1

Section 8.6Testing a claim about a

standard deviation

Objective

For a population with standard deviation σ, use a sample too test a claim about the standard deviation.

Tests of a standard deviation use the 2-distribution

2

Notation

3

Notation

4

(1) The sample is a simple random sample

(2) The population is normally distributed

Very strict condition!!!

Requirements

5

Test StatisticDenoted 2 (as in 2-score) since the test uses the 2 -distribution.

n Sample size

s Sample standard deviation

σ0 Claimed standard deviation

6

Critical Values

Right-tailed test “>“ Needs one critical value (right tail)

Use StatCrunch: Chi-Squared Calculator

7

Critical Values

Left-tailed test “<” Needs one critical value (left tail)

Use StatCrunch: Chi-Squared Calculator

8

Critical Values

Two-tailed test “≠“ Needs two critical values (right and left tail)

Use StatCrunch: Chi-Squared Calculator

9

Statisics Test Scores

Tests scores in the author’s previous statistic classes have followed a normal distribution with a standard deviation equal to 14.1. His current class has 27 tests scores with a standard deviation of 9.3.

Use a 0.01 significance level to test the claim that this class has less variation than the past classes.

Example 1

What we know: σ0 = 14.1 n = 27 s = 9.3

Claim: σ < 14.1 using α = 0.01

Note: Test conditions are satisfied since population is normally distributed

Problem 14, pg 449

10

What we know: σ0 = 14.1 n = 27 s = 9.3

Claim: σ < 14.1 using α = 0.01

H0 : σ = 14.1

H1 : σ < 14.1 Left-tailed

Using Critical RegionsExample 1

2 in critical region

(df = 26)

Initial Conclusion: Since 2 in critical region, Reject H0

Final Conclusion: Accept the claim that the new class has less variance than the past classes

2

2L

Test statistic:

Critical value:

11

Calculating P-value for a Variance

Stat → Variance → One sample → with summary

12

Then hit Next

Enter the Sample variance (s2)

Sample size (n)

Calculating P-value for a Variance

s2 = 9.32 = 86.49

NOTE: Must use Variance

13

Then hit Calculate

Select Hypothesis Test

Enter the Null:variance (σ02)

Select Alternative (“<“, “>”, or “≠”)

Calculating P-value for a Variance

σ02 = 14.12 = 198.81

14

Test statistic (2)

P-value

The resulting table shows both the test statistic (2) and the P-value

Calculating P-value for a Variance

15

What we know: σ0 = 14.1 n = 27 s = 9.3

Claim: σ < 14.1 using α = 0.01

Using Critical RegionsExample 1

Using StatCrunch

Initial Conclusion: Since P-value < α (α = 0.01), Reject H0

Final Conclusion: Accept the claim that the new class has less variance than the past classes

We are 99.44% confident the claim holds

Stat → Variance→ One sample → With summary

Null: proportion=

Alternative

Sample variance:

Sample size:

86.49

27

198.81

<

● Hypothesis Test

H0 : σ = 14.1

H1 : σ < 14.1

P-value = 0.0056s2 = 86.49

σ02 = 198.81

16

BMI for Miss America

Listed below are body mass indexes (BMI) for recent Miss America winners. In the 1920s and 1930s, distribution of the BMIs formed a normal distribution with a standard deviation of 1.34.

Use a 0.01 significance level to test the claim that recent Miss America winners appear to have variation that is different from that of the 1920s and 1930s.

Example 2

What we know: σ0 = 1.34 n = 10 s = 1.186

Claim: σ ≠ 1.34 using α = 0.01

Note: Test conditions are satisfied since population is normally distributed

Problem 17, pg 449

19.5 20.3 19.6 20.2 17.8 17.9 19.1 18.8 17.6 16.8

Using StatCrunch: s = 1.1862172

17

Using Critical RegionsExample 2

2 not in critical region(df = 26)

Test statistic:

Critical values: 2

0.005

2R

2L

Initial Conclusion: Since 2 not in critical region, Accept H0

Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s

Since H0 accepted, the observed significance isn’t useful.

What we know: σ0 = 1.34 n = 10 s = 1.186

Claim: σ ≠ 1.34 using α = 0.01

H0 : σ = 1.34

H1 : σ ≠ 1.34 two-tailed

18

Using P-valueExample 2

Using StatCrunch

Stat → Variation → One sample → With summary

Null: proportion=

Alternative

Sample variance:

Sample size:

1.407

10

1.796

<

● Hypothesis Test

P-value = 0.509

Initial Conclusion: Since P-value ≥ α (α = 0.01), Accept H0

Final Conclusion: Reject the claim recent winners have a different variations than in the 20s and 30s

Since H0 accepted, the observed significance isn’t useful.

s2 = 1.407

σ02 = 1.796

What we know: σ0 = 1.34 n = 10 s = 1.186

Claim: σ ≠ 1.34 using α = 0.01

H0 : σ2 = 1.796

H1 : σ2 < 1.796