TANSCHE SMTPT31 Mathematical Statistics Unit 5

SMTPT31

Mathematical Statistics
(TANSCHE Syllabus)

Madurai Kamaraj University
II- B.Sc., Mathematics, III - Semester

ARO Study Circle

Contents

Chapter 5
Tests of Significance for Small Samples

Syllabus

Using the chi-square distribution - Student’s t- distribution - F- distribution
Chapter 13 sections 13-13.5, Chapter 14 sections 14-14.2.10 and Sections 14.5 – 14.5.5

5.1 Chi-Square Distribution

The Chi-square distribution (pronounced as "Ki-square") is one of the most important distributions in statistics. It is widely used for testing hypotheses, especially when dealing with categorical data. The chi-square distribution was first discovered by Karl Pearson in 1900.

Definition 5.1.1 (Chi-Square Variate). The square of a standard normal variate is called a chi-square variate with 1 degree of freedom (d.f.). If \( X \sim N(\mu , \sigma ^2) \), then: \[ Z = \frac {X - \mu }{\sigma } \sim N(0,1) \] Then: \[ Z^2 = \left ( \frac {X - \mu }{\sigma } \right )^2 \] is a chi-square variate with 1 d.f. In general, if \( X_i \) for \( i = 1, 2, \dots , n \) are \( n \) independent normal variates with mean \( \mu _i \) and variance \( \sigma _i^2 \), then: \[ \chi ^2 = \sum _{i=1}^{n} \left ( \frac {X_i - \mu _i}{\sigma _i} \right )^2 \] is a chi-square variate with \( n \) degrees of freedom.

5.2 Derivation of Chi-Square Distribution

Method of Moment Generating Function

Let \( X_i, (i = 1, 2, \dots , n) \) be independent \( N(\mu _i, \sigma _i^2) \) variates. We want the distribution of: \[ \chi ^2 = \sum _{i=1}^{n} \left ( \frac {X_i - \mu _i}{\sigma _i} \right )^2 = \sum _{i=1}^{n} U_i^2 \] where \( U_i = \frac {X_i - \mu _i}{\sigma _i} \sim N(0,1) \). Since \( X_i \)’s are independent, \( U_i \)’s are also independent. Therefore: \[ M_{\chi ^2}(t) = M_{\sum U_i^2}(t) = \prod _{i=1}^{n} M_{U_i^2}(t) = [M_{U_i^2}(t)]^n \] Now: \begin {align*} M_{U_i^2}(t) &= E(e^{t U_i^2}) \\ &= \int _{-\infty }^{\infty } e^{t u_i^2} \frac {1}{\sqrt {2\pi }} e^{-u_i^2/2} du_i \\ &= \frac {1}{\sqrt {2\pi }} \int _{-\infty }^{\infty } e^{-\frac {1}{2}(1-2t)u_i^2} du_i \\ &= (1 - 2t)^{-1/2} \end {align*} Therefore: \[ M_{\chi ^2}(t) = (1 - 2t)^{-n/2} \] This is the m.g.f. of a Gamma variate with parameters \( \frac {1}{2} \) and \( \frac {n}{2} \). Hence, by the uniqueness theorem of m.g.f.’s: \[ \chi ^2 = \sum _{i=1}^{n} \left ( \frac {X_i - \mu _i}{\sigma _i} \right )^2 \] is a Gamma variate with parameters \( \frac {1}{2} \) and \( \frac {n}{2} \). Thus, the p.d.f. of chi-square distribution with \( n \) degrees of freedom is: \[ f(\chi ^2) = \frac {1}{2^{n/2} \Gamma (n/2)} e^{-\chi ^2/2} (\chi ^2)^{(n/2)-1}, \quad 0 \leq \chi ^2 < \infty \]

Remark 5.2.1. If a random variable \( X \) follows a chi-square distribution with \( n \) degrees of freedom, we write \( X \sim \chi ^2_{(n)} \).
Remark 5.2.2.
1.
If \( X_i; i = 1, 2, \dots , n \) are \( n \) independent normal variates with mean \( \mu _i \) and S.D. \( \sigma _i \), then \( \sum _{i=1}^{n} \left ( \frac {X_i - \mu _i}{\sigma _i} \right )^2 \) is a chi-square variate with \( n \) d.f.
2.
Normal distribution is a particular case of chi-square distribution when \( n = 1 \).

5.3 Moment Generating Function of Chi-Square Distribution

Let \( X \sim \chi ^2_{(n)} \). Then: \[ M_X(t) = E(e^{tX}) = \int _0^{\infty } e^{tx} f(x) \, dx \] \begin {align*} M_X(t) &= \frac {1}{2^{n/2} \Gamma (n/2)} \int _0^{\infty } e^{tx} e^{-x/2} x^{(n/2)-1} \, dx \\ &= \frac {1}{2^{n/2} \Gamma (n/2)} \int _0^{\infty } e^{-\left ( \frac {1-2t}{2} \right )x} x^{(n/2)-1} \, dx \\ &= \frac {1}{2^{n/2} \Gamma (n/2)} \cdot \frac {\Gamma (n/2)}{\left ( \frac {1-2t}{2} \right )^{n/2}} \\ &= (1 - 2t)^{-n/2}, \quad |t| < \frac {1}{2} \end {align*}

5.3.1 Cumulant Generating Function of Chi-Square Distribution

If \( X \sim \chi ^2_{(n)} \), then: \[ K_X(t) = \log M_X(t) = -\frac {n}{2} \log (1 - 2t) \] Expanding: \[ K_X(t) = \frac {n}{2} \left [ 2t + \frac {(2t)^2}{2} + \frac {(2t)^3}{3} + \frac {(2t)^4}{4} + \cdots \right ] \] Comparing coefficients: \begin {align*} \kappa _1 &= n \quad \text {(Mean)} \\ \kappa _2 &= 2n \quad \text {(Variance)} \\ \kappa _3 &= 8n \\ \kappa _4 &= 48n \end {align*} In general: \[ \kappa _r = n \cdot 2^{r-1} (r-1)! \] Therefore: \begin {align*} \text {Mean} &= n \\ \text {Variance} &= 2n \\ \beta _1 &= \frac {\mu _3^2}{\mu _2^3} = \frac {(8n)^2}{(2n)^3} = \frac {8}{n} \\ \beta _2 &= \frac {\mu _4}{\mu _2^2} = \frac {48n + 12n^2}{4n^2} = 3 + \frac {12}{n} \end {align*}

5.3.2 Limiting Form of Chi-Square Distribution for Large Degrees of Freedom

If \( X \sim \chi ^2_{(n)} \), then \( M_X(t) = (1 - 2t)^{-n/2} \). Let \( Z = \frac {X - \mu }{\sigma } \) be the standard chi-square variate, where \( \mu = n \) and \( \sigma ^2 = 2n \). Then: \[ M_Z(t) = e^{-\mu t / \sigma } M_X(t/\sigma ) = e^{-nt/\sqrt {2n}} (1 - 2t/\sqrt {2n})^{-n/2} \] Taking logarithm: \begin {align*} K_Z(t) &= \log M_Z(t) \\ &= -t\sqrt {\frac {n}{2}} - \frac {n}{2} \log \left ( 1 - t\sqrt {\frac {2}{n}} \right ) \\ &= -t\sqrt {\frac {n}{2}} + \frac {n}{2} \left [ t\sqrt {\frac {2}{n}} + \frac {t^2}{2} \cdot \frac {2}{n} + O(n^{-3/2}) \right ] \\ &= \frac {t^2}{2} + O(n^{-1/2}) \end {align*} Taking the limit as \( n \to \infty \): \[ \lim _{n \to \infty } M_Z(t) = e^{t^2/2} \] This is the m.g.f. of the standard normal distribution. Hence, as \( n \to \infty \), the chi-square distribution tends to the normal distribution.

5.3.3 Mode and Skewness of Chi-Square Distribution

The p.d.f. of chi-square distribution is: \[ f(x) = \frac {1}{2^{n/2} \Gamma (n/2)} e^{-x/2} x^{(n/2)-1}, \quad 0 \leq x < \infty \] Taking logarithm: \[ \log f(x) = \text {constant} - \frac {x}{2} + \left ( \frac {n}{2} - 1 \right ) \log x \] Differentiating: \[ \frac {f'(x)}{f(x)} = -\frac {1}{2} + \frac {n/2 - 1}{x} = \frac {n - 2 - x}{2x} \] For the mode, set \( f'(x) = 0 \): \[ x = n - 2 \] Thus, the mode of the chi-square distribution with \( n \) d.f. is \( (n - 2) \). Karl Pearson’s Coefficient of Skewness: \[ \text {Skewness} = \frac {\text {Mean} - \text {Mode}}{\text {S.D.}} = \frac {n - (n-2)}{\sqrt {2n}} = \sqrt {\frac {2}{n}} \] Since the coefficient of skewness is positive, the chi-square distribution is positively skewed. As \( n \) increases, the skewness decreases, and the distribution tends to symmetry.

5.3.4 Additive Property of Chi-Square Variates

Theorem 5.3.1 (Additive Property). If \( X_1, X_2, \dots , X_k \) are independent chi-square variates with \( n_1, n_2, \dots , n_k \) d.f. respectively, then: \[ \sum _{i=1}^{k} X_i \sim \chi ^2_{(n_1 + n_2 + \cdots + n_k)} \]
Proof : For each \( X_i \), the m.g.f. is: \[ M_{X_i}(t) = (1 - 2t)^{-n_i/2} \] Since the variables are independent: \[ M_{\sum X_i}(t) = \prod _{i=1}^{k} (1 - 2t)^{-n_i/2} = (1 - 2t)^{-(n_1 + n_2 + \cdots + n_k)/2} \] This is the m.g.f. of a chi-square variate with \( \sum n_i \) d.f. Hence, by the uniqueness theorem of m.g.f.’s: \[ \sum _{i=1}^{k} X_i \sim \chi ^2_{(n_1 + n_2 + \cdots + n_k)} \] □
Remark 5.3.2.
1.
The converse of the additive property is also true.
2.
Another useful version: If \( X \) and \( Y \) are independent non-negative variates such that \( X + Y \) follows chi-square with \( n_1 + n_2 \) d.f., and \( X \) is chi-square with \( n_1 \) d.f., then \( Y \) is chi-square with \( n_2 \) d.f.

5.3.5 Chi-Square Probability Curve

The chi-square probability curve has the following characteristics: From the derivative: \[ f'(x) = \frac {n - 2 - x}{2x} f(x) \]

1.
For \( n = 1 \) and \( n = 2 \), the curve is monotonically decreasing.
2.
For \( n > 2 \):
  • \( f'(x) > 0 \) for \( 0 < x < n-2 \) (curve rises)
  • \( f'(x) < 0 \) for \( x > n-2 \) (curve falls)
3.
The curve attains its maximum at \( x = n-2 \).
4.
The curve is positively skewed.
5.
The \( x \)-axis is an asymptote to the curve.
Theorem 5.3.3 (Theorem 13.1). If \( \chi _1^2 \) and \( \chi _2^2 \) are two independent chi-square variates with \( n_1 \) and \( n_2 \) d.f. respectively, then: \[ \frac {\chi _1^2}{\chi _2^2} \sim \beta _2\left ( \frac {n_1}{2}, \frac {n_2}{2} \right ) \]
Proof : The joint probability differential is: \[ dP(\chi _1^2, \chi _2^2) = \frac {1}{2^{(n_1+n_2)/2} \Gamma (n_1/2) \Gamma (n_2/2)} e^{-(\chi _1^2 + \chi _2^2)/2} (\chi _1^2)^{(n_1/2)-1} (\chi _2^2)^{(n_2/2)-1} d\chi _1^2 d\chi _2^2 \] Let: \[ u = \frac {\chi _1^2}{\chi _2^2}, \quad v = \chi _2^2 \] Then \( \chi _1^2 = uv \), \( \chi _2^2 = v \). The Jacobian is: \[ |J| = v \] Thus: \[ dG(u,v) = \frac {1}{2^{(n_1+n_2)/2} \Gamma (n_1/2) \Gamma (n_2/2)} e^{-(1+u)v/2} u^{(n_1/2)-1} v^{(n_1+n_2)/2-1} du \, dv \] Integrating over \( v \): \[ dG_1(u) = \frac {1}{B(n_1/2, n_2/2)} \cdot \frac {u^{(n_1/2)-1}}{(1+u)^{(n_1+n_2)/2}} du \] Hence: \[ U = \frac {\chi _1^2}{\chi _2^2} \sim \beta _2\left ( \frac {n_1}{2}, \frac {n_2}{2} \right ) \] □
Theorem 5.3.4 (Theorem 13.2). If \( \chi _1^2 \) and \( \chi _2^2 \) are independent chi-square variates with \( n_1 \) and \( n_2 \) d.f. respectively, then: \[ U = \frac {\chi _1^2}{\chi _1^2 + \chi _2^2} \sim \beta _1\left ( \frac {n_1}{2}, \frac {n_2}{2} \right ) \] and \[ V = \chi _1^2 + \chi _2^2 \sim \chi ^2_{(n_1 + n_2)} \] are independently distributed.
Proof : Let: \[ u = \frac {\chi _1^2}{\chi _1^2 + \chi _2^2}, \quad v = \chi _1^2 + \chi _2^2 \] Then \( \chi _1^2 = uv \), \( \chi _2^2 = (1-u)v \). The Jacobian is: \[ |J| = v \] The joint distribution becomes: \[ dG(u,v) = \frac {1}{2^{(n_1+n_2)/2} \Gamma (n_1/2) \Gamma (n_2/2)} u^{(n_1/2)-1} (1-u)^{(n_2/2)-1} e^{-v/2} v^{(n_1+n_2)/2-1} du \, dv \] This can be written as the product: \[ dG(u,v) = \left [ \frac {1}{B(n_1/2, n_2/2)} u^{(n_1/2)-1} (1-u)^{(n_2/2)-1} du \right ] \times \left [ \frac {1}{2^{(n_1+n_2)/2} \Gamma ((n_1+n_2)/2)} e^{-v/2} v^{(n_1+n_2)/2-1} dv \right ] \] Since the joint distribution factors into the product of marginal distributions, \( U \) and \( V \) are independent. Therefore: \[ U \sim \beta _1\left ( \frac {n_1}{2}, \frac {n_2}{2} \right ), \quad V \sim \chi ^2_{(n_1 + n_2)} \] □
Theorem 5.3.5 (Theorem 13.3). In a random and large sample: \[ \chi ^2 = \sum _{i=1}^{k} \frac {(O_i - E_i)^2}{E_i} \] follows a chi-square distribution approximately with \( (k-1) \) degrees of freedom, where \( O_i \) is the observed frequency and \( E_i \) is the corresponding expected frequency of the \( i \)-th class.
Proof : Let us consider a random sample of size \( n \), whose members are distributed at random in \( k \) classes. Let \( p_i \) be the probability that a sample observation will fall in the \( i \)-th cell. Then the probability of having \( n_i \) members in the \( i \)-th cell is given by the multinomial probability law: \[ P = \frac {n!}{n_1! n_2! \cdots n_k!} p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k} \] Using Stirling’s approximation for large \( n \), we get: \[ P \approx C \prod _{i=1}^{k} \left ( \frac {np_i}{n_i} \right )^{n_i + 1/2} \] Let \( \lambda _i = np_i \) be the expected frequency for the \( i \)-th cell. Define: \[ \xi _i = \frac {n_i - \lambda _i}{\sqrt {\lambda _i}} \] Then \( n_i = \lambda _i + \xi _i \sqrt {\lambda _i} \). Taking logarithms and expanding: \[ \log (P/C) \approx -\frac {1}{2} \sum _{i=1}^{k} \xi _i^2 \] Also: \[ \sum _{i=1}^{k} \xi _i \sqrt {\lambda _i} = \sum _{i=1}^{k} (n_i - \lambda _i) = n - n = 0 \] Thus, \( \xi _i \) are distributed as independent standard normal variates with one linear constraint. Therefore: \[ \sum _{i=1}^{k} \xi _i^2 = \sum _{i=1}^{k} \frac {(n_i - \lambda _i)^2}{\lambda _i} \] is a chi-square variate with \( (k-1) \) degrees of freedom. □

5.3.6 Summary of Key Results for Chi-Square Distribution

Property Result
Probability Density Function \( f(x) = \dfrac {1}{2^{n/2} \Gamma (n/2)} e^{-x/2} x^{(n/2)-1}, \quad 0 \leq x < \infty \)
Mean \( n \)
Variance \( 2n \)
Mode \( n - 2 \text { (for } n \geq 2\text {)} \)
Moment Generating Function \( M_X(t) = (1 - 2t)^{-n/2}, \quad |t| < \frac {1}{2} \)
Cumulants \( \kappa _r = n \cdot 2^{r-1} (r-1)! \)
Skewness \( \beta _1 = \dfrac {8}{n} \)
Kurtosis \( \beta _2 = 3 + \dfrac {12}{n} \)
Additive Property \( \chi ^2_{n_1} + \chi ^2_{n_2} = \chi ^2_{n_1 + n_2} \)
Limiting Form As \( n \to \infty \), tends to normal distribution
Test Statistic \( \chi ^2 = \displaystyle \sum _{i=1}^{k} \dfrac {(O_i - E_i)^2}{E_i} \)
Degrees of Freedom for Goodness of Fit\( k - 1 \) (or \( k - p - 1 \) if parameters are estimated)

5.3.7 Important Remarks

Remark 5.3.6.
1.
The chi-square test is a non-parametric test because it does not make any assumptions about the population parameters.
2.
The test is always a one-tailed test because the chi-square distribution is positively skewed and the critical region lies in the right tail.
3.
When applying the chi-square test, the total of observed frequencies must equal the total of expected frequencies.
4.
The degrees of freedom are reduced by the number of parameters estimated from the data.
5.
For the validity of the chi-square test:
  • The sample observations should be independent.
  • The total frequency \( N \) should be reasonably large (usually greater than 50).
  • No theoretical cell frequency should be less than 5.
  • If any cell has expected frequency less than 5, it should be pooled with adjacent cells.

5.4 Student’s t-Distribution

The entire large sample theory was based on the application of the "Normal Test" (cf. Section 12.9). However, if the sample size \( n \) is small, the distribution of various statistics, e.g., \( Z = \frac {\bar {x} - \mu }{\sigma / \sqrt {n}} \) or \( Z = \frac {X - nP}{\sqrt {nPQ}} \), etc., are far from normality and as such the ’normal test’ cannot be applied if \( n \) is small. In such cases, exact sample tests, pioneered by W.S. Gosset (1908) who wrote under the pen name of Student, and later developed and extended by Prof. R.A. Fisher (1926), are used.

Definition 5.4.1 (Student’s t). Let \( x_i, (i = 1, 2, \dots , n) \) be a random sample of size \( n \) from a normal population with mean \( \mu \) and variance \( \sigma ^2 \). Then Student’s \( t \) is defined by: \[ t = \frac {\bar {x} - \mu }{S / \sqrt {n}} \] where: \[ \bar {x} = \frac {1}{n} \sum _{i=1}^{n} x_i \] is the sample mean, and: \[ S^2 = \frac {1}{n-1} \sum _{i=1}^{n} (x_i - \bar {x})^2 \] is an unbiased estimate of the population variance \( \sigma ^2 \). This statistic follows Student’s t-distribution with \( \nu = (n-1) \) degrees of freedom, with probability density function: \[ f(t) = \frac {1}{\sqrt {\nu } B\left (\frac {1}{2}, \frac {\nu }{2}\right )} \cdot \frac {1}{\left [ 1 + \frac {t^2}{\nu } \right ]^{(\nu +1)/2}}, \quad -\infty < t < \infty \]
Remark 5.4.2.
1.
A statistic \( t \) following Student’s t-distribution with \( n \) d.f. is denoted as \( t \sim t_n \).
2.
If we take \( \nu = 1 \), the t-distribution reduces to the standard Cauchy distribution: \[ f(t) = \frac {1}{\pi (1+t^2)}, \quad -\infty < t < \infty \]
3.
The t-distribution is symmetric about zero.

5.4.1 Derivation of Student’s t-Distribution

The expression for \( t \) can be rewritten as: \[ \frac {t^2}{n-1} = \frac {(\bar {x} - \mu )^2 / (\sigma ^2/n)}{(n s^2 / \sigma ^2)} \] Since \( x_i \)’s are from a normal population: \[ \bar {x} \sim N(\mu , \sigma ^2/n) \Rightarrow \frac {(\bar {x} - \mu )}{\sigma /\sqrt {n}} \sim N(0,1) \] Thus, \( \frac {(\bar {x} - \mu )^2}{\sigma ^2/n} \) is a chi-square variate with 1 d.f. Also, \( \frac {n s^2}{\sigma ^2} \) is a chi-square variate with \( (n-1) \) d.f. (by Theorem 13.5). Since \( \bar {x} \) and \( s^2 \) are independently distributed, \( \frac {t^2}{n-1} \) is the ratio of two independent chi-square variates with 1 and \( (n-1) \) d.f. respectively. Hence, it is a \( \beta _2\left (\frac {1}{2}, \frac {n-1}{2}\right ) \) variate. Therefore, the p.d.f. of \( t \) is: \[ f(t) = \frac {1}{\sqrt {\nu } B\left (\frac {1}{2}, \frac {\nu }{2}\right )} \cdot \frac {1}{\left [ 1 + \frac {t^2}{\nu } \right ]^{(\nu +1)/2}}, \quad -\infty < t < \infty \]

5.4.2 Fisher’s Definition of t

Definition 5.4.3 (Fisher’s t). Fisher’s \( t \) is defined as the ratio of a standard normal variate to the square root of an independent chi-square variate divided by its degrees of freedom. If \( \xi \sim N(0,1) \) and \( \chi ^2 \) is an independent chi-square variate with \( n \) d.f., then: \[ t = \frac {\xi }{\sqrt {\chi ^2 / n}} \] and it follows Student’s t-distribution with \( n \) degrees of freedom.

5.4.3 Distribution of Fisher’s t

Since \( \xi \) and \( \chi ^2 \) are independent, their joint probability differential is: \[ dF(\xi , \chi ^2) = \frac {1}{\sqrt {2\pi }} e^{-\xi ^2/2} \cdot \frac {e^{-\chi ^2/2} (\chi ^2)^{(n/2)-1}}{2^{n/2} \Gamma (n/2)} d\xi d\chi ^2 \] Let: \[ t = \frac {\xi }{\sqrt {\chi ^2/n}}, \quad u = \chi ^2 \] Then \( \xi = t\sqrt {u/n} \). The Jacobian is: \[ |J| = \sqrt {\frac {u}{n}} \] The joint distribution becomes: \[ dG(t,u) = \frac {1}{\sqrt {2\pi } \, 2^{n/2} \Gamma (n/2) \sqrt {n}} e^{-u(1+t^2/n)/2} u^{(n-1)/2} du \, dt \] Integrating over \( u \): \[ dG_1(t) = \frac {1}{\sqrt {n} B\left (\frac {1}{2}, \frac {n}{2}\right )} \cdot \frac {dt}{\left [ 1 + \frac {t^2}{n} \right ]^{(n+1)/2}}, \quad -\infty < t < \infty \] This is the same as the p.d.f. of Student’s t-distribution with \( n \) d.f.

Remark 5.4.4. Student’s t may be regarded as a particular case of Fisher’s t.

5.4.4 Constants of t-Distribution

Since \( f(t) \) is symmetric about \( t = 0 \), all moments of odd order vanish: \[ \mu _{2r+1} = 0, \quad r = 0, 1, 2, \dots \] In particular, the mean is 0. Hence central moments coincide with moments about origin. The moments of even order are given by: \[ \mu _{2r} = n^r \frac {(2r-1)(2r-3)\cdots 3 \cdot 1}{(n-2)(n-4)\cdots (n-2r)}, \quad n > 2r \] In particular: \[ \mu _2 = \frac {n}{n-2}, \quad n > 2 \] \[ \mu _4 = \frac {3n^2}{(n-2)(n-4)}, \quad n > 4 \] Thus: \[ \beta _1 = 0 \quad \text {and} \quad \beta _2 = \frac {3(n-2)}{(n-4)} \]

Remark 5.4.5.
1.
As \( n \to \infty \), \( \beta _1 = 0 \) and \( \beta _2 = 3 \), so the t-distribution tends to the normal distribution.
2.
The moment generating function of the t-distribution does not exist because moments of order \( 2r \geq n \) do not exist.
Example 5.4.6. Express the constants \( y_0 \), \( a \) and \( m \) of the distribution: \[ dF(x) = y_0 \left [ 1 - \frac {x^2}{a^2} \right ]^m dx, \quad -a \leq x \leq a \] in terms of its \( \mu _2 \) and \( \beta _2 \). Show that if \( x \) is related to a variable \( t \) by the equation: \[ x = \frac {at}{\left [ 2(m+1) + t^2 \right ]^{1/2}} \] then \( t \) has Student’s distribution with \( 2(m+1) \) degrees of freedom. Use the transformation to calculate the probability that \( t \geq 2 \) when the degrees of freedom are 2 and also when 4.
Solution. First, determine the constant \( y_0 \) from the condition that total probability is unity: \[ \int _{-a}^{a} y_0 \left ( 1 - \frac {x^2}{a^2} \right )^m dx = 1 \] Since the integrand is even: \[ 2y_0 \int _{0}^{a} \left ( 1 - \frac {x^2}{a^2} \right )^m dx = 1 \] Let \( x = a \sin \theta \). Then \( dx = a \cos \theta \, d\theta \): \[ 2y_0 a \int _{0}^{\pi /2} \cos ^{2m+1} \theta \, d\theta = 1 \] Using the Beta integral: \[ 2 \int _{0}^{\pi /2} \sin ^{p} \theta \cos ^{q} \theta \, d\theta = B\left ( \frac {p+1}{2}, \frac {q+1}{2} \right ) \] We get: \[ y_0 a B\left ( m+1, \frac {1}{2} \right ) = 1 \] \[ y_0 = \frac {1}{a B\left ( m+1, \frac {1}{2} \right )} \] Since the distribution is symmetric about \( x = 0 \): \[ \mu _{2r+1} = 0, \quad r = 0, 1, 2, \dots \] The moments of even order are: \[ \mu _{2r} = a^{2r} \frac {B\left ( r + \frac {1}{2}, m+1 \right )}{B\left ( m+1, \frac {1}{2} \right )} \] In particular: \[ \mu _2 = \frac {a^2}{2m+3} \Rightarrow a^2 = (2m+3)\mu _2 \] Also: \[ \mu _4 = \frac {3a^4}{(2m+5)(2m+3)} \] Therefore: \[ \beta _2 = \frac {\mu _4}{\mu _2^2} = \frac {3(2m+3)}{2m+5} \] Solving for \( m \): \[ m = \frac {9 - 5\beta _2}{2(\beta _2 - 3)} \] Now, consider the transformation: \[ x = \frac {at}{\sqrt {n + t^2}}, \quad n = 2m + 2 \] Then: \[ 1 - \frac {x^2}{a^2} = \frac {n}{n + t^2} = \left ( 1 + \frac {t^2}{n} \right )^{-1} \] Also: \[ dx = \frac {a n}{(n + t^2)^{3/2}} dt \] The p.d.f. of \( t \) becomes: \[ dF(t) = y_0 \frac {1}{\left ( 1 + \frac {t^2}{n} \right )^m} \cdot \frac {a n}{(n + t^2)^{3/2}} dt \] Substituting \( y_0 = \frac {1}{a B(m+1, 1/2)} \): \[ dF(t) = \frac {1}{\sqrt {n} B\left ( \frac {n}{2}, \frac {1}{2} \right )} \cdot \frac {dt}{\left [ 1 + \frac {t^2}{n} \right ]^{(n+1)/2}} \] This is the p.d.f. of Student’s t-distribution with \( n = 2(m+1) \) d.f. For 2 d.f.: \( n = 2 \Rightarrow m = 0 \) Then \( x = \frac {at}{\sqrt {2 + t^2}} \). When \( t = 2 \), \( x = \frac {2a}{\sqrt {6}} = a\sqrt {\frac {2}{3}} \) \[ P(t \geq 2) = \int _{a\sqrt {2/3}}^{a} \frac {1}{2a} dx = \frac {1}{2} \left ( 1 - \sqrt {\frac {2}{3}} \right ) = \frac {\sqrt {3} - \sqrt {2}}{2\sqrt {3}} \] For 4 d.f.: \( n = 4 \Rightarrow m = 1 \) \[ P(t \geq 2) = \frac {1}{2} - \frac {5\sqrt {2}}{16} \]

5.4.5 Limiting Form of t-Distribution

As \( n \to \infty \), the p.d.f. of the t-distribution tends to the standard normal p.d.f.: \[ f(t) = \frac {1}{\sqrt {2\pi }} e^{-t^2/2}, \quad -\infty < t < \infty \]

Proof : \begin {align*} \lim _{n \to \infty } \frac {1}{\sqrt {n} B\left (\frac {1}{2}, \frac {n}{2}\right )} &= \frac {1}{\sqrt {2\pi }} \\ \lim _{n \to \infty } \left ( 1 + \frac {t^2}{n} \right )^{-(n+1)/2} &= e^{-t^2/2} \end {align*} □

5.4.6 Graph of t-Distribution

The t-distribution has the following characteristics:

1.
It is symmetrical about \( t = 0 \).
2.
As \( t \) increases, \( f(t) \) decreases rapidly and tends to zero as \( t \to \infty \).
3.
The \( t \)-axis is an asymptote to the curve.
4.
For small \( n \), the tails of the t-distribution have greater probability than the tails of the standard normal distribution.
5.
As \( n \to \infty \), the t-distribution tends to the standard normal distribution.

5.4.7 Critical Values of t

The critical values of \( t \) at level of significance \( \alpha \) and d.f. \( \nu \) for a two-tailed test are given by: \[ P(|t| > t_\nu (\alpha )) = \alpha \] Since the t-distribution is symmetric: \[ P(t > t_\nu (\alpha )) = \alpha /2 \] \[ P(t > t_\nu (2\alpha )) = \alpha \] Thus, the significant value of \( t \) for a single-tailed test at level \( \alpha \) is the same as the value for a two-tailed test at level \( 2\alpha \).

5.4.8 Applications of t-Distribution

The t-distribution has many applications in statistics:

1.
To test if the sample mean differs significantly from a hypothetical population mean.
2.
To test the significance of the difference between two sample means.
3.
To test the significance of an observed sample correlation coefficient.
4.
To test the significance of regression coefficients.

5.4.9 t-Test for Single Mean

Suppose we want to test whether a random sample \( x_i, (i = 1, 2, \dots , n) \) has been drawn from a normal population with a specified mean \( \mu _0 \). Under the null hypothesis \( H_0: \mu = \mu _0 \), the statistic: \[ t = \frac {\bar {x} - \mu _0}{S / \sqrt {n}} \] follows Student’s t-distribution with \( (n-1) \) d.f. where: \[ S^2 = \frac {1}{n-1} \sum _{i=1}^{n} (x_i - \bar {x})^2 \]

Remark 5.4.7. For numerical problems, if \( \bar {x} \) is fractional, use the step deviation method: \[ S^2 = \frac {1}{n-1} \left [ \sum d_i^2 - \frac {(\sum d_i)^2}{n} \right ] \] where \( d_i = x_i - A \) (A is any arbitrary constant).
Example 5.4.8. A machinist is making engine parts with axle diameters of 0.700 inch. A random sample of 10 parts shows a mean diameter of 0.742 inch with a standard deviation of 0.040 inch. Compute the statistic you would use to test whether the work is meeting the specifications.
Solution. We are given: \[ \mu = 0.700, \quad \bar {x} = 0.742, \quad s = 0.040, \quad n = 10 \] Null Hypothesis: \( H_0: \mu = 0.700 \) Alternative Hypothesis: \( H_1: \mu \neq 0.700 \) Test Statistic: \[ t = \frac {\bar {x} - \mu }{S/\sqrt {n}} = \frac {\bar {x} - \mu }{\sqrt {s^2/(n-1)}} \sim t_{(n-1)} \] \begin {align*} t &= \frac {0.742 - 0.700}{\sqrt {(0.040)^2/9}} \\ &= \frac {0.042}{0.01333} = 3.15 \end {align*} The calculated t value is 3.15 with 9 d.f. Compare with tabulated t at 5
Example 5.4.9. The mean weekly sales of soap bars in departmental stores was 146.3 bars per store. After an advertising campaign, the mean weekly sales in 22 stores for a typical week increased to 153.7 and showed a standard deviation of 17.2. Was the advertising campaign successful?
Solution. Given: \[ n = 22, \quad \bar {x} = 153.7, \quad s = 17.2 \] Null Hypothesis: \( H_0: \mu = 146.3 \) Alternative Hypothesis: \( H_1: \mu > 146.3 \) (Right-tailed) Test Statistic: \[ t = \frac {\bar {x} - \mu }{\sqrt {s^2/(n-1)}} = \frac {153.7 - 146.3}{\sqrt {(17.2)^2/21}} = \frac {7.4}{3.754} = 1.97 \] Tabulated \( t_{0.05} \) for 21 d.f. (single-tail) is 1.72. Since \( t = 1.97 > 1.72 \), reject \( H_0 \). The advertising campaign was successful.
Example 5.4.10. A random sample of 10 boys had the following I.Q.’s: 70, 120, 110, 101, 88, 83, 95, 98, 107, 100. Do these data support the assumption of a population mean I.Q. of 100?
Solution.
\( X \) \( d = X - 90 \) \( d^2 \)
70 -20 400
120 30 900
110 20 400
101 11 121
88 -2 4
83 -7 49
95 5 25
98 8 64
107 17 289
100 10 100
Total 72 2352
\[ \bar {x} = 90 + \frac {72}{10} = 97.2 \]\[ S^2 = \frac {1}{9} \left [ 2352 - \frac {(72)^2}{10} \right ] = 203.73 \] Test Statistic: \[ t = \frac {97.2 - 100}{\sqrt {203.73/10}} = \frac {-2.8}{4.51} = -0.62 \] Tabulated \( t_{0.05} \) for 9 d.f. is 2.262. Since \( |t| < 2.262 \), accept \( H_0 \). The data support the assumption of mean I.Q. of 100.
Example 5.4.11. The heights of 10 males of a given locality are found to be 70, 67, 62, 68, 61, 68, 70, 64, 64, 66 inches. Is it reasonable to believe that the average height is greater than 64 inches? Test at 5
Solution. \[ \bar {x} = 66, \quad \sum (x - \bar {x})^2 = 90, \quad S^2 = 10 \] Null Hypothesis: \( H_0: \mu = 64 \) Alternative Hypothesis: \( H_1: \mu > 64 \) (Right-tailed) Test Statistic: \[ t = \frac {66 - 64}{\sqrt {10/10}} = 2 \] Tabulated t for 9 d.f. at 5
Example 5.4.12. A random sample of 16 values from a normal population showed a mean of 41.5 inches and the sum of squares of deviations from this mean equal to 135 square inches. Show that the assumption of a mean of 43.5 inches for the population is not reasonable. Obtain 95
Solution. Given: \[ n = 16, \quad \bar {x} = 41.5, \quad \sum (x-\bar {x})^2 = 135 \] \[ S^2 = \frac {135}{15} = 9, \quad S = 3 \] Null Hypothesis: \( H_0: \mu = 43.5 \) Test Statistic: \[ t = \frac {41.5 - 43.5}{3/\sqrt {16}} = \frac {-2}{0.75} = -2.667 \] Tabulated \( t_{0.05} \) for 15 d.f. is 2.131. Since \( |t| = 2.667 > 2.131 \), reject \( H_0 \). The assumption is not reasonable. 95\[ \bar {x} \pm t_{0.05} \frac {S}{\sqrt {n}} = 41.5 \pm 2.131 \times 0.75 = 41.5 \pm 1.598 = (39.902, 43.098) \] 99\[ \bar {x} \pm t_{0.01} \frac {S}{\sqrt {n}} = 41.5 \pm 2.947 \times 0.75 = 41.5 \pm 2.210 = (39.290, 43.710) \]

5.4.10 t-Test for Difference of Means

Suppose we want to test if two independent samples \( x_i \) and \( y_j \) have been drawn from two normal populations with means \( \mu _X \) and \( \mu _Y \). Under \( H_0: \mu _X = \mu _Y \) and assuming \( \sigma _X^2 = \sigma _Y^2 = \sigma ^2 \), the statistic: \[ t = \frac {\bar {x} - \bar {y}}{S \sqrt {\frac {1}{n_1} + \frac {1}{n_2}}} \] follows Student’s t-distribution with \( (n_1 + n_2 - 2) \) d.f. where: \[ S^2 = \frac {1}{n_1 + n_2 - 2} \left [ \sum (x_i - \bar {x})^2 + \sum (y_j - \bar {y})^2 \right ] \]

Remark 5.4.13. Assumptions for t-test for difference of means:
1.
Parent populations are normal.
2.
Population variances are equal and unknown.
3.
Samples are random and independent.

5.4.11 Paired t-Test

When the sample sizes are equal and observations are paired, we use the paired t-test. Let \( d_i = x_i - y_i \) be the differences. Under \( H_0: \mu _d = 0 \), the statistic: \[ t = \frac {\bar {d}}{S/\sqrt {n}} \] follows Student’s t-distribution with \( (n-1) \) d.f. where: \[ \bar {d} = \frac {1}{n} \sum d_i, \quad S^2 = \frac {1}{n-1} \sum (d_i - \bar {d})^2 \]

Example 5.4.14. Below are given the gain in weights (in lbs.) of pigs fed on two diets A and B. Diet A: 25, 32, 30, 34, 24, 14, 32, 24, 30, 31, 35, 25 Diet B: 44, 34, 22, 10, 47, 31, 40, 30, 32, 35, 18, 21, 35, 29, 22 Test if the two diets differ significantly.
Solution. For Diet A: \( n_1 = 12, \bar {x} = 28, \sum (x-\bar {x})^2 = 380 \) For Diet B: \( n_2 = 15, \bar {y} = 30, \sum (y-\bar {y})^2 = 1410 \) \[ S^2 = \frac {380 + 1410}{12+15-2} = 71.6 \] \[ t = \frac {28 - 30}{\sqrt {71.6 \left ( \frac {1}{12} + \frac {1}{15} \right )}} = \frac {-2}{\sqrt {10.74}} = -0.61 \] Tabulated t for 25 d.f. at 5
Example 5.4.15. Samples of two types of electric light bulbs were tested for length of life and following data were obtained: Sample I: \( n_1 = 8, \bar {x}_1 = 1234, s_1 = 36 \) Sample II: \( n_2 = 7, \bar {x}_2 = 1036, s_2 = 40 \) Is the difference in the means sufficient to warrant that type I is superior to type II?
Solution. \[ S^2 = \frac {1}{13} [8(36)^2 + 7(40)^2] = 1659.08 \] \[ t = \frac {1234 - 1036}{\sqrt {1659.08 \left ( \frac {1}{8} + \frac {1}{7} \right )}} = \frac {198}{21.09} = 9.39 \] Tabulated t for 13 d.f. at 5
Example 5.4.16. The heights of six randomly chosen sailors are in inches: 63, 65, 68, 69, 71, 72. Those of 10 randomly chosen soldiers are 61, 62, 65, 66, 69, 69, 70, 71, 72, 73. Discuss the light that these data throw on the suggestion that sailors are on the average taller than soldiers.
Solution. For Sailors: \( n_1 = 6, \bar {x} = 68, \sum (x-\bar {x})^2 = 60 \) For Soldiers: \( n_2 = 10, \bar {y} = 67.8, \sum (y-\bar {y})^2 = 153 - 6 = 147 \) \[ S^2 = \frac {60 + 147}{6+10-2} = \frac {207}{14} = 14.786 \] \[ t = \frac {68 - 67.8}{\sqrt {14.786 \left ( \frac {1}{6} + \frac {1}{10} \right )}} = \frac {0.2}{\sqrt {3.94}} = \frac {0.2}{1.985} = 0.10 \] Tabulated t for 14 d.f. at 5
Example 5.4.17. A certain stimulus administered to each of 12 patients resulted in the following increase of blood pressure: 5, 2, 8, -1, 3, 0, -2, 1, 5, 0, 4, 6. Can it be concluded that the stimulus will, in general, be accompanied by an increase in blood pressure?
Solution. Here, the data are paired. Compute \( \bar {d} \) and \( S^2 \). \[ \sum d = 31, \quad n = 12, \quad \bar {d} = \frac {31}{12} = 2.583 \] \[ \sum d^2 = 185, \quad S^2 = \frac {1}{11} \left [ 185 - \frac {(31)^2}{12} \right ] = 9.538 \] Test Statistic: \[ t = \frac {\bar {d}}{S/\sqrt {n}} = \frac {2.583}{\sqrt {9.538/12}} = \frac {2.583}{0.892} = 2.89 \] Tabulated t for 11 d.f. at 5
Example 5.4.18. In a certain experiment to compare two types of pig foods A and B, the following results of increase in weights were observed in pigs:
Pig number 1 2 3 4 5 6 7 8 Total
Food A 4953515247505253 407
Food B 5255525350545453 423
(i) Assuming that the two samples of pigs are independent, can we conclude that food B is better than food A? (ii) Also examine the case when the same set of eight pigs were used in both the foods.
Solution. (i) Independent samples: For Food A: \( n_1 = 8, \bar {x} = 50.875, \sum (x-\bar {x})^2 = 30.875 \) For Food B: \( n_2 = 8, \bar {y} = 52.875, \sum (y-\bar {y})^2 = 16.875 \) \[ S^2 = \frac {30.875 + 16.875}{14} = 3.41 \] \[ t = \frac {50.875 - 52.875}{\sqrt {3.41 \left ( \frac {1}{8} + \frac {1}{8} \right )}} = \frac {-2}{0.923} = -2.17 \] Tabulated t for 14 d.f. at 5 (ii) Paired samples: \[ d: -3, -2, -1, -1, -3, -4, -2, 0 \] \[ \sum d = -16, \quad \bar {d} = -2, \quad \sum d^2 = 44 \] \[ S^2 = \frac {1}{7} \left [ 44 - \frac {(-16)^2}{8} \right ] = \frac {1}{7} [44 - 32] = 1.714 \] \[ t = \frac {-2}{\sqrt {1.714/8}} = \frac {-2}{0.463} = -4.32 \] Tabulated t for 7 d.f. at 5

5.4.12 Summary of Key Results for t-Distribution

Property Result
Probability Density Function \( f(t) = \dfrac {1}{\sqrt {\nu } B\left (\frac {1}{2}, \frac {\nu }{2}\right )} \cdot \dfrac {1}{\left [ 1 + \frac {t^2}{\nu } \right ]^{(\nu +1)/2}}, \quad -\infty < t < \infty \)
Mean \( 0 \)
Variance \( \dfrac {\nu }{\nu - 2}, \quad \nu > 2 \)
Skewness \( \beta _1 = 0 \)
Kurtosis \( \beta _2 = \dfrac {3(\nu - 2)}{(\nu - 4)}, \quad \nu > 4 \)
Moment Generating Function Does not exist
Limiting Form As \( \nu \to \infty \), tends to \( N(0,1) \)
Test Statistic (Single Mean) \( t = \dfrac {\bar {x} - \mu }{S/\sqrt {n}} \sim t_{(n-1)} \)
Test Statistic (Difference of Means) \( t = \dfrac {\bar {x} - \bar {y}}{S\sqrt {\frac {1}{n_1} + \frac {1}{n_2}}} \sim t_{(n_1+n_2-2)} \)
Test Statistic (Paired Observations) \( t = \dfrac {\bar {d}}{S/\sqrt {n}} \sim t_{(n-1)} \)

5.4.13 Important Remarks

Remark 5.4.19.
1.
The t-test is used when the sample size is small (usually \( n < 30 \)).
2.
The basic assumption of the t-test is that the parent population is normal.
3.
The t-distribution is symmetric and has heavier tails than the normal distribution.
4.
For single-tailed tests, use the critical value at \( 2\alpha \) from the two-tailed tables.
5.
The sample variance \( S^2 \) is an unbiased estimate of the population variance \( \sigma ^2 \).
6.
Before applying the t-test for difference of means, it is desirable to test the equality of variances using the F-test.

5.5 F-Distribution

The F-distribution was discovered by Sir Ronald A. Fisher and is named after him. It plays a very important role in statistical inference, particularly in the analysis of variance (ANOVA) and in testing the equality of variances from two normal populations.

Definition 5.5.1 (F-Statistic). If \( X \) and \( Y \) are two independent chi-square variates with \( \nu _1 \) and \( \nu _2 \) degrees of freedom respectively, then the F-statistic is defined as: \[ F = \frac {X / \nu _1}{Y / \nu _2} \] In other words, \( F \) is the ratio of two independent chi-square variates divided by their respective degrees of freedom.
Remark 5.5.2.
1.
The sampling distribution of \( F \) does not involve any population parameters and depends only on the degrees of freedom \( \nu _1 \) and \( \nu _2 \).
2.
A statistic \( F \) following Snedecor’s F-distribution with \( (\nu _1, \nu _2) \) d.f. is denoted as \( F \sim F(\nu _1, \nu _2) \).
3.
\( \nu _1 \) is called the numerator degrees of freedom and \( \nu _2 \) is called the denominator degrees of freedom.

5.5.1 Derivation of Snedecor’s F-Distribution

Since \( X \) and \( Y \) are independent chi-square variates with \( \nu _1 \) and \( \nu _2 \) d.f. respectively, their joint probability differential is: \[ dF(x,y) = \frac {1}{2^{(\nu _1+\nu _2)/2} \Gamma (\nu _1/2) \Gamma (\nu _2/2)} e^{-(x+y)/2} x^{(\nu _1/2)-1} y^{(\nu _2/2)-1} dx \, dy \] for \( 0 \leq x, y < \infty \). Let us make the following transformation: \[ F = \frac {x / \nu _1}{y / \nu _2}, \quad u = y \] Then: \[ x = \frac {\nu _1}{\nu _2} F u, \quad y = u \] The Jacobian of transformation is: \[ |J| = \frac {\nu _1}{\nu _2} u \] The joint distribution of the transformed variables is: \[ dG(F,u) = \frac {(\nu _1/\nu _2)^{\nu _1/2}}{2^{(\nu _1+\nu _2)/2} \Gamma (\nu _1/2) \Gamma (\nu _2/2)} e^{-u(1+\nu _1 F/\nu _2)/2} u^{(\nu _1+\nu _2)/2-1} F^{(\nu _1/2)-1} du \, dF \] Integrating out \( u \) over the range 0 to \( \infty \): \[ g(F) = \frac {(\nu _1/\nu _2)^{\nu _1/2}}{B(\nu _1/2, \nu _2/2)} \cdot \frac {F^{(\nu _1/2)-1}}{\left [ 1 + \frac {\nu _1}{\nu _2} F \right ]^{(\nu _1+\nu _2)/2}}, \quad 0 \leq F < \infty \] This is the probability density function of the F-distribution with \( (\nu _1, \nu _2) \) degrees of freedom.

5.5.2 Constants of F-Distribution

The \( r \)-th moment about origin of the F-distribution is: \[ \mu _r' = E(F^r) = \left ( \frac {\nu _2}{\nu _1} \right )^r \frac {\Gamma \left ( r + \frac {\nu _1}{2} \right ) \Gamma \left ( \frac {\nu _2}{2} - r \right )}{\Gamma \left ( \frac {\nu _1}{2} \right ) \Gamma \left ( \frac {\nu _2}{2} \right )}, \quad \nu _2 > 2r \] In particular: \[ \mu _1' = \frac {\nu _2}{\nu _2 - 2}, \quad \nu _2 > 2 \] Thus, the mean of the F-distribution is independent of \( \nu _1 \). \[ \mu _2' = \frac {\nu _2^2(\nu _1 + 2)}{\nu _1(\nu _2 - 2)(\nu _2 - 4)}, \quad \nu _2 > 4 \] \[ \mu _2 = \mu _2' - (\mu _1')^2 = \frac {2\nu _2^2(\nu _1 + \nu _2 - 2)}{\nu _1(\nu _2 - 2)^2(\nu _2 - 4)}, \quad \nu _2 > 4 \]

Remark 5.5.3. As \( \nu _1 \) and \( \nu _2 \) become large, the F-distribution tends to a normal distribution with mean 1 and variance \( 2\left ( \frac {1}{\nu _1} + \frac {1}{\nu _2} \right ) \).

5.5.3 Mode and Points of Inflexion of F-Distribution

Taking logarithm of the p.d.f.: \[ \log f(F) = C + \left ( \frac {\nu _1}{2} - 1 \right ) \log F - \frac {\nu _1 + \nu _2}{2} \log \left ( 1 + \frac {\nu _1}{\nu _2} F \right ) \] Differentiating and setting to zero: \[ \frac {\nu _1 - 2}{2F} - \frac {\nu _1(\nu _1 + \nu _2)}{2(\nu _2 + \nu _1 F)} = 0 \] \[ F = \frac {\nu _2(\nu _1 - 2)}{\nu _1(\nu _2 + 2)} \] Thus, the mode of the F-distribution is: \[ \text {Mode} = \frac {\nu _2(\nu _1 - 2)}{\nu _1(\nu _2 + 2)} \]

Remark 5.5.4.
1.
Since \( F > 0 \), the mode exists if and only if \( \nu _1 > 2 \).
2.
The mode is always less than unity.
3.
The points of inflexion of the F-distribution exist when \( \nu _1 > 4 \) and are equidistant from the mode.
4.
The F-distribution is positively skewed since the mean is greater than the mode.

5.5.4 Important Results on F-Distribution

Theorem 5.5.5. If \( X \sim F(n_1, n_2) \), then \( 1/X \sim F(n_2, n_1) \).
Proof : Since: \[ F = \frac {X/\nu _1}{Y/\nu _2} \] Therefore: \[ \frac {1}{F} = \frac {Y/\nu _2}{X/\nu _1} \sim F(n_2, n_1) \] □
Corollary 5.5.6. \[ P[F(n_1, n_2) \geq c] = P\left [ F(n_2, n_1) \leq \frac {1}{c} \right ] \]

Thus, the \((1-\alpha )\) significant points of \( F(n_2, n_1) \) distribution are the reciprocals of the \(\alpha \)-significant points of \( F(n_1, n_2) \) distribution.

Example 5.5.7. When \( \nu _1 = 2 \), show that the significance level of \( F \) corresponding to a significance probability \( p \) is: \[ F = \frac {\nu _2}{2} \left [ p^{-2/\nu _2} - 1 \right ] \]
Solution. When \( \nu _1 = 2 \), the p.d.f. of F-distribution becomes: \[ g(F) = \frac {(2/\nu _2)}{B(1, \nu _2/2)} \cdot \frac {1}{\left ( 1 + \frac {2F}{\nu _2} \right )^{(\nu _2+2)/2}} \] The probability that \( F \) exceeds a given value is: \[ p = \int _F^{\infty } g(F) dF \] Let \( t = 1 + \frac {2F}{\nu _2} \). Then: \[ p = \int _{1+\frac {2F}{\nu _2}}^{\infty } \frac {dt}{t^{(\nu _2+2)/2}} = \left ( 1 + \frac {2F}{\nu _2} \right )^{-\nu _2/2} \] Therefore: \[ p^{-2/\nu _2} = 1 + \frac {2F}{\nu _2} \] \[ F = \frac {\nu _2}{2} \left [ p^{-2/\nu _2} - 1 \right ] \]
Example 5.5.8. If \( F(n_1, n_2) \) represent an F-variate with \( n_1 \) and \( n_2 \) d.f., prove that \( F(n_2, n_1) \) is distributed as \( 1/F(n_1, n_2) \). Deduce that: \[ P[F(n_1, n_2) \geq c] = P\left [ F(n_2, n_1) \leq \frac {1}{c} \right ] \] Show how the probability points of \( F(n_2, n_1) \) can be obtained from those of \( F(n_1, n_2) \).
Solution. Let \( X \) and \( Y \) be independent chi-square variates with \( n_1 \) and \( n_2 \) d.f. respectively. Then: \[ F = \frac {X/n_1}{Y/n_2} \sim F(n_1, n_2) \] Therefore: \[ \frac {1}{F} = \frac {Y/n_2}{X/n_1} \sim F(n_2, n_1) \] Now: \[ P[F(n_1, n_2) \geq c] = P\left [ \frac {1}{F(n_1, n_2)} \leq \frac {1}{c} \right ] = P\left [ F(n_2, n_1) \leq \frac {1}{c} \right ] \] Let \( c \) be the upper \(\alpha \)-significant point of \( F(n_1, n_2) \). Then: \[ P[F(n_1, n_2) \geq c] = \alpha \] Therefore: \[ P\left [ F(n_2, n_1) \leq \frac {1}{c} \right ] = \alpha \] Thus: \[ P\left [ F(n_2, n_1) \geq \frac {1}{c} \right ] = 1 - \alpha \] Hence, the \((1-\alpha )\) significant points of \( F(n_2, n_1) \) are the reciprocals of the \(\alpha \)-significant points of \( F(n_1, n_2) \).
Example 5.5.9. Prove that if \( n_1 = n_2 \), the median of the F-distribution is at \( F = 1 \) and that the quartiles \( Q_1 \) and \( Q_3 \) satisfy \( Q_1 Q_3 = 1 \).
Solution. Let \( n_1 = n_2 = n \). The median \( M \) of \( F(n,n) \) is given by: \[ P[F(n,n) \leq M] = 0.5 \] Using \( 1/F(n,n) \sim F(n,n) \): \[ P\left [ F(n,n) \geq \frac {1}{M} \right ] = 0.5 \] Since the distribution is continuous: \[ P\left [ F(n,n) \leq \frac {1}{M} \right ] = 0.5 \] Thus: \[ M = \frac {1}{M} \Rightarrow M = 1 \] Similarly, for quartiles: \[ P[F(n,n) \leq Q_1] = 0.25 \] \[ P[F(n,n) \geq Q_3] = 0.25 \] Then: \[ P\left [ F(n,n) \leq \frac {1}{Q_3} \right ] = 0.25 \] Therefore: \[ Q_1 = \frac {1}{Q_3} \Rightarrow Q_1 Q_3 = 1 \]
Example 5.5.10. Let \( X_1, X_2, \dots , X_n \) be a random sample from \( N(0,1) \). Define: \[ \bar {X}_k = \frac {1}{k} \sum _{1}^{k} X_i, \quad \bar {X}_{n-k} = \frac {1}{n-k} \sum _{k+1}^{n} X_i \] Find the distribution of: (a) \( \frac {1}{2}(\bar {X}_k + \bar {X}_{n-k}) \) (b) \( k\bar {X}_k^2 + (n-k)\bar {X}_{n-k}^2 \) (c) \( X_1^2 / X_2^2 \) (d) \( X_1 / X_2 \)
Solution. (a) Since \( X_i \sim N(0,1) \), we have: \[ \bar {X}_k \sim N\left ( 0, \frac {1}{k} \right ), \quad \bar {X}_{n-k} \sim N\left ( 0, \frac {1}{n-k} \right ) \] Since the two samples are independent, \( \bar {X}_k \) and \( \bar {X}_{n-k} \) are independent. Therefore: \[ \frac {1}{2}(\bar {X}_k + \bar {X}_{n-k}) \sim N\left ( 0, \frac {1}{4k} + \frac {1}{4(n-k)} \right ) \] \[ = N\left ( 0, \frac {n}{4k(n-k)} \right ) \] (b) Since: \[ \sqrt {k}\bar {X}_k \sim N(0,1), \quad \sqrt {n-k}\bar {X}_{n-k} \sim N(0,1) \] \[ k\bar {X}_k^2 \sim \chi ^2_{(1)}, \quad (n-k)\bar {X}_{n-k}^2 \sim \chi ^2_{(1)} \] By the additive property of chi-square: \[ k\bar {X}_k^2 + (n-k)\bar {X}_{n-k}^2 \sim \chi ^2_{(2)} \] (c) Since \( X_1^2 \sim \chi ^2_{(1)} \) and \( X_2^2 \sim \chi ^2_{(1)} \) are independent: \[ \frac {X_1^2/1}{X_2^2/1} \sim F(1,1) \] (d) \( X_1 / X_2 \) is the ratio of two independent standard normal variates, which follows the standard Cauchy distribution.

5.5.5 Applications of F-Distribution

The F-distribution has the following applications in statistical theory:

1.
To test the equality of population variances (F-test for homogeneity of variances).
2.
In analysis of variance (ANOVA) to test the equality of several population means.
3.
In regression analysis to test the overall significance of the regression model.
4.
To test the significance of the multiple correlation coefficient.

5.5.6 F-Test for Equality of Population Variances

Suppose we want to test:

1.
Whether two independent samples have been drawn from normal populations with the same variance \( \sigma ^2 \).
2.
Whether two independent estimates of the population variance are homogeneous or not.

Under the null hypothesis \( H_0: \sigma _X^2 = \sigma _Y^2 = \sigma ^2 \), the statistic: \[ F = \frac {S_X^2}{S_Y^2} \] follows Snedecor’s F-distribution with \( (n_1-1, n_2-1) \) degrees of freedom. where: \[ S_X^2 = \frac {1}{n_1-1} \sum _{i=1}^{n_1} (x_i - \bar {x})^2, \quad S_Y^2 = \frac {1}{n_2-1} \sum _{j=1}^{n_2} (y_j - \bar {y})^2 \]

Remark 5.5.11.
1.
The greater of the two variances should be placed in the numerator.
2.
\( n_1 \) corresponds to the greater variance.
3.
The F-test assumes that the populations are normal.

5.5.7 Critical Values of F-Distribution

The available F-tables give the critical values for the right-tailed test. The significant value \( F_\alpha (n_1, n_2) \) at level of significance \( \alpha \) and \( (n_1, n_2) \) d.f. is determined by: \[ P[F > F_\alpha (n_1, n_2)] = \alpha \] The reciprocal relation between upper and lower significant points: \[ F_\alpha (n_1, n_2) = \frac {1}{F_{1-\alpha }(n_2, n_1)} \]

Example 5.5.12. Pumpkins were grown under two experimental conditions. Two random samples of 11 and 9 pumpkins show the sample standard deviations of their weights as 0.8 and 0.5 respectively. Assuming that the weight distributions are normal, test the hypothesis that the true variances are equal, against the alternative that they are not, at the 10[Assume that \( P(F_{10,8} \geq 3.35) = 0.05 \) and \( P(F_{8,10} \geq 3.07) = 0.05 \)]
Solution. Null Hypothesis: \( H_0: \sigma _X^2 = \sigma _Y^2 \) Alternative Hypothesis: \( H_1: \sigma _X^2 \neq \sigma _Y^2 \) (Two-tailed) Given: \[ n_1 = 11, \quad n_2 = 9, \quad s_X = 0.8, \quad s_Y = 0.5 \] \[ S_X^2 = \frac {n_1}{n_1-1} s_X^2 = \frac {11}{10} (0.8)^2 = 0.704 \] \[ S_Y^2 = \frac {n_2}{n_2-1} s_Y^2 = \frac {9}{8} (0.5)^2 = 0.28125 \] \[ F = \frac {0.704}{0.28125} = 2.5 \] For a two-tailed test at \( \alpha = 0.10 \), the critical values are: \[ F > F_{10,8}(0.05) = 3.35 \quad \text {and} \quad F < F_{10,8}(0.95) = 0.326 \] Since the calculated \( F = 2.5 \) lies between 0.326 and 3.35, it is not significant. Accept \( H_0 \). The population variances are equal.
Example 5.5.13. In one sample of 8 observations, the sum of the squares of deviations of the sample values from the sample mean was 84.4 and in the other sample of 10 observations it was 102.6. Test whether this difference is significant at 5
Solution. Given: \[ n_1 = 8, \quad n_2 = 10, \quad \sum (x-\bar {x})^2 = 84.4, \quad \sum (y-\bar {y})^2 = 102.6 \] \[ S_X^2 = \frac {84.4}{7} = 12.057 \] \[ S_Y^2 = \frac {102.6}{9} = 11.4 \] Null Hypothesis: \( H_0: \sigma _X^2 = \sigma _Y^2 \) Test Statistic: \[ F = \frac {12.057}{11.4} = 1.057 \] Tabulated \( F_{0.05} \) for (7,9) d.f. is 3.29. Since \( F < 3.29 \), accept \( H_0 \). The variances are not significantly different.
Example 5.5.14. Two random samples gave the following results:
SampleSizeSample MeanSum of squares of deviations from the mean
1 10 15 90
2 12 14 108
Test whether the samples come from the same normal population at 5[Given: \( F_{0.05}(9,11) = 2.90, F_{0.05}(11,9) = 3.10 \) (approx.)]
Solution. A normal population has two parameters: mean \( \mu \) and variance \( \sigma ^2 \). To test if the samples come from the same normal population, we must test:
1.
Equality of population means
2.
Equality of population variances
First, test equality of variances using F-test: \[ S_1^2 = \frac {90}{9} = 10, \quad S_2^2 = \frac {108}{11} = 9.82 \] \[ F = \frac {10}{9.82} = 1.018 \] Tabulated \( F_{0.05}(9,11) = 2.90 \). Since \( F < 2.90 \), accept equality of variances. Now, test equality of means using t-test: \[ S^2 = \frac {90 + 108}{10 + 12 - 2} = \frac {198}{20} = 9.9 \] \[ t = \frac {15 - 14}{\sqrt {9.9 \left ( \frac {1}{10} + \frac {1}{12} \right )}} = \frac {1}{\sqrt {1.815}} = 0.742 \] Tabulated \( t_{0.05} \) for 20 d.f. is 2.086. Since \( |t| < 2.086 \), accept equality of means. Since both hypotheses are accepted, the samples come from the same normal population.

5.5.8 Summary of Key Results for F-Distribution

Property Result
Probability Density Function \( f(F) = \dfrac {(\nu _1/\nu _2)^{\nu _1/2}}{B(\nu _1/2, \nu _2/2)} \cdot \dfrac {F^{(\nu _1/2)-1}}{\left [ 1 + \frac {\nu _1}{\nu _2} F \right ]^{(\nu _1+\nu _2)/2}}, \quad 0 \leq F < \infty \)
Mean \( \dfrac {\nu _2}{\nu _2 - 2}, \quad \nu _2 > 2 \)
Variance \( \dfrac {2\nu _2^2(\nu _1 + \nu _2 - 2)}{\nu _1(\nu _2 - 2)^2(\nu _2 - 4)}, \quad \nu _2 > 4 \)
Mode \( \dfrac {\nu _2(\nu _1 - 2)}{\nu _1(\nu _2 + 2)}, \quad \nu _1 > 2 \)
Reciprocal Property \( \dfrac {1}{F} \sim F(\nu _2, \nu _1) \)
Test Statistic (Variance Ratio) \( F = \dfrac {S_1^2}{S_2^2} \sim F(n_1-1, n_2-1) \)
Critical Values \( F_\alpha (\nu _1, \nu _2) = \dfrac {1}{F_{1-\alpha }(\nu _2, \nu _1)} \)

5.5.9 Important Remarks

Remark 5.5.15.
1.
The F-test is used to test the equality of variances from two normal populations.
2.
The greater variance should always be placed in the numerator.
3.
The F-distribution is positively skewed.
4.
The F-test is a one-tailed test for testing equality of variances.
5.
For a two-tailed test, use \( F > F_{\alpha /2} \) and \( F < F_{1-\alpha /2} \).
6.
Before applying the t-test for difference of means, the F-test should be used to test the equality of variances.

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