BUSINESS ECONOMICS

BUSINESS ECONOMICS

PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

Subject BUSINESS ECONOMICS

Paper No and Title 2, Applied Business Statistics

Module No and Title 14, Sampling Distribution: F-distribution

Module Tag BSE_P2_M14

BUSINESS ECONOMICS

PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

TABLE OF CONTENTS

1. Learning Outcomes 2. Introduction

3. F-Distribution

3.1 Construction of F Random Variable 3.2 Properties of the F distribution

3.3 Critical values under the F-Density curve 4. Applications of the F distribution

4.1 F-test for Equality of Variances

4.1.1 Significance of F-test for Equality of Variances 4.2 Confidence Interval Estimation for

𝟐 𝛔_𝟐^𝟐

4.3 Caution regarding Normality Assumption 5 Summary

6 Appendix

BUSINESS ECONOMICS

PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION 1. Learning Outcome

Often, it is essential to shed light on the difference in the variability of two population distributions. This is one of the primary reasons, you shall gain knowledge about the F- distribution in this module.

• You shall learn the construction of F random variable, its parameters and how, for a given F value, the probability under the F density curve is determined.

• You shall also know its two important applications; Hypothesis testing for variability of two population distributions and Confidence interval estimates of the ratio of two population variances.

• You shall also understand the significance of the F test. On the basis of the outcome of this test, regarding the equality of the variances of the two population distributions, the decision regarding the suitability of the ‘two-sample t procedure’ or ‘pooled t procedure’ for hypothesis testing, concerning the difference between the two sample means, is taken.

2. Introduction

In the earlier modules, we learned how the analysis concerning confidence interval and hypothesis testing for single mean and single proportion is extended to two sample means and proportions. Apart from obtaining information about the difference between two population means and proportions, it is often the case that investigator also needs to compare the variability (variances or standard deviations) of the two population distributions Sometimes engineers are required to compare inherent variability in the two or more processes or scientists need to explore whether the ferritin distribution in the younger adults had a larger variability than in the elderly population. For making such comparisons, we introduce a new family of distributions, named, the F-distribution. It is also extensively applied in comparing variability of more than two samples.

3. F- Distribution

In addition to the t-distribution, the F distribution also plays an important role in connection with sampling from a population that is normally distributed. While doing confidence interval estimation and hypothesis testing that involve two different population distributions, the investigator necessarily needs to know whether the two population distributions have equal spread or not. Because depending upon the variability of the two distributions, the investigator decides whether to use ‘pooled variance’ or ‘two-sample procedure’ for making inferences related to two sample means. If the two distributions have equal variances, then a common (also called pooled) variance must be estimated from the sample data before proceeding with final analysis. It is the F-distribution that concerns testing of variability of the two independent population distributions around their respective means.

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PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

3.1 Construction of F Random Variable

Suppose X1 and X2 are two independent random variables drawn from two normal populations, with respective sizes n1 and n2, having chi-squared distributions with ν1 (= n1−1) and ν2 (= n2−1) degrees of freedom (df). Then the F random variable is defined as a ratio of the two chi-squared variables divided by their respective degrees of freedom, i.e.

F = ^{𝑋1 / 𝜈1}

𝑋_{2 / 𝜈2} (1) Here, we know, X₁= 𝜒₁ ² =^{(𝑛1−1)𝑆}_𝜎 ¹²

12 andX₂= 𝜒₂ ² =^{(𝑛2−1)𝑆}_𝜎 ²²

22 . Substituting the values of X1 and X2 in equation (1) we obtain, F = ^{𝑆12⁄𝜎12}

𝑆₂²⁄𝜎₂² =^𝑆12𝜎22

𝑆₂²𝜎₁² (2)

The random variable so constructed has an F distribution with ν1 (= n1−1) and ν2 (= n2−1) degrees of freedom. On account of its formula containing ratios of two variances, it is also called variance-ratio distribution. Its complicated density function will not be used in our analysis, so we have avoided writing here.

3.2 Properties of the F distribution

 The F distribution has two parameters, ν1 and ν2. The parameter ν1 (= n1−1) is called the number of degrees of freedom of the numerator andν2 (= n2−1) is the number of degrees of freedom of the denominator. Both ν1 and ν2 are positive integers. The curves of a typical F- distribution are shown in Figure 1. The F distribution is said to be completely identified if ν1

and ν2 are known.

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

 The F random variable cannot assume a negative value. Since it is a ratio of two chi-squared variables and a chi-squared variable itself, by definition, is a ratio of only positive numbers.

Therefore, F random variable is always a positive number. Unlike the t- and Z random variables that range from -∞ to +∞, the F variable starts with minimum value of zero on the horizontal axis and it extends to any positive number.

 The density curve of F-distribution is not symmetric in nature, rather it is skewed to the right (Figure 1). As a result, unlike t and Z random variables, both the upper and lower tail areas must be calculated separately for F-distribution.

3.3 Critical values under the F-Density curve

Considering its importance, F-distribution has been extensively tabulated. Similar to the notation tα,ν for t-distribution, we use 𝐹_{𝛼,𝜈1,𝜈2}for the value of F random variable, on the horizontal axis that captures α area to the right side of the F density curve with ν₁ and ν₂df for the numerator and denominator respectively. That is P (F ≥𝐹_{𝛼,𝜈1,𝜈2}) = 𝛼. The Table A.4, titled ‘Critical Values for F Distribution’ tabulates 𝐹_{𝛼,𝜈1,𝜈2} for four values of 𝛼 =.10, .05, .01, and .001 and various values of ν₁and ν₂. The df of numerator is tabulated in various columns and df of denominator is shown in various groups of rows in the Table. Corresponding to each value of the df of the denominator, there is a group of four rows exhibiting four values of 𝛼 mentioned above. So for each combination of𝜈1and 𝜈2 there are four critical values of 𝐹𝛼,𝜈₁,𝜈₂ that capture .10, .05, .01, and .001 area under the F curve in the upper tail (Figure 2). For example, to find value Fα,6,10, we look at the intersection of column corresponding to 6 df of numerator and a group of 4 rows corresponding to 10 df of the denominator. This cell contains four critical F values, namely, 2.46, 3.22, 5.39, and 9.93 corresponding to four values of 𝛼 = 0.10, 0.05, 0.01, and 0.001 respectively.

This can be re-written as,

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PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

F.10,6,10= 2.46, F.05,6,10 = 3.22, F.01,6,10 = 5.39, and F.001,6,10 = 9.93 To interpret F.10,6,10= 2.46, for an F-distribution with 𝜈₁= 6 and 𝜈₂= 10, area under the density curve to the right of value 2.46 on the horizontal axis is 10% .

Now, how to find F value with 0.10 of area on its left tail? Since the distribution is not symmetric in nature, does this mean that the left-tailed critical values must also be tabulated? However, this is not required because of the relationship between 𝐹_{𝛼,𝜈1,𝜈2} and 𝐹_{(1−𝛼),𝜈1,𝜈2} given as

𝐹_{(1−𝛼),𝜈1,𝜈2} = ¹

𝐹_{𝛼,𝜈2,𝜈1} (3)

This can be explained as follows; the critical value F.10,6,10 gives 10% area to the right which implies 90% is to the right. Likewise, F.90,6,10 gives 90% area to the right, and therefore 10% area to the left is captured for F-distribution with 𝜈₁and 𝜈₂ df for numerator and denominator respectively. We know from equation (3), F.90,6,10=F(1-.10), 6,10= 1

F_.10,10,6

⁄ . From the Table we see that four values lie at the intersection of the column corresponding to 10 numerator df and 6 denominator df. And further, within this group of 4 values, the first value = 2.94 corresponds to the 10% area to the right.

Therefore, F_.90,6,10= 1

F_.10,10,6

⁄ = 1 2.94⁄ = .340. This implies on the F density curve with 6 numerator and 10 denominator df, 10% area lies to the left of the F value = 0.340.

Similarly, the critical value F.95,6,10 captures 5% area to the left tail of the F-distribution.

F_.95,6,10 = F(1-.05), 6,10= 1

F_.05,10,6

⁄ = 1 4.06⁄ = 0.246

To generalize, in order to obtain the critical value F_𝜈1,𝜈2 with α area in the left tail [meaning (1-α) area in the right tail], we need to compute reciprocal of F_{𝛼,𝜈2,𝜈1}from the F table i.e. the same α area in the F density curve with order of degrees of freedom being reversed.

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

Example 1.

Compute the following F values: (a) F_.05,6,10 (b) F_.99,8,12 (c) F_.95,5,12(d) F_.10,10,6

Solution:

To obtain these probabilities we use Table A.4

(a) F_.05,6,10= 3.22 (intersection of column ν1 =6 and group of rows ν2 =10, second value in the cell)

(b) F_.99,8,12= 1

F_.01,12,8

⁄ = 1 5.67⁄ = .1764 (intersection of column ν1 =12 and group of rows ν2

=8, reciprocal of the third value in the cell) (c) F_.95,5,12= 1

F_.05,12,5

⁄ = 1 4.68⁄ = .214 (intersection of column ν1 =12 and group of rows ν2

=5, reciprocal of the second value in the cell)

(d) F_.10,10,6 = 2.94 (intersection of column ν1 =10 and group of rows ν2 = 6, first value in the cell)

Example 2.

Compute the following probabilities:

(a) P (F ≤ 3.33) for ν1 = 5 and ν2 =10 (b) P (F ≤ 4.33) for ν1 = 5 and ν2 =10 (c) P (.253 ≤ F ≤ 7.01) for ν1 = 4 and ν2 = 8

Solution:

Using Table A.4 we obtain these probabilities

(a) For this we look at the four values given in cell that corresponds to the intersection of column ν1 =5 and group of rows ν2 =10. The second value is 3.33 and this corresponds to α=.05 area in the upper tail. Thus for ν1 = 5 and ν2 =10, P (F ≤ 3.33) = .95

(b) Since ν1 = 5 and ν2 =10 are the same as in part (a), so we look at the same four values. We can see value 4.33 does not appear anywhere. Rather it lies between the second and third values which are 3.33 (for α=.05) and 5.64 (for α=.01) respectively. Thus we cannot find the exact probability however, we can say that P (F ≤ 4.33) for ν1 = 5 and ν2 =10 would lie between .95 and .99.

(d) P (.253 ≤ F ≤ 7.01) = P (F ≤ 7.01) − P (F ≤ 0.253). From the Table we find that with combination ν1 = 4 and ν2 =8, the F=7.01 appears corresponding to α=.01 area in the upper tail. This ⇒ P (F≤ 7.01) = 0.99. To find the area on the left hand side, we need to find reciprocal of the four values that appear corresponding to the combination ν1 = 8 and ν2 = 4 (when order of the df is reversed). The reciprocal of the four values are 0.253 (=

1⁄3.95), 0.166(= 1 6.04⁄ ), 0.068(= 1 14.80⁄ )and 0.0204 (= 1 49.00⁄ ) that correspond to respectively 10%, 5%, 1% and 0.1% area in the left tail. Thus P (F ≤.253) = 0.10 (Figure 3).

Hence, P (0.253 ≤ F ≤ 7.01) = P (F ≤ 7.01) − P (F ≤ 0.253) = 0.99 − 0.10= 0.89.

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PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

Percentiles of the F-Distribution

As the Table A.4, gives information on four 𝛼 areas to the right side of the F density curve for a given values of 𝜈₁ and 𝜈₂, we can find the following percentiles only; 10^th and 90^th, 5^th and 95^th, 1^st and 99^th, and .1^st and 99.9^th.

Example 3.

Compute the following:

(a) The 95^th percentile of the F distribution with ν1 =10and ν2 =15 (b) The 1^st percentile of the F distribution with ν1 =11and ν2 =20

Solution:

To obtain these quantities we use Table A.4

(a) The 95^th percentile is a value of F above which 5% of the values fall and below which 95% of the observations lie, i.e. we require 5% area on the upper tail. F_.05,10,15= 2.54

(b) The 1^st percentile of the F distribution with ν1 =11and ν2 =20 is a value of F above which 99% of the values fall and below which 1% of the observation lie, i.e. we require 99% area on the upper tail. F_.99,11,20= 1

F_.01,20,11

⁄ = 1 4.10⁄ = 0.244

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION 4. Applications of F-distribution

The two most important applications of F distribution are related to computing interval estimates for the ratio of population variances (^σ1

2

σ₂²) and comparing variability of two samples drawn independently from the two different normally distributed populations. The applications are covered in the other modules, however for completeness, we provide two examples to briefly explain the use of F-distribution.

4.1 F-test for Equality of Variances

𝑆₂²⁄𝜎₂² =^𝑆12𝜎22

𝑆₂²𝜎₁²is a random variable having F distribution with ν1

(= n1−1) and ν2 (= n2−1)degrees of freedom for the numerator and denominator respectively.

Here, the hypothesis is to check whether the two populations from which the samples are drawn have equal variances or not i.e., the null and alternative hypothesis are: H0: σ₁²= σ₂² ; and Ha:σ₁²≠ σ₂² . Since the F random variable involves a ratio of variances, both the hypotheses can be expressed as a ratio of the population variances; that is H0 : ^𝛔𝟏

𝟐

𝛔_𝟐^𝟐 = 𝟏; Ha = ^𝛔𝟏

𝟐

𝛔_𝟐^𝟐 ≠ 𝟏.

Accordingly, the F-test statistic becomes = ^𝑆12

𝑆₂² , with ν1 (= n1−1) and ν2 (= n2−1) degrees of freedom for numerator and denominator respectively. Further, since the alternative hypothesis involves sign of not equal to one, the two-tailed rejection region is used. Assuming level of significance to be α, then α/2 area on both the sides of the distribution would be included in the rejection region. i.e., if calculated F value is either ≥ 𝐹_{𝛼 2⁄ ,𝜈1,𝜈2} or it is ≤ 𝐹_{(1−𝛼 2)}⁄ ,𝜈₁,𝜈₂; we reject the null in favour of the alternative hypothesis.

Example 4:

A lecturer in the department of Economics asserts that there is no difference in the variability in the marks of students studying Economics as a major subject (group 1) and in the marks of students who studies Economics as an inter-disciplinary subject (group 2). To test his assertion, two random samples from both the groups were taken with the following results:

Sample size of group1 = n1= 16 SD of group1 = 𝑆₁² = 260 Sample size of group2 = n2 = 21 SD of group2 = 𝑆₂² = 300 Test the assertion of the lecturer at 10% significance level

Solution:

To test the claim we form the following hypothesis of no difference in the variability or significant difference in variability in marks of the two groups of students, i.e.:

H0: ^σ1

2

σ₂² = 1; Ha : ^σ1

2

σ₂² ≠ 1.

Now the F-test statistic is: F = ^𝑆12

𝑆₂² = ²⁶⁰

300 = 0.87

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Given the alternative hypothesis, it is a two-tailed test. To find the rejection region we require the two critical values of F density curve with numerator df =15 (16-1) and denominator df =20 (21- 1), with 5% (α/2) of area in both the tails, i.e. we require, F_.05,15,20 and F_.95,15,20.

Using Table A.4, we find F_.05,15,20 = 2.20 and F_.95,15,20= 1/F_.05,20,15= 1/2.33 = 0.45. The non- rejection region consists of values of F statistic, lying between 0.45 and 2.20 (Figure 4).

Since, the calculated F lies in the non-rejection region, (0.45 ≤ F=.87 ≤ 2.20), the null hypothesis in not rejected. Hence, on the basis of the given data, we do not negate the claim of the lecturer that there is no significant difference in the variability in the marks of students studying Economics as a major subject (group 1) and in the marks of students who studies Economics as an inter-disciplinary subject (group 2).

We must understand that if the alternative hypothesis had been Ha = ^σ1

2

σ₂²> 1 or Ha = ^σ1

2

σ₂²< 1. We would have used one-tailed hypothesis, with rejection region consisting of only right-tailed or left-tailed area (For details, refer the module on ‘Two sample inferences’).

4.1.1

Significance of F-test for Equality of Variances

The problems involving population variances arise much less frequently, however F-test for equality of variances is important whenever we are concerned with inferences regarding difference between means of two different population distributions i.e. (µ1 -µ2).

While making inferences regarding (µ1 -µ2), using sample information, t-test procedure, is used under the following conditions;

 when the populations from which the two sample are drawn, are known to be normal,

 both or one of the two samples is small, and

 population variances are unknown

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

The t-statistic used is =^{𝑋̅−𝑌̅}

𝑆𝐸_{𝑋̅−𝑌̅}= ^{𝑋̅−𝑌̅}

√^𝑆12

𝑛1+ ^𝑆22

𝑛2

(4)

Where n1 and n2 are sizes of the two random samples X and Y. Apart from the above assumptions, this formula is also based on an important assumption that the two population distributions have unequal variances. Here, since the two sample variances are separately used, this procedure is also called the two-sample t procedure or the separate variance t test.

Before carrying out the above t-test, we need to test whether this assumption is valid or not for the populations under observation. For this, we conduct the F-test. If the null hypothesis, H0: σ₁²=σ₂² is rejected in favour of the alternative hypothesis, Ha:σ₁²≠σ₂², we continue to use the above formula. However, if the null is not rejected, means the two populations have equal variance; we use the pooled t procedure. In this procedure, a common variance must be estimated from the information based on both the samples, denoted as 𝑆_𝑝². Keeping in view the difference in the two sample sizes, a weighted average of the two sample variances is used, with weights being the respective proportions of the degrees of freedom of each sample, in the total degrees of freedom ( n1 + n2 −2) when the two samples are combined. We know, the first sample contributes (n₁− 1) df and the second sample contributes (n₂− 1) df , in the total degrees of freedom = n₁+ n₂− 2, when both the samples are combined.

Therefore, S_p² = ⁿ¹⁻¹

n₁+n₂−2S₁²+ ⁿ²⁻¹

n₁+n₂−2S₂² In the t-statistic formula in equation (4), S₁²and S₂ ² are replaced by S_p².

Therefore the pooled t statistic is = ^{X ̅ −Y̅}

√^Sp2

n1 + ^Sp

2 n2

= ^{X ̅ −Y̅}

√S_p²(¹

n1+ ¹

n2)

= ^{X ̅ −Y̅}

S_p√(¹

n1+ ¹

n2)

(5)

Recent research has shown that when σ₁²=σ₂², the use of ‘pooled t test’ does perform better than the ‘two sample t-test’. However, it can give incorrect results when applied to the case where the variances of the two populations are unequal ( σ₁²≠σ₂²). As a result, statistician suggest the use of the two-sample t procedure, until there are convincing evidence that variance are equal.

4.2 Confidence Interval Estimation for

𝟐 𝛔_𝟐^𝟐

Another important application of F-distribution is to compute confidence interval estimates for the ratio of the two population variances (^𝛔𝟏

𝟐

𝛔_𝟐^𝟐). From the above discussion we know;

P (𝐹_{(1−𝛼 2)⁄ ,𝜈1,𝜈2}≤ 𝐹 ≤ 𝐹_{𝛼 2⁄ ,𝜈1,𝜈2} = 1− α (See Figure 5)

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

Replacing F with ^𝑆1

2𝜎₂²

𝑆₂²𝜎₁² from equation (2) and manipulating the terms such that the ratio(^𝜎22

𝜎₁²) remains in the middle we obtain,

P(^𝑆22

𝑆₁²𝐹_{(1−𝛼 2)⁄ ,𝜈1,𝜈2}≤^𝜎22

𝜎₁²≤^𝑆22

𝑆₁²𝐹_{𝛼 2⁄ ,𝜈1,𝜈2}) = 1− α Further, P (^𝑆_𝑆²²

12

1

𝐹_{𝛼 2⁄ ,𝜈2,𝜈1} ≤^𝜎_𝜎²²

12≤^𝑆_𝑆²²

12𝐹𝛼 2⁄ ,𝜈₁,𝜈₂) = 1− α [using equation (3)] (6) Since, we need interval estimate for the ratio(^𝜎_𝜎²²

12), inverting each term in equation (6) we obtain P (^𝑆12

𝑆₂² 1

𝐹_{𝛼 2⁄ ,𝜈1,𝜈2} ≤^𝜎12

𝜎₂²≤^𝑆12

𝑆₂²𝐹_{𝛼 2,⁄ 𝜈2,𝜈1}) = 1− α

Therefore, if two samples of sizes n1 and n2 are drawn from the two normal populations, then

𝑆₁² 𝑆₂²

1

𝐹_{𝛼 2⁄ ,𝜈1,𝜈2} and ^𝑆1

2

𝑆₂² 𝐹_{𝛼 2,⁄ 𝜈2,𝜈1} give the respective lower and upper limits for 100(1−α) % confidence interval for the ratio of population variances (^σ_σ¹²

22). An interval estimate for the ratio of the population standard deviations (^σ1

σ₂) is obtained by taking the square root of each limit.

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

Example 5:

On the basis of the given information, obtain 98% confidence interval estimate for both the ratio of population variances(^σ_σ¹²

22) and standard deviations(^σ1

σ₂).

n1 = 26 n2 = 16 𝑆₁² = 4 𝑆₂² = 2.90

Solution:

For a 98% confidence interval, 1−α =.98 ⇒ α= .02 ⇒ 𝛼 2⁄ = .01. And 𝜈₁= 25 and𝜈₂ =15.

Further, the two respective F values, giving cut-off areas of 1% on the left and right tails, are 𝐹_.01,25,15 and 𝐹_.99,25,15.

We know 𝐹_.99,25,15= ¹

𝐹_.01,15.25 = ¹

2.85 = 0.351 and 𝐹_.01,25,15 = 3.28 (using Table A.4) Therefore, the 98% confidence interval estimate for (^σ_σ¹²

22) is [_2.90⁴ (0.351) ≤^σ_σ¹²

22 ≤ _2.90⁴ 3.28 ].

Solving this we obtain 0.484 and 4.52, as the lower and upper limits respectively within which the ratio (^σ_σ¹²

22)is expected to lie with 98% confidence. Further, taking square root of both the limits we obtain 0.696 and 2.13, as the two limits of the confidence interval for the ratio (^σ1

σ₂).

4.3 Caution regarding Normality Assumption

For application of F-distribution, we assume that both the population distributions are normally distributed. Here, it must be kept in mind that the F-test is very sensitive to the assumption of normality. Therefore even if there is a mild deviation from normality for either of the two population distributions, we should avoid using F-distribution, and in that case a non-parametric approach should be followed.

The t-test mentioned in the earlier module is also based on the assumption of normal population.

However, sensitivity of the F-test is much more than that of the t-distribution.

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION 5. Summary

 In this module we have learned about the F-distribution which finds extensive application in comparing variability of two sample variances and determining confidence interval estimates of the ratios of the two population variances.

 The calculation of the F test statistic is based on the ratio of two sample variances.

Accordingly, the degrees of freedom of the numerator and of denominator are the two parameters of the F-distribution.

 The F-distribution is not a symmetric curve, therefore both the left- and right-tailed critical values are to be calculated separately for inferential statistics.

 One of the main reason for the testing the difference between the two population variances is to find out the appropriateness of the ‘two-sample t procedure’ or ‘pooled t procedure’ for testing of hypothesis for the difference between the two sample means.

 Similar to the t-distribution, the F-distribution also concerns with two independent populations that are normally distributed. Therefore, before using F-distribution it must be ensured that both the sampled population distributions are independent and are normally distributed.

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

6. Appendix

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PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

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PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

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PAPER No. : 2, APPLIED BUSINESS STATISTICS

MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION

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MODULE No. : 14, SAMPLING DISTRIBUTION: F-DISTRIBUTION