Introduction to Chi Square Distribution Lecture 34

 Introduction to Chi Square Distribution 

&

Statistical Inference

Lecture 34

Let X1, X2,..., Xi,..., Xn be the observations of a random sample selected from a normal population having mean "μ" and standard deviation "σ.".

We define a new z as defined as:

Where:

E(z) = 0 and Var(z) = 1.

Definition: The Chi square variable is the sum of squares of standard normal variables.

Where:

This new variable ( ) is called chi square variable with n degree of freedom and its probability function is called chi square distribution with n degree of freedom, given by

Properties of the chi square distribution

The chi square distribution has the following properties:

i. The chi square is a continuous distribution ranging from 0 to plus infinity.

ii. The mean of the chi square distribution is equal to the degree of freedom.

 E(χ2) = n

iii. The variance of the chi square distribution is equal to twice the degree of freedom.

Var (χ2) = 2n

iv. The moment-generating function about the origin is given by

v. The moment ratios of chi square with n degree of freedom is given by

Confidence interval estimation of variance of a normal population

Let X1, X2,..., Xi,..., Xn be the observations of a random sample selected from a normal population having mean "μ" and standard deviation "σ.". The sampling distribution of sample variance approaches the chi square distribution with (n - 1) degrees of freedom.

That's

Now a 100 (1-α)% confidence interval estimate for population variance is given by

Example 9.1: A random sample of size 25 selected from a normal population having mean μ and standard deviation σ. The following observations were observed:

Find a 95% confidence interval for population variance.

Solution:

Example 9.2: A random sample of ten scores obtained by the students in a math test are as follows: The following numbers were chosen from a normal population with a mean of μ and a standard deviation of σ: 2, 16, 3, 10, 11, 4, 6, 7, 9, 12. What will be the 90% confidence limits for the whole class variance? 

Solution:

Testing Hypothesis about Variance / Standard deviation of a Normal Population

Let S2 be${\mathrm{ the\; estimate\; of\; population\; variance\; computed\; from\; the\; values\; of\; a\; random\; sample\; of\; size\; n\; selected\; from\; a\; normal\; population\; having\; mean }}^{}$ μ and a standard deviation σ. The sampling distribution of sample variance approaches the chi square distribution with (n - 1) degrees of freedom.

That's

Testing Procedure:

i. State null & alternative hypotheses:

ii. The significance level: α

iii. The test statistic:

vi. Critical Region:

Reject H0, when  χ2 calculatd ≤ χ2 1-α/2(n-1)  OR  χ2 calculatd ≥ χ 2 α/2(n-1)

Reject H0, when  χ2 calculatd ≥ χ 2 α(n-1)

Reject H0, when  χ2 calculatd ≤ χ2 1-α(n-1) 

v. Computation

vi. Remarks

Example 9.3: A company that makes hard safety hats for construction workers is worried about the average and variance of the force that the helmet’s wearer experiences when exposed to an outside force. As per the manufacturer’s design, the average force that the helmets transmit to workers is 800 pounds, with a standard deviation up to 40 pounds. A random sample of 30 helmets was used for the tests, and the results showed that the sample mean and sample standard deviation were 825 pounds and 48.5 pounds, respectively. At the 5% level, is there enough evidence in the data to draw the conclusion that the population standard deviation is equal to 40 pounds?

 

Solution:

i. The null and alternative hypotheses may be stated as:

ii. The significance level: α = 0.05

iii. The test statistic:

iv. Critical Region:

Reject H0, when  χ2  ≤ χ20.975(29) = 16.047   OR  χ2 ≥ χ 20.25(29) = 45.772

v. Computation:

vi. Remarks: The computed value of chi square falling in the acceptance region, the sample data, does not provide sufficient evidence to reject the null hypothesis. Thus, it is concluded that the company claim is validated.

Example 9.4: A forester desires to use a mist blower to apply an herbicide treatment to a dense, undestroyed strip of striped maple that is preventing the desired hardwood regeneration. She wants to ensure that the treatment has a low variability of less than 0.06 gal or a consistent application rate. When she collects the sample data on this kind of mist blower (n = 11), she obtains a sample variance of 0.064 gal. test the hypothesis at 5% significance level that the variance is significantly higher than 0.06 gal.

Solution:

i. State null & alternative hypotheses:

ii. The significance level:  α = 0.05

iii. The test statistic:

iv. Critical Region:

Reject H0, when χ2 ≥ χ 20.25(10) = 18.307

v. Computation:

vi. Remarks: The chi square computed value falls in the acceptance region; the sample data does not provide sufficient evidence to reject the null hypothesis at the 5% significance level. Thus, it is concluded the variance is 0.06.

Example 9.5: The weights of a random sample of 10 packs of corn flour selected from a normal distribution are 14.2, 13.7, 14.1, 14.3, 14.1, 13.8, 14.4, 14.8, 13.9, and 14.3. Test the hypothesis that the variance of the normal population from which the sample is drawn is less than 100 at the 1% significance level.

Solution:

i. State null & alternative hypotheses:

ii. The significance level:  α = 0.01

iii. The test statistic:

iv. Critical Region:

Reject H0, when  χ2  ≤ χ20.99(9) = 2.088

v. Computation:

vi. Remarks: The chi square computed value falls in the acceptance region; the sample data does not provide sufficient evidence to reject the null hypothesis at the 1% significance level. Thus, it is concluded the variance is 50.

• Read More: Barlrtt's Test & Confidence Interval for population variances