗ :

n

Now solve for n:

(2.000)

n ≈ (1.960)(12)

n ≈

n ≈

I have rounded to four significant digits, since that is all we have for z ∗ .

We want a margin of error of at most 2.000 for the 95% confidence

interval. We got n ≈ 138. 3.

What is wrong with this?

Even in the world of Harry Potter, it seems unlikely that one can choose a

simple random sample of 138.3 spiral-horned snorkacks, and measure

138 .3 horn lengths.

We want a margin of error of at most 2.000 for the 95% confidence

interval. We got n ≈ 138. 3.

What is wrong with this?

Even in the world of Harry Potter, it seems unlikely that one can choose a

simple random sample of 138.3 spiral-horned snorkacks, and measure

138 .3 horn lengths.

The sample size must be an integer, and must give a margin of error of at

most 2. 000 , so we choose the next larger integer.