copyright © 1998, triola, elementary statistics addison wesley longman 1 testing a claim about a...

Copyright © 1998, Triola, Elementary Statistics

Addison Wesley Longman1

Testing a Claim about a Testing a Claim about a Standard Deviation or Standard Deviation or

VarianceVarianceSection 7-6Section 7-6

M A R I O F. T R I O L ACopyright © 1998, Triola, Elementary Statistics

Addison Wesley Longman



Assumption• In testing hypothesis made about a population

standard deviation or variance 2, we assume that the population has values that are normally distributed.



Chi-Square Distribution




X 2 =(n – 1) s 2

2

Test Statistic




n = sample size

s 2 = sample variance

2 = population variance (given in null hypothesis)

X 2 =(n – 1) s 2

2

Test Statistic



Figure 7-14 Figure 7-15There is a different distribution for eachnumber of degrees of freedom.

Properties of the Chi-Square Distribution

Not symmetric

All values are nonnegative

0

Chi-Square Distribution for 10 and 20 Degrees of Freedom

All values are nonnegative

0



Critical Value for Chi-Square Distribution

Table A-4

Formula card

Appendix

Degrees of freedom (df ) = n – 1



When using Table A-4, it is essential to note that each critical value separates an area to the right that responds to the value given in the top row.




Right-tailed test




Right-tailed test Left-tailed

test

1 –




Right-tailed test Left-tailed

test

Two-tailed test

1 –

2 2

1 – /2/2



All three methods 1) Traditional method 2) P-value method 3) Confidence intervals and the testing procedure Step 1 to Step 8 in Section 7-3 are still valid, except that the test statistic is a chi-square test statistic



P-Value Method

Use Table A-4 to identify limits that contain the P-value, similar to method in section 7-4.



Figure 7-17 Testing

a Claim about a Mean,

Proportion, Standard

Deviation, or Variance



ExampleShown below are birth weights (in kilograms) ofmale babies born to mothers on a special vitamin supplement. Test the claim that this sample comes from a population with a standard deviation equal to 0.470kg (which is the standard deviation for male birth weights in general).

3.73 4.37 3.73 4.33 3.39 3.68 4.68 3.52 3.02 4.09 2.47 4.13 4.47 3.22 3.42 2.54



Example

Solution

Step 1: = 0.470

Step 2: = 0.470

Step 4: Select = 0.05 (significance level)

Step 5: The sample standard deviation s is relevant to this test ---- Use chi-square test statistic

Step 3: H0: = 0.470 versus H1: = 0.470



X2 Test Statistic

Assume the conjecture is true!

Test Statistic:

(Step 6)

X 2 =(n – 1) s 2

2

Two-tailed test

2 2 = 0.025

1 – /2/2

n = 16, x = 3.675 s2 = .432

Critical values: 6.262 & 24.788 Critical Region: X2 < 6.262

or: X2 > 27.488

X2 =27.488X2 =6.262



X2 Test Statistic


Test Statistic:

(Step 6)

X 2 =(n – 1) s 2

2

Two-tailed test

2 2 = 0.025

1 – /2/2

n = 16 x = 3.675 s2 = .432 X2 = 15 * .432 / .4702 = 29.339

Critical values: 6.262 & 24.788 Critical Region: X2 < 6.262

or: X2 > 27.488

X2 =27.488X2 =6.262

Sample data: X2=29.339



Example

Conclusion: Based on the reported sample measurements, there is enough evidence to rejection H0: = 0.470. Therefore, the vitaminsupplement does appear to affect the variation among birth weights. (The effect on babies are not the same!)

Step 8:



P-Value Method When n is large, X 2 center (50% quantile)

is close to df = (n -1). So, in a two side test if a p-value method is used, then

(i) if sample X 2 > n - 1,

p-value = 2 * (right tail area)

(ii) if sample X 2 < n - 1,

p-value = 2 * (left tail area)



Example (p-value method)

Solution

Step 1: = 0.470

Step 2: = 0.470

Step 4: Select = 0.05 (significance level)

Step 5: The sample standard deviation s is relevant to this test ---- Use chi-square test statistic

Step 3: H0: = 0.470 versus H1: = 0.470



X2 Test Statistic


Test Statistic:

(Step 6)

X 2 =(n – 1) s 2

2

Two-tailed test !!

n = 16 x = 3.675 s2 = .432 X2 = 15 * .432 / .4702 = 29.339


15X2=29.339 > df =15

p-value = 2 *( right tail area)



X2 Test Statistic


Test Statistic:

(Step 6)

X 2 =(n – 1) s 2

2

Two-tailed test !!


15

p-value = 2*( right tail area)

p-value limits: 2 * 0.01 = 0.02 (>) 2 * 0.025 = 0.05 (<)

P-value < 0.05 Reject Null Hypothesis



Example

Conclusion: Based on the reported sample measurements, there is enough evidence to rejection H0: = 0.470. Therefore, the vitaminsupplement does appear to affect the variation among birth weights. (The effect on babies are not the same!)

Step 8:

copyright © 1998, triola, elementary statistics addison wesley longman 1 testing a claim about a...

Documents