TANSCHE SMTPT31 Mathematical Statistics Unit 4

Mathematical Statistics
(TANSCHE Syllabus)

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

ARO Study Circle

Contents

4 Sampling Distribution and Test of Significance
4.1 Sampling - Introduction
4.1.1 What is Population?
4.1.2 Why Do We Need Sampling?
4.2 Types of Sampling
4.2.1 Purposive Sampling (or Judgment Sampling)
4.2.2 Random Sampling
4.2.3 Simple Sampling
4.2.4 Stratified Sampling
4.2.5 Summary of Sampling Types
4.3 Parameter and Statistic
4.3.1 Sampling Distribution of a Statistic
4.3.2 Standard Error
4.3.3 Standard Errors for Common Statistics
4.3.4 Why is Standard Error Useful?
4.4 Tests of Significance
4.4.1 Why Large Samples Use Normal Test
4.5 Null Hypothesis
4.5.1 Alternative Hypothesis
4.5.2 One-Tailed and Two-Tailed Tests
4.6 Errors in Sampling
4.6.1 Type I Error - Producer’s Risk
4.6.2 Type II Error - Consumer’s Risk
4.6.3 Summary of Errors
4.7 Critical Region and Level of Significance
4.7.1 Level of Significance
4.7.2 One-Tailed and Two-Tailed Tests
4.7.3 Critical Values (Significant Values)
4.7.4 Critical Values from Normal Distribution
4.7.5 Table of Critical Values of \(Z\)
4.7.6 Decision Rule for Hypothesis Testing
4.7.7 Summary of Steps in Hypothesis Testing
4.8 Test of Significance for Large Samples
4.8.1 Area Properties of Normal Distribution
4.8.2 Decision Rules for Large Sample Tests
4.9 Sampling of Attributes
4.9.1 Binomial Distribution and Sampling of Attributes
4.9.2 Test for Single Proportion
4.9.3 Properties of Sample Proportion
4.9.4 Test for Difference of Proportions
4.9.5 Difference of Proportions when Population Proportions are Known
4.9.6 Summary of Tests for Proportions
4.10 Sampling of Variables
4.11 Unbiased Estimate for Population Mean and Variance
4.12 Standard Error of Sample Mean
4.13 Test of Significance for Single Mean
4.13.1 Confidence Limits for Population Mean
4.14 Test of Significance for Difference of Means
4.15 Test of Significance for Difference of Standard Deviations

Chapter 4
Sampling Distribution and Test of Significance

Syllabus

Sampling - Tests of significance - Null Hypothesis - Tests of significance for large samples.
Chapter 12

4.1 Sampling - Introduction

4.1.1 What is Population?

In statistics, when we study something, we are usually interested in a particular group of people or things. For example, we might want to know the average income of people in a city, or the average height of students in a college. This entire group that we want to study is called a population or universe. A population can be:

• Finite: Like all students in a particular college (we can count them).
• Infinite: Like all possible outcomes when tossing a coin (we cannot count them).

4.1.2 Why Do We Need Sampling?

If we want to know something about a population, the best way would be to examine every single member. This is called a complete enumeration or census. But in real life, this is often impossible or impractical. Here’s why:

1.

Time and Cost: It takes a lot of time and money to check everyone. For example, finding the average income of all people in India would be extremely expensive.

2.

Destructive Testing: Sometimes, testing destroys the item. For example, if we want to test how strong a firecracker is, we have to burst it. We cannot test all firecrackers!

3.

Infinite Population: If the population is infinite, we simply cannot study everyone.

Because of these problems, we use a sample.

Definition 4.1.1 (Sample). A sample is a finite subset of the population that we actually study. The number of items in the sample is called the sample size.

Definition 4.1.2 (Sampling). The process of selecting a sample from a population is called sampling.

By studying the sample, we try to guess or estimate the characteristics of the whole population. The difference between the true population value and our estimate from the sample is called the sampling error. This error is always present in any sampling method.

Example 4.1.3. When you go to buy sugar from a shop, you don’t check every grain of sugar in the bag. Instead, you take a small amount (a sample) in your hand, check its quality, and then decide whether to buy the whole bag.

4.2 Types of Sampling

Some of the commonly known and frequently used types of sampling are:

1.

Purposive sampling,

2.

Random sampling,

3.

Stratified sampling,

4.

Systematic Sampling.

4.2.1 Purposive Sampling (or Judgment Sampling)

In this method, the researcher selects the sample based on his or her own judgment or a specific purpose.

Example 4.2.1. If someone wants to show that the standard of living in Delhi has increased, they might only visit rich areas like South Delhi and ignore the poorer areas. This sample is biased and does not represent the whole city.

Remark 4.2.2. The main problem with purposive sampling is bias. The researcher might intentionally or unintentionally choose a sample that supports their own view, making the results unreliable.

4.2.2 Random Sampling

To avoid the bias of purposive sampling, we use random sampling. In this method, every member of the population has an equal chance of being selected in the sample. Suppose a population has \(N\) items and we want to select a sample of \(n\) items. The total number of possible samples is \(^N C_n\). In random sampling, each of these \(^N C_n\) possible samples has an equal chance of being selected.

Example 4.2.3 (Lottery Method). Suppose we have 100 students and we want to select 10 of them randomly. We write the numbers 1 to 100 on 100 identical slips of paper, put them in a box, mix them well, and then draw 10 slips without looking. The students whose numbers are drawn will form our random sample.

Remark 4.2.4. To ensure randomness, we can also use:

• Tippet’s Random Number Tables: These are tables of random digits. We can pick any page and any row or column and use the numbers to select our sample.
• Throwing a dice or a lottery system.

4.2.3 Simple Sampling

Simple sampling is a special type of random sampling. It has two important conditions:

1.

Every unit in the population has the same chance of being selected.

2.

The selection of one unit does not affect the selection of another unit. In other words, each draw is independent.

For example, if we toss a coin, the outcome of one toss does not affect the outcome of another toss. This is independent.

Remark 4.2.5. Random sampling does NOT always mean simple sampling.

Example 4.2.6. Suppose a bag has 5 white balls and 5 black balls. If we draw one ball and do NOT put it back, the probability of drawing a white ball in the second draw changes. The draws are not independent. So, even though every ball has an equal chance in each draw, the sampling is random but not simple.

To make it a simple sample, we must put the ball back (sampling with replacement) so that the conditions remain the same for each draw.

4.2.4 Stratified Sampling

Sometimes, a population is not uniform. For example, a city might have rich, middle-class, and poor areas. If we take a random sample, we might accidentally miss one group. In stratified sampling, we first divide the population into smaller, more uniform groups called strata (singular: stratum). These groups are based on some characteristic (like income, age, gender, etc.). Then, we take a random sample from each stratum. The sample size from each stratum is usually proportional to its size in the population.

Example 4.2.7. To find the average income in a city, we might divide the city into:

• Stratum 1: Rich areas
• Stratum 2: Middle-class areas
• Stratum 3: Poor areas

Then we take a random sample from each area. This gives a more accurate picture of the whole city.

Remark 4.2.8. Stratified sampling is considered one of the best sampling methods because it ensures that all parts of the population are represented, giving a more reliable estimate.

4.2.5 Summary of Sampling Types

Type of Sampling	How it works	Main Problem
Purposive	Researcher chooses based on judgment	High chance of bias
Random	Every unit has equal chance	Can be difficult to do properly
Simple	Random sampling with independent draws	Requires sampling with replacement for finite populations
Stratified	Population divided into groups, then random samples from each	More complex to organize

4.3 Parameter and Statistic

In statistics, we use two important terms that are often confused: parameter and statistic. Let us understand them clearly.

Definition 4.3.1 (Parameter). A parameter is a numerical value that describes some characteristic of a population. For example, the population mean \((\mu )\) and population variance \((\sigma ^2)\) are parameters. These values are fixed (but usually unknown to us).

Definition 4.3.2 (Statistic). A statistic is a numerical value computed from a sample. For example, the sample mean \((\bar {x})\) and sample variance \((s^2)\) are statistics. Since they are based on samples, their values change from one sample to another.

Remark 4.3.3. We use the following notation:

• Population mean = \(\mu \)
• Population variance = \(\sigma ^2\)
• Sample mean = \(\bar {x}\)
• Sample variance = \(s^2\)

Definition 4.3.4 (Unbiased Estimate). A statistic \(t = t(x_1, x_2, \dots , x_n)\) is an unbiased estimate of the population parameter \(\theta \) if: \[ E(t) = \theta \] In words, "the average value of the statistic (over many samples) equals the true population parameter."

Example 4.3.5. The sample mean \(\bar {x}\) is an unbiased estimate of the population mean \(\mu \) because: \[ E(\bar {x}) = \mu \] This means that if we take many samples and calculate their means, the average of all these sample means will be equal to the true population mean.

4.3.1 Sampling Distribution of a Statistic

When we draw samples from a population, each sample gives us a different value of the statistic. The collection of all possible values of a statistic (from all possible samples of a given size) forms its sampling distribution. Suppose we have a finite population of size \(N\). The total number of possible samples of size \(n\) is: \[ ^N C_n = \frac {N!}{n!(N-n)!} = k \text { (say)} \] The mean of this sampling distribution is: \[ \bar {t} = \frac {1}{k} \sum _{i=1}^{k} t_i \] The variance of this sampling distribution is: \[ \operatorname {Var}(t) = \frac {1}{k} \sum _{i=1}^{k} (t_i - \bar {t})^2 \]

4.3.2 Standard Error

The standard error (S.E.) of a statistic is simply the standard deviation of its sampling distribution.

Definition 4.3.6 (Standard Error). The standard deviation of the sampling distribution of a statistic is called its Standard Error.

Remark 4.3.7. Standard error is very important in statistics because:

1.

It tells us how much a statistic varies from sample to sample.

2.

A smaller standard error means the statistic is more precise.

3.

It helps us test hypotheses and construct confidence intervals.

4.3.3 Standard Errors for Common Statistics

For large samples, the standard errors of some common statistics are given below. Here:

• \(n\) = sample size
• \(\sigma ^2\) = population variance
• \(P\) = population proportion
• \(Q = 1 - P\)
• \(n_1, n_2\) = sizes of two independent samples

Statistic	Standard Error
Sample mean \(\bar {x}\)	\(\dfrac {\sigma }{\sqrt {n}}\)
Sample proportion \(p\)	\(\sqrt {\dfrac {PQ}{n}}\)
Sample standard deviation \(s\)	\(\sqrt {\dfrac {\sigma ^2}{2n}}\)
Sample variance \(s^2\)	\(\sigma ^2 \sqrt {\dfrac {2}{n}}\)
Sample median	\(\dfrac {1.25331 \sigma }{\sqrt {n}}\)
Sample quartiles	\(\dfrac {1.36263 \sigma }{\sqrt {n}}\)
Difference of two means \((\bar {x}_1 - \bar {x}_2)\)	\(\sqrt {\dfrac {\sigma _1^2}{n_1} + \dfrac {\sigma _2^2}{n_2}}\)
Difference of two proportions \((p_1 - p_2)\)	\(\sqrt {\dfrac {P_1 Q_1}{n_1} + \dfrac {P_2 Q_2}{n_2}}\)

4.3.4 Why is Standard Error Useful?

Remark 4.3.8.

1.

Precision of Estimate: The smaller the standard error, the more precise our estimate. The reciprocal of the standard error \((1/\text {S.E.})\) is a measure of precision.

2.

Hypothesis Testing: For large samples, the statistic: \[ Z = \frac {t - E(t)}{\text {S.E.}(t)} \sim N(0,1) \] This is the basis for many large-sample tests.

3.

Confidence Limits: Standard error helps us find the range in which the population parameter is likely to lie. For example, the limits for population proportion are: \[ p \pm 3\sqrt {\frac {pq}{n}} \]

4.

Sample Size Determination: To reduce the standard error to half, we must increase the sample size four times. This helps in planning surveys.

4.4 Tests of Significance

Definition 4.4.1 (Test of Significance). A test of significance is a procedure to determine whether the difference between:

1.

A sample statistic and a hypothetical population parameter, or

2.

Two sample statistics from independent samples

is significant or could be attributed to chance.

4.4.1 Why Large Samples Use Normal Test

For large samples (usually \(n > 30\)), almost all distributions (Binomial, Poisson, etc.) can be approximated by the normal distribution. Therefore, we use the normal test for large samples.

Theorem 4.4.2. If \(X \sim N(\mu , \sigma ^2)\), then: \[ Z = \frac {X - \mu }{\sigma } \sim N(0,1) \] where \(Z\) is called the standard normal variate.

From normal probability tables: \begin {align*} P(|Z| \leq 1.96) &= 0.95 \\ P(|Z| \leq 2.58) &= 0.99 \\ P(|Z| \leq 3) &= 0.9973 \end {align*} This gives us the critical values of \(Z\) at different significance levels.

Test Type	1% Level	5% Level	10% Level
Two-tailed	2.58	1.96	1.645
Right-tailed	2.33	1.645	1.28
Left-tailed	-2.33	-1.645	-1.28

4.5 Null Hypothesis

Definition 4.5.1 (Null Hypothesis). A null hypothesis is a hypothesis which is tested for possible rejection under the assumption that it is true. It is usually a statement that there is no significant difference.

Example 4.5.2.

• For testing a single mean: \(H_0: \mu = \mu _0\) (the sample mean does not differ significantly from the population mean).
• For comparing two means: \(H_0: \mu _1 = \mu _2\) (the two sample means do not differ significantly).
• For comparing proportions: \(H_0: P_1 = P_2\) (the two proportions are equal).

4.5.1 Alternative Hypothesis

The alternative hypothesis (denoted by \(H_1\)) is any hypothesis that is complementary to the null hypothesis.

Definition 4.5.3 (Alternative Hypothesis). The alternative hypothesis is the statement that we accept if we reject the null hypothesis.

For example, if \(H_0: \mu = \mu _0\), then the alternative could be:

1.

\(H_1: \mu \neq \mu _0\) (two-tailed test)

2.

\(H_1: \mu > \mu _0\) (right-tailed test)

3.

\(H_1: \mu < \mu _0\) (left-tailed test)

4.5.2 One-Tailed and Two-Tailed Tests

The choice of alternative hypothesis determines whether we use a one-tailed or two-tailed test.

Definition 4.5.4 (Two-Tailed Test). When the alternative hypothesis is of the form \(H_1: \mu \neq \mu _0\), we use a two-tailed test. The critical region is split between both tails of the distribution.

Definition 4.5.5 (One-Tailed Test). When the alternative hypothesis is of the form \(H_1: \mu > \mu _0\) or \(H_1: \mu < \mu _0\), we use a one-tailed test. The critical region lies entirely in one tail.

Example 4.5.6. Suppose a company claims that its new bulb lasts longer than the old one. If we want to test this claim:

• \(H_0: \mu _1 = \mu _2\) (no difference)
• \(H_1: \mu _1 < \mu _2\) (new bulb lasts longer - left-tailed test)

4.6 Errors in Sampling

Definition 4.6.1 (Type I Error). Rejecting the null hypothesis when it is actually true. \[ P(\text {Reject } H_0 \mid H_0 \text { is true}) = \alpha \]

Definition 4.6.2 (Type II Error). Accepting the null hypothesis when it is actually false (i.e., when the alternative hypothesis is true). \[ P(\text {Accept } H_0 \mid H_1 \text { is true}) = \beta \]

4.6.1 Type I Error - Producer’s Risk

In quality control, rejecting a good product or a good lot is called producer’s risk. This is the same as Type I error. \[ \text {Producer's Risk} = P(\text {Reject a good lot}) = \alpha \]

4.6.2 Type II Error - Consumer’s Risk

Accepting a bad product or a bad lot is called consumer’s risk. This is the same as Type II error. \[ \text {Consumer's Risk} = P(\text {Accept a bad lot}) = \beta \]

4.6.3 Summary of Errors

	Decision: Accept \(H_0\)	Decision: Reject \(H_0\)
\(H_0\)is True	Correct Decision	Type I Error (\(\alpha \))
\(H_0\)is False	Type II Error (\(\beta \))	Correct Decision

Remark 4.6.3.

1.

The level of significance \(\alpha \) is the maximum probability of committing a Type I error. It is usually set at 5% or 1% before collecting the data.

2.

We try to keep both \(\alpha \) and \(\beta \) as small as possible, but reducing one usually increases the other.

3.

Increasing the sample size helps to reduce both types of errors.

4.7 Critical Region and Level of Significance

When we perform a hypothesis test, we divide the entire sample space into two regions:

Definition 4.7.1 (Critical Region (Rejection Region)). The set of all values of the test statistic that lead to rejection of the null hypothesis \(H_0\) is called the critical region or region of rejection. It is usually denoted by \(\omega \).

Definition 4.7.2 (Acceptance Region). The set of all values of the test statistic that lead to acceptance of the null hypothesis is called the acceptance region. It is denoted by \(\overline {\omega }\).

Remark 4.7.3. The two regions together make up the entire sample space: \[ \omega \cup \overline {\omega } = S \quad \text {and} \quad \omega \cap \overline {\omega } = \phi \]

4.7.1 Level of Significance

Definition 4.7.4 (Level of Significance). The probability that a random value of the test statistic falls in the critical region (i.e., the probability of rejecting \(H_0\) when it is true) is called the level of significance. It is denoted by \(\alpha \). \[ P(t \in \omega \mid H_0) = \alpha \]

Remark 4.7.5.

1.

The level of significance is the maximum probability of committing a Type I error.

2.

It is always fixed in advance before collecting the sample data.

3.

Commonly used levels are:

• 5% level (\(\alpha = 0.05\))
• 1% level (\(\alpha = 0.01\))
• 10% level (\(\alpha = 0.10\))

4.

The confidence coefficient is \((1 - \alpha )\). For example, at 5% level, the confidence coefficient is 95%.

4.7.2 One-Tailed and Two-Tailed Tests

Definition 4.7.6 (Two-Tailed Test). A test where the alternative hypothesis is of the form \(H_1: \theta \neq \theta _0\) is called a two-tailed test. The critical region is divided between both tails of the distribution.

Definition 4.7.7 (Right-Tailed Test). A test where the alternative hypothesis is of the form \(H_1: \theta > \theta _0\) is called a right-tailed test. The critical region lies entirely in the right tail.

Definition 4.7.8 (Left-Tailed Test). A test where the alternative hypothesis is of the form \(H_1: \theta < \theta _0\) is called a left-tailed test. The critical region lies entirely in the left tail.

Example 4.7.9. Suppose we are comparing two brands of bulbs:

• To test if they are different: \(H_1: \mu _1 \neq \mu _2\) → Two-tailed test
• To test if brand A is better: \(H_1: \mu _1 < \mu _2\) → Left-tailed test
• To test if brand B is better: \(H_1: \mu _1 > \mu _2\) → Right-tailed test

4.7.3 Critical Values (Significant Values)

Definition 4.7.10 (Critical Value). The value of the test statistic that separates the critical region from the acceptance region is called the critical value or significant value. It depends on:

1.

The level of significance \(\alpha \)

2.

The alternative hypothesis (one-tailed or two-tailed)

For large samples, the test statistic follows a standard normal distribution: \[ Z = \frac {t - E(t)}{\text {S.E.}(t)} \sim N(0,1) \]

4.7.4 Critical Values from Normal Distribution

For a two-tailed test at level \(\alpha \): \[ P(|Z| > z_{\alpha }) = \alpha \] This means the total area in both tails is \(\alpha \). Since the normal curve is symmetric, each tail has area \(\alpha /2\). For a right-tailed test: \[ P(Z > z_{\alpha }) = \alpha \] For a left-tailed test: \[ P(Z < -z_{\alpha }) = \alpha \]

4.7.5 Table of Critical Values of \(Z\)

Type of Test	Level of Significance (\(\alpha \))
	1%	5%	10%
Two-tailed	\(|z_{\alpha }| = 2.58\)	\(|z_{\alpha }| = 1.96\)	\(|z_{\alpha }| = 1.645\)
Right-tailed	\(z_{\alpha } = 2.33\)	\(z_{\alpha } = 1.645\)	\(z_{\alpha } = 1.28\)
Left-tailed	\(z_{\alpha } = -2.33\)	\(z_{\alpha } = -1.645\)	\(z_{\alpha } = -1.28\)

Remark 4.7.11. The critical value for a one-tailed test at level \(\alpha \) is the same as the critical value for a two-tailed test at level \(2\alpha \).

4.7.6 Decision Rule for Hypothesis Testing

1.

Compute the test statistic \(Z\) under \(H_0\).

2.

Compare \(|Z|\) with the critical value \(z_{\alpha }\).

3.

If \(|Z| < z_{\alpha }\): The result is not significant. The difference could be due to chance. Accept \(H_0\).

4.

If \(|Z| > z_{\alpha }\): The result is significant. The difference is real. Reject \(H_0\).

Remark 4.7.12. Sometimes, if \(|Z| > 3\), the result is considered highly significant and \(H_0\) is rejected at any level.

4.7.7 Summary of Steps in Hypothesis Testing

1.

Null Hypothesis: Set up \(H_0\).

2.

Alternative Hypothesis: Set up \(H_1\) to decide between one-tailed and two-tailed tests.

3.

Level of Significance: Choose \(\alpha \) (usually 5% or 1%).

4.

Test Statistic: Compute: \[ Z = \frac {t - E(t)}{\text {S.E.}(t)} \]

5.

Conclusion: Compare \(|Z|\) with \(z_{\alpha }\) and accept or reject \(H_0\).

4.8 Test of Significance for Large Samples

For large samples (usually \(n > 30\)), the sampling distribution of almost all statistics is approximately normal. This is due to the Central Limit Theorem. Therefore, we can use the normal distribution to test hypotheses.

Theorem 4.8.1. If \(X \sim N(\mu , \sigma ^2)\), then: \[ Z = \frac {X - \mu }{\sigma } \sim N(0,1) \] This property also holds approximately for large samples from non-normal populations.

4.8.1 Area Properties of Normal Distribution

From normal probability tables: \begin {align*} P(|Z| \leq 1.96) &= 0.95 \quad \text {(5\% significance level)} \\ P(|Z| \leq 2.58) &= 0.99 \quad \text {(1\% significance level)} \\ P(|Z| \leq 3) &= 0.9973 \end {align*} Therefore: \begin {align*} P(|Z| > 1.96) &= 0.05 \\ P(|Z| > 2.58) &= 0.01 \\ P(|Z| > 3) &= 0.0027 \end {align*}

Remark 4.8.2.

4.8.2 Decision Rules for Large Sample Tests

1.

If \(|Z| > 3\): Always reject \(H_0\) (highly significant).

2.

If \(|Z| > 1.96\): Reject \(H_0\) at 5% level of significance.

3.

If \(|Z| > 2.58\): Reject \(H_0\) at 1% level of significance.

4.

For one-tailed tests, use 1.645 (5%) and 2.33 (1%).

4.9 Sampling of Attributes

When we classify population members into two categories (e.g., possessing a certain attribute or not), we are dealing with attributes. Examples include:

• Smokers vs. non-smokers
• Defective vs. non-defective items
• Voters in favour vs. against a proposal

In such cases, we are interested in the proportion of individuals possessing the attribute.

4.9.1 Binomial Distribution and Sampling of Attributes

If we take a sample of size \(n\) from a population where the proportion of successes is \(P\) (and failures \(Q = 1 - P\)), then the number of successes \(X\) follows a binomial distribution: \[ P(X = x) = \binom {n}{x} P^x Q^{n-x}, \quad x = 0, 1, 2, \ldots , n \] For large \(n\), this binomial distribution can be approximated by a normal distribution.

4.9.2 Test for Single Proportion

Let \(X\) be the number of successes in a sample of size \(n\). Then: \begin {align*} E(X) &= nP \\ V(X) &= nPQ \end {align*} The sample proportion is: \[ p = \frac {X}{n} \] For large samples, the normal test statistic is: \[ Z = \frac {p - P}{\sqrt {PQ/n}} \sim N(0,1) \]

Remark 4.9.1.

4.9.3 Properties of Sample Proportion

1.

The sample proportion \(p\) is an unbiased estimate of the population proportion \(P\): \[ E(p) = P \]

2.

The standard error of the sample proportion is: \[ \text {S.E.}(p) = \sqrt {\frac {PQ}{n}} \]

3.

If \(P\) is unknown, we estimate it using the sample proportion: \[ \hat {P} = p, \quad \hat {Q} = q = 1-p \]

4.

The probable limits for the population proportion are: \[ p \pm 3\sqrt {\frac {pq}{n}} \]

5.

The \(95\%\) confidence limits for \(P\) are: \[ p \pm 1.96\sqrt {\frac {pq}{n}} \]

6.

The \(99\%\) confidence limits for \(P\) are: \[ p \pm 2.58\sqrt {\frac {pq}{n}} \]

4.9.4 Test for Difference of Proportions

Suppose we have two independent samples from two populations. Let:

• \(n_1, n_2\) = sample sizes
• \(p_1, p_2\) = sample proportions
• \(P_1, P_2\) = population proportions

For large samples: \[ p_1 \sim N\left (P_1, \frac {P_1Q_1}{n_1}\right ), \quad p_2 \sim N\left (P_2, \frac {P_2Q_2}{n_2}\right ) \] Under \(H_0: P_1 = P_2 = P\) (say), the test statistic is: \[ Z = \frac {p_1 - p_2}{\sqrt {PQ\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}} \sim N(0,1) \] The pooled estimate of \(P\) is: \[ \hat {P} = \frac {n_1 p_1 + n_2 p_2}{n_1 + n_2} = \frac {X_1 + X_2}{n_1 + n_2} \] Then: \[ Z = \frac {p_1 - p_2}{\sqrt {\hat {P}\hat {Q}\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}} \sim N(0,1) \]

4.9.5 Difference of Proportions when Population Proportions are Known

If the population proportions \(P_1\) and \(P_2\) are known and we want to test if the difference is likely to be hidden in sampling: \[ H_0: p_1 = p_2 \quad \text {(the samples will not reveal the difference)} \] The test statistic is: \[ |Z| = \frac {|P_1 - P_2|}{\sqrt {\frac {P_1Q_1}{n_1} + \frac {P_2Q_2}{n_2}}} \sim N(0,1) \]

4.9.6 Summary of Tests for Proportions

Remark 4.9.2.

1.

Single Proportion: \[ Z = \frac {p - P}{\sqrt {PQ/n}} \]

2.

Difference of Proportions (Unknown \(P\)): \[ Z = \frac {p_1 - p_2}{\sqrt {\hat {P}\hat {Q}\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}}, \quad \hat {P} = \frac {n_1 p_1 + n_2 p_2}{n_1 + n_2} \]

3.

Difference of Proportions (Known \(P_1, P_2\)): \[ |Z| = \frac {|P_1 - P_2|}{\sqrt {\frac {P_1Q_1}{n_1} + \frac {P_2Q_2}{n_2}}} \]

Solved Examples

Example 4.9.3. A dice is thrown 9,000 times and a throw of 3 or 4 is observed 3,240 times. Show that the dice cannot be regarded as an unbiased one and find the limits between which the probability of a throw of 3 or 4 lies.

Solution. Here, success = getting 3 or 4. We are given: \[ n = 9000, \quad X = 3240 \] Sample proportion: \[ p = \frac {X}{n} = \frac {3240}{9000} = 0.36 \] Null Hypothesis \(H_0\): The dice is unbiased, so: \[ P = \frac {1}{3}, \quad Q = \frac {2}{3} \] Alternative Hypothesis \(H_1: P \neq \frac {1}{3}\) (Two-tailed). Test Statistic: \begin {align*} Z& = \frac {p - P}{\sqrt {PQ/n}} \sim N(0,1) \\ Z &= \frac {0.36 - 0.3333}{\sqrt {(0.3333 \times 0.6667)/9000}} \\ &= \frac {0.0267}{\sqrt {0.00002469}} = \frac {0.0267}{0.00497} = 5.37 \end {align*} Conclusion: Since \(|Z| = 5.37 > 3\), we reject \(H_0\). The dice is biased. Limits for Population Proportion: \begin {align*} p \pm 3\sqrt {\frac {pq}{n}} &= 0.36 \pm 3\sqrt {\frac {0.36 \times 0.64}{9000}} \\ &= 0.36 \pm 3\sqrt {\frac {0.2304}{9000}} \\ &= 0.36 \pm 3 \times 0.00506 \\ &= 0.36 \pm 0.01518 \\ &= (0.34482, 0.37518) \end {align*} Hence, the probability of getting 3 or 4 almost certainly lies between 0.345 and 0.375.

Example 4.9.4. A random sample of 500 pineapples was taken from a large consignment and 65 were found to be bad. Show that the S.E. of the proportion of bad ones in a sample of this size is 0.015 and deduce that the percentage of bad pineapples in the consignment almost certainly lies between 8.5 and 17.5.

Solution. We are given: \[ n = 500, \quad X = 65 \] Sample proportion: \[ p = \frac {65}{500} = 0.13, \quad q = 1 - p = 0.87 \] Since \(P\) is unknown, we use: \[ \hat {P} = p = 0.13, \quad \hat {Q} = q = 0.87 \] Standard Error: \begin {align*} \text {S.E.}(p) &= \sqrt {\frac {\hat {P}\hat {Q}}{n}} \\ &= \sqrt {\frac {0.13 \times 0.87}{500}} \\ &= \sqrt {\frac {0.1131}{500}} = \sqrt {0.0002262} = 0.01504 \approx 0.015 \end {align*} Limits for population proportion: \begin {align*} p \pm 3 \times \text {S.E.}(p) &= 0.13 \pm 3(0.015) \\ &= 0.13 \pm 0.045 \\ &= (0.085, 0.175) \end {align*} Hence, the percentage of bad pineapples lies almost certainly between 8.5% and 17.5%.

Example 4.9.5. A random sample of 500 apples was taken from a large consignment and 60 were found to be bad. Obtain the \(98\%\) confidence limits for the percentage of bad apples in the consignment. \[ \left [\int _0^{2.33} \phi (t) dt = 0.49 \text { nearly}\right ] \]

Solution. We are given: \[ n = 500, \quad X = 60 \] Sample proportion: \[ p = \frac {60}{500} = 0.12, \quad q = 0.88 \] For \(98\%\) confidence level, the significant value of \(Z\) is 2.33. Confidence limits for population proportion: \begin {align*} p \pm 2.33 \sqrt {\frac {pq}{n}} &= 0.12 \pm 2.33 \sqrt {\frac {0.12 \times 0.88}{500}} \\ &= 0.12 \pm 2.33 \sqrt {\frac {0.1056}{500}} \\ &= 0.12 \pm 2.33 \times 0.01453 \\ &= 0.12 \pm 0.03385 \\ &= (0.08615, 0.15385) \end {align*} Hence, \(98\%\) confidence limits for the percentage of bad apples are (8.62%, 15.38%).

Example 4.9.6. In a sample of 1,000 people in Maharashtra, 540 are rice eaters and the rest are wheat eaters. Can we assume that both rice and wheat are equally popular in this State at \(1\%\) level of significance?

Solution. We are given: \[ n = 1000, \quad X = 540 \] Sample proportion: \[ p = \frac {540}{1000} = 0.54 \] Null Hypothesis \(H_0: P = 0.5\) (Both are equally popular). Alternative Hypothesis \(H_1: P \neq 0.5\) (Two-tailed). Test Statistic: \begin {align*} Z &= \frac {p - P}{\sqrt {PQ/n}} \\ &= \frac {0.54 - 0.50}{\sqrt {0.5 \times 0.5 / 1000}} \\ &= \frac {0.04}{0.01581} = 2.53 \end {align*} Conclusion: At \(1\%\) level, critical value is 2.58. Since \(|Z| = 2.53 < 2.58\), we accept \(H_0\). Rice and wheat are equally popular in Maharashtra.

Example 4.9.7. Twenty people were attacked by a disease and only 18 survived. Will you reject the hypothesis that the survival rate, if attacked by this disease, is \(85\%\) in favour of the hypothesis that it is more, at \(5\%\) level? (Use Large Sample Test.)

Solution. We are given: \[ n = 20, \quad X = 18 \] Sample proportion: \[ p = \frac {18}{20} = 0.90 \] Null Hypothesis \(H_0: P = 0.85\). Alternative Hypothesis \(H_1: P > 0.85\) (Right-tailed). Test Statistic: \begin {align*} Z &= \frac {p - P}{\sqrt {PQ/n}} \\ &= \frac {0.90 - 0.85}{\sqrt {0.85 \times 0.15 / 20}} \\ &= \frac {0.05}{0.0798} = 0.626 \end {align*} Conclusion: For a right-tailed test at 5%, critical value is 1.645. Since \(Z = 0.626 < 1.645\), we accept \(H_0\). The survival rate is not significantly more than 85%.

Example 4.9.8. Random samples of 400 men and 600 women were asked whether they would like to have a flyover near their residence. 200 men and 325 women were in favour of the proposal. Test the hypothesis that proportions of men and women in favour of the proposal are same against that they are not, at \(5\%\) level.

Solution. We are given: \[ n_1 = 400, \quad X_1 = 200, \quad p_1 = \frac {200}{400} = 0.5 \] \[ n_2 = 600, \quad X_2 = 325, \quad p_2 = \frac {325}{600} = 0.5417 \] Null Hypothesis \(H_0: P_1 = P_2\). Alternative Hypothesis \(H_1: P_1 \neq P_2\) (Two-tailed). Pooled estimate: \[ \hat {P} = \frac {X_1 + X_2}{n_1 + n_2} = \frac {200 + 325}{400 + 600} = \frac {525}{1000} = 0.525 \] \[ \hat {Q} = 1 - 0.525 = 0.475 \] Test Statistic: \begin {align*} Z &= \frac {p_1 - p_2}{\sqrt {\hat {P}\hat {Q}\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}} \\ &= \frac {0.5 - 0.5417}{\sqrt {0.525 \times 0.475 \times \left (\frac {1}{400} + \frac {1}{600}\right )}} \\ &= \frac {-0.0417}{\sqrt {0.001039}} = \frac {-0.0417}{0.03223} = -1.29 \end {align*} Conclusion: Since \(|Z| = 1.29 < 1.96\), we accept \(H_0\) at 5% level. There is no significant difference between men and women in their opinion.

Example 4.9.9. A company has the head office at Calcutta and a branch at Bombay. The personnel director wanted to know if the workers at the two places would like the introduction of a new plan of work and a survey was conducted for this purpose. Out of a sample of 500 workers at Calcutta, \(62\%\) favoured the new plan. At Bombay out of a sample of 400 workers, \(41\%\) were against the new plan. Is there any significant difference between the two groups in their attitude towards the new plan at \(5\%\) level?

Solution. We are given: \[ n_1 = 500, \quad p_1 = 0.62 \] \[ n_2 = 400, \quad p_2 = 1 - 0.41 = 0.59 \] Null Hypothesis \(H_0: P_1 = P_2\). Alternative Hypothesis \(H_1: P_1 \neq P_2\) (Two-tailed). Pooled estimate: \[ \hat {P} = \frac {n_1 p_1 + n_2 p_2}{n_1 + n_2} = \frac {500(0.62) + 400(0.59)}{900} = \frac {310 + 236}{900} = 0.6067 \] \[ \hat {Q} = 1 - 0.6067 = 0.3933 \] Test Statistic: \begin {align*} Z &= \frac {p_1 - p_2}{\sqrt {\hat {P}\hat {Q}\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}} \\ &= \frac {0.62 - 0.59}{\sqrt {0.6067 \times 0.3933 \times \left (\frac {1}{500} + \frac {1}{400}\right )}} \\ &= \frac {0.03}{\sqrt {0.001069}} = \frac {0.03}{0.0327} = 0.917 \end {align*} Conclusion: Since \(|Z| = 0.917 < 1.96\), we accept \(H_0\) at 5% level. There is no significant difference between the two groups.

Example 4.9.10. Before an increase in excise duty on tea, 800 persons out of a sample of 1,000 persons were found to be tea drinkers. After an increase in duty, 800 people were tea drinkers in a sample of 1,200 people. Using standard error of proportion, state whether there is a significant decrease in the consumption of tea after the increase in excise duty?

Solution. We are given: \[ n_1 = 1000, \quad X_1 = 800, \quad p_1 = \frac {800}{1000} = 0.80 \] \[ n_2 = 1200, \quad X_2 = 800, \quad p_2 = \frac {800}{1200} = 0.6667 \] Null Hypothesis \(H_0: P_1 = P_2\). Alternative Hypothesis \(H_1: P_1 > P_2\) (Right-tailed). Pooled estimate: \[ \hat {P} = \frac {X_1 + X_2}{n_1 + n_2} = \frac {800 + 800}{1000 + 1200} = \frac {1600}{2200} = 0.7273 \] \[ \hat {Q} = 1 - 0.7273 = 0.2727 \] Test Statistic: \begin {align*} Z &= \frac {p_1 - p_2}{\sqrt {\hat {P}\hat {Q}\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}} \\ &= \frac {0.80 - 0.6667}{\sqrt {0.7273 \times 0.2727 \times \left (\frac {1}{1000} + \frac {1}{1200}\right )}} \\ &= \frac {0.1333}{\sqrt {0.0003636}} = \frac {0.1333}{0.01907} = 6.99 \end {align*} Conclusion: Since \(Z = 6.99 > 2.33\), we reject \(H_0\) at 1% level. There is a significant decrease in tea consumption after the increase in excise duty.

Example 4.9.11. A cigarette manufacturing firm claims that its brand A of the cigarettes outsells its brand B by \(8\%\). If it is found that 42 out of a sample of 200 smokers prefer brand A and 18 out of another random sample of 100 smokers prefer brand B, test whether the \(8\%\) difference is a valid claim. (Use \(5\%\) level of significance.)

Solution. We are given: \[ n_1 = 200, \quad X_1 = 42, \quad p_1 = \frac {42}{200} = 0.21 \] \[ n_2 = 100, \quad X_2 = 18, \quad p_2 = \frac {18}{100} = 0.18 \] Null Hypothesis \(H_0: P_1 - P_2 = 0.08\) (The 8% difference is valid). Alternative Hypothesis \(H_1: P_1 - P_2 \neq 0.08\) (Two-tailed). Pooled estimate: \[ \hat {P} = \frac {X_1 + X_2}{n_1 + n_2} = \frac {42 + 18}{200 + 100} = \frac {60}{300} = 0.20 \] \[ \hat {Q} = 0.80 \] Test Statistic: \begin {align*} Z &= \frac {(p_1 - p_2) - (P_1 - P_2)}{\sqrt {\hat {P}\hat {Q}\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}} \\ &= \frac {(0.21 - 0.18) - 0.08}{\sqrt {0.20 \times 0.80 \times \left (\frac {1}{200} + \frac {1}{100}\right )}} \\ &= \frac {-0.05}{\sqrt {0.0024}} = \frac {-0.05}{0.0490} = -1.02 \end {align*} Conclusion: Since \(|Z| = 1.02 < 1.96\), we accept \(H_0\) at 5% level. The 8% difference is a valid claim.

Example 4.9.12. On the basis of their total scores, 200 candidates of a civil service examination are divided into two groups, the upper 30 per cent and the remaining 70 per cent. Consider the first question of this examination. Among the first group, 40 had the correct answer, whereas among the second group, 80 had the correct answer. On the basis of these results, can one conclude that the first question is no good at discriminating ability of the type being examined here?

Solution. We have: \[ n = 200, \quad n_1 = 60, \quad n_2 = 140 \] \[ X_1 = 40, \quad X_2 = 80 \] \[ p_1 = \frac {40}{60} = 0.6667, \quad p_2 = \frac {80}{140} = 0.5714 \] Null Hypothesis \(H_0: P_1 = P_2\) (The question does not discriminate). Alternative Hypothesis \(H_1: P_1 \neq P_2\) (Two-tailed). Pooled estimate: \[ \hat {P} = \frac {40 + 80}{60 + 140} = \frac {120}{200} = 0.60, \quad \hat {Q} = 0.40 \] Test Statistic: \begin {align*} Z &= \frac {p_1 - p_2}{\sqrt {\hat {P}\hat {Q}\left (\frac {1}{n_1} + \frac {1}{n_2}\right )}} \\ &= \frac {0.6667 - 0.5714}{\sqrt {0.60 \times 0.40 \times \left (\frac {1}{60} + \frac {1}{140}\right )}} \\ &= \frac {0.0953}{\sqrt {0.007619}} = \frac {0.0953}{0.0873} = 1.09 \end {align*} Conclusion: Since \(|Z| = 1.09 < 1.96\), we accept \(H_0\). The question is not good at discriminating ability.

Example 4.9.13. In a year there are 956 births in a town A, of which \(52.5\%\) were males, while in towns A and B combined, this proportion in a total of 1,406 births was 0.496. Is there any significant difference in the proportion of male births in the two towns?

Solution. We have: \[ n_1 = 956, \quad p_1 = 0.525 \] \[ n_1 + n_2 = 1406 \Rightarrow n_2 = 450 \] \[ \hat {P} = 0.496 \] Find \(p_2\): \begin {align*} \frac {956(0.525) + 450(p_2)}{1406} &= 0.496 \\ 501.9 + 450p_2 &= 697.376 \\ 450p_2 &= 195.476 \\ p_2 &= 0.4344 \end {align*} Null Hypothesis \(H_0: P_1 = P_2\). Alternative Hypothesis \(H_1: P_1 \neq P_2\) (Two-tailed). \[ \hat {Q} = 1 - 0.496 = 0.504 \] Test Statistic: \begin {align*} Z &= \frac {0.525 - 0.4344}{\sqrt {0.496 \times 0.504 \times \left (\frac {1}{956} + \frac {1}{450}\right )}} \\ &= \frac {0.0906}{\sqrt {0.001521}} = \frac {0.0906}{0.0390} = 2.32 \end {align*} Conclusion: Since \(|Z| = 2.32 > 1.96\), we reject \(H_0\) at 5% level. There is significant difference in the proportion of male births.

Example 4.9.14. In two large populations, there are 30 and 25 per cent respectively of blue-eyed people. Is this difference likely to be hidden in samples of 1,200 and 900 respectively from the two populations?

Solution. We are given: \[ n_1 = 1200, \quad P_1 = 0.30, \quad Q_1 = 0.70 \] \[ n_2 = 900, \quad P_2 = 0.25, \quad Q_2 = 0.75 \] Null Hypothesis \(H_0: p_1 = p_2\) (The samples will not reveal the difference). Test Statistic: \begin {align*} |Z| &= \frac {|P_1 - P_2|}{\sqrt {\frac {P_1Q_1}{n_1} + \frac {P_2Q_2}{n_2}}} \\ &= \frac {0.05}{\sqrt {\frac {0.21}{1200} + \frac {0.1875}{900}}} \\ &= \frac {0.05}{\sqrt {0.000175 + 0.000208}} \\ &= \frac {0.05}{\sqrt {0.000383}} = \frac {0.05}{0.01957} = 2.55 \end {align*} Conclusion: Since \(|Z| = 2.55 > 1.96\), we reject \(H_0\) at 5% level. The difference is unlikely to be hidden in sampling.

Example 4.9.15. In a random sample of 400 students of the university teaching departments, it was found that 300 students failed in the examination. In another random sample of 500 students of the affiliated colleges, the number of failures in the same examination was found to be 300. Find out whether the proportion of failures in the university teaching departments is significantly greater than the proportion of failures in the university teaching departments and affiliated colleges taken together.

Solution. We have: \[ n_1 = 400, \quad p_1 = \frac {300}{400} = 0.75 \] \[ n_2 = 500, \quad p_2 = \frac {300}{500} = 0.60 \] Pooled proportion: \[ \hat {p} = \frac {400(0.75) + 500(0.60)}{400 + 500} = \frac {300 + 300}{900} = 0.6667 \] \[ \hat {q} = 0.3333 \] Standard Error of \((\hat {p} - p_1)\): \begin {align*} \text {S.E.}(\hat {p} - p_1) &= \sqrt {\frac {\hat {p}\hat {q}}{n_1 + n_2}} \times \frac {n_2}{n_1} \\ &= \sqrt {\frac {0.6667 \times 0.3333}{900}} \times \frac {500}{400} \\ &= \sqrt {0.000247} \times 1.25 = 0.0157 \times 1.25 = 0.0196 \end {align*} Null Hypothesis \(H_0: \hat {p} = p_1\). Test Statistic: \[ Z = \frac {\hat {p} - p_1}{\text {S.E.}(\hat {p} - p_1)} = \frac {0.6667 - 0.75}{0.0196} = \frac {-0.0833}{0.0196} = -4.25 \] Conclusion: Since \(|Z| = 4.25 > 3\), we reject \(H_0\). There is significant difference between \(p_1\) and \(\hat {p}\).

Example 4.9.16. If for one-half of \(n\) events, the chance of success is \(p\) and the chance of failure is \(q\), while for the other half the chance of success is \(q\) and the chance of failure is \(p\), show that the standard deviation of the number of successes is the same as if the chance of success were \(p\) in all the cases, i.e., \(\sqrt {npq}\) but that the mean of the number of successes is \(n/2\) and not \(np\).

Solution. Let \(X_1\) and \(X_2\) denote the number of successes in the first half and second half of \(n\) events respectively. For the first half: \[ X_1 \sim B\left (\frac {n}{2}, p\right ) \] For the second half: \[ X_2 \sim B\left (\frac {n}{2}, q\right ) \] Now: \begin {align*} E(X_1) &= \frac {n}{2}p, \quad V(X_1) = \frac {n}{2}pq \\ E(X_2) &= \frac {n}{2}q, \quad V(X_2) = \frac {n}{2}pq \end {align*} Let \(X = X_1 + X_2\) be the total number of successes. Mean: \begin {align*} E(X) &= E(X_1) + E(X_2) \\ &= \frac {n}{2}p + \frac {n}{2}q \\ &= \frac {n}{2}(p + q) = \frac {n}{2} \end {align*} Variance (since \(X_1\) and \(X_2\) are independent): \begin {align*} V(X) &= V(X_1) + V(X_2) \\ &= \frac {n}{2}pq + \frac {n}{2}pq \\ &= npq \end {align*} Thus: \[ \text {S.D.}(X) = \sqrt {V(X)} = \sqrt {npq} \] Hence, the mean is \(n/2\) (not \(np\)) and the standard deviation is \(\sqrt {npq}\), which is the same as if the probability of success were \(p\) in all cases.

4.10 Sampling of Variables

Now, we turn to variables where each individual provides a numerical value. Examples include:

• Height of students
• Weight of apples
• Age of workers
• Income of families

In the case of sampling of variables, each member of the population provides a value of the variable, and the aggregate of these values forms the frequency distribution of the population. From the population, a random sample of size \( n \) can be drawn, which is the same as choosing \( n \) values from the distribution.

4.11 Unbiased Estimate for Population Mean and Variance

Definition 4.11.1 (Sample Mean and Variance). Let \( x_1, x_2, \dots , x_n \) be a random sample of size \( n \) from a population with mean \( \mu \) and variance \( \sigma ^2 \). Then: \[ \bar {x} = \frac {1}{n} \sum _{i=1}^{n} x_i \quad \text {(sample mean)} \] \[ s^2 = \frac {1}{n} \sum _{i=1}^{n} (x_i - \bar {x})^2 \quad \text {(sample variance)} \]

Theorem 4.11.2 (Sample Mean is Unbiased). The sample mean \( \bar {x} \) is an unbiased estimate of the population mean \( \mu \): \[ E(\bar {x}) = \mu \]

Proof : Since each \( x_i \) is a sample observation from the population, it can take any value from the population with equal probability: \[ E(x_i) = \frac {1}{N} \sum _{i=1}^{N} X_i = \mu \] Therefore: \begin {align*} E(\bar {x}) &= E\left ( \frac {1}{n} \sum _{i=1}^{n} x_i \right ) \\ &= \frac {1}{n} \sum _{i=1}^{n} E(x_i) \\ &= \frac {1}{n} \sum _{i=1}^{n} \mu = \mu \end {align*} □

Theorem 4.11.3 (Sample Variance is Biased). The sample variance \( s^2 \) is not an unbiased estimate of the population variance \( \sigma ^2 \): \[ E(s^2) = \frac {n-1}{n} \sigma ^2 \]

Proof : \begin {align*} E(s^2) &= E\left [ \frac {1}{n} \sum _{i=1}^{n} (x_i - \bar {x})^2 \right ] \\ &= \frac {1}{n} \sum _{i=1}^{n} E(x_i - \mu )^2 - E(\bar {x} - \mu )^2 \\ &= \frac {1}{n} \sum _{i=1}^{n} \sigma ^2 - \frac {\sigma ^2}{n} \\ &= \frac {n-1}{n} \sigma ^2 \end {align*} □

Definition 4.11.4 (Unbiased Estimate of Population Variance). An unbiased estimate of \( \sigma ^2 \) is given by: \[ S^2 = \frac {1}{n-1} \sum _{i=1}^{n} (x_i - \bar {x})^2 \] Because: \[ E(S^2) = \sigma ^2 \]

Remark 4.11.5.

1.

For large samples (\( n \to \infty \)), \( s^2 \approx S^2 \), so we can use \( s^2 \) as an estimate of \( \sigma ^2 \).

2.

The relationship between \( s^2 \) and \( S^2 \) is: \[ s^2 = \frac {n-1}{n} S^2 \]

4.12 Standard Error of Sample Mean

Theorem 4.12.1 (Standard Error of Mean). The standard error of the sample mean \( \bar {x} \) for a random sample of size \( n \) from a population with variance \( \sigma ^2 \) is: \[ \text {S.E.}(\bar {x}) = \frac {\sigma }{\sqrt {n}} \]

Proof : \begin {align*} V(\bar {x}) &= V\left ( \frac {1}{n} \sum _{i=1}^{n} x_i \right ) \\ &= \frac {1}{n^2} \sum _{i=1}^{n} V(x_i) \quad (\text {since observations are independent}) \\ &= \frac {1}{n^2} \cdot n\sigma ^2 = \frac {\sigma ^2}{n} \end {align*} Therefore: \[ \text {S.E.}(\bar {x}) = \sqrt {V(\bar {x})} = \frac {\sigma }{\sqrt {n}} \] □

Remark 4.12.2. If the population standard deviation \( \sigma \) is unknown, we use its estimate: \[ \hat {\sigma } = s \quad \text {(for large samples)} \]

4.13 Test of Significance for Single Mean

Theorem 4.13.1 (Test for Single Mean). For large samples, under the null hypothesis \( H_0: \mu = \mu _0 \), the test statistic is: \[ Z = \frac {\bar {x} - \mu }{\sigma / \sqrt {n}} \sim N(0,1) \] If \( \sigma \) is unknown, we use \( \hat {\sigma } = s \): \[ Z = \frac {\bar {x} - \mu }{s / \sqrt {n}} \sim N(0,1) \]

4.13.1 Confidence Limits for Population Mean

Definition 4.13.2 (Confidence Limits for Mean). The \( 100(1-\alpha )\% \) confidence limits for the population mean \( \mu \) are: \[ \bar {x} \pm z_{\alpha /2} \frac {\sigma }{\sqrt {n}} \] In particular:

• \( 95\% \) confidence limits: \( \bar {x} \pm 1.96 \frac {\sigma }{\sqrt {n}} \)
• \( 99\% \) confidence limits: \( \bar {x} \pm 2.58 \frac {\sigma }{\sqrt {n}} \)
• \( 98\% \) confidence limits: \( \bar {x} \pm 2.33 \frac {\sigma }{\sqrt {n}} \)

Remark 4.13.3. For sampling from a finite population of size \( N \), the standard error is multiplied by the finite population correction factor: \[ \text {S.E.}(\bar {x}) = \frac {\sigma }{\sqrt {n}} \sqrt {\frac {N-n}{N-1}} \]

Example 4.13.4 (Example 12.15). A sample of 900 members has a mean 3.4 cms. and s.d. 2.61 cms. Is the sample from a large population of mean 3.25 cms. and s.d. 2.61 cms.? If the population is normal and its mean is unknown, find the 95% and 98% fiducial limits of true mean.

Solution. We are given: \[ n = 900, \quad \bar {x} = 3.4, \quad \sigma = 2.61, \quad \mu = 3.25 \] Null Hypothesis \( H_0: \mu = 3.25 \). Alternative Hypothesis \( H_1: \mu \neq 3.25 \) (Two-tailed). Test Statistic: \begin {align*} Z &= \frac {\bar {x} - \mu }{\sigma / \sqrt {n}} \\ &= \frac {3.40 - 3.25}{2.61 / \sqrt {900}} \\ &= \frac {0.15}{2.61/30} = \frac {0.15}{0.087} = 1.72 \end {align*} Conclusion: Since \( |Z| = 1.72 < 1.96 \), we accept \( H_0 \). The sample could have come from the population. 95% Fiducial Limits: \begin {align*} \bar {x} \pm 1.96 \frac {\sigma }{\sqrt {n}} &= 3.40 \pm 1.96 \times \frac {2.61}{30} \\ &= 3.40 \pm 0.1705 = (3.2295, 3.5705) \end {align*} 98% Fiducial Limits: \begin {align*} \bar {x} \pm 2.33 \frac {\sigma }{\sqrt {n}} &= 3.40 \pm 2.33 \times \frac {2.61}{30} \\ &= 3.40 \pm 0.2027 = (3.1973, 3.6027) \end {align*}

Example 4.13.5 (Example 12.16). An insurance agent has claimed that the average age of policyholders who insure through him is less than the average for all agents, which is 30.5 years. A random sample of 100 policyholders who had insured through him gave the following age distribution: \[ \begin {array}{c|ccccc} \text {Age} & 16-20 & 21-25 & 26-30 & 31-35 & 36-40 \\ \hline \text {No. of persons} & 12 & 22 & 20 & 30 & 16 \end {array} \] Calculate the arithmetic mean and standard deviation of this distribution and use these values to test his claim at the 5% level of significance. You are given that \( Z(1.645) = 0.95 \).

Solution. First, compute the sample mean and standard deviation: \[ \begin {array}{c|c|c|c|c|c} \text {Age} & f & x & d = \frac {x-28}{5} & fd & fd^2 \\ \hline 16-20 & 12 & 18 & -2 & -24 & 48 \\ 21-25 & 22 & 23 & -1 & -22 & 22 \\ 26-30 & 20 & 28 & 0 & 0 & 0 \\ 31-35 & 30 & 33 & 1 & 30 & 30 \\ 36-40 & 16 & 38 & 2 & 32 & 64 \\ \hline \text {Total} & 100 & & & 16 & 164 \end {array} \] \[ \bar {x} = 28 + 5 \times \frac {16}{100} = 28.8 \text { years} \] \[ s = 5 \times \sqrt {\frac {164}{100} - \left (\frac {16}{100}\right )^2} = 5 \times \sqrt {1.64 - 0.0256} = 5 \times 1.270 = 6.35 \text { years} \] Null Hypothesis \( H_0: \mu = 30.5 \). Alternative Hypothesis \( H_1: \mu < 30.5 \) (Left-tailed). Test Statistic: \[ Z = \frac {\bar {x} - \mu }{s/\sqrt {n}} = \frac {28.8 - 30.5}{6.35/\sqrt {100}} = \frac {-1.7}{0.635} = -2.68 \] Conclusion: For a left-tailed test at 5%, critical value is -1.645. Since \( Z = -2.68 < -1.645 \), we reject \( H_0 \). The agent’s claim is valid.

Example 4.13.6 (Example 12.17). As an application of Central Limit Theorem, show that if \( E \) is such that \( P(|\bar {X} - \mu | < E) > 0.95 \), then the minimum sample size \( n \) is given by \( n = \frac {(1.96)^2 \sigma ^2}{E^2} \), where \( \mu \) and \( \sigma ^2 \) are the mean and variance respectively of the population and \( \bar {X} \) is the mean of the random sample.

Solution. By Central Limit Theorem, \( \bar {X} \sim N(\mu , \sigma ^2/n) \) asymptotically. \[ Z = \frac {\bar {X} - \mu }{\sigma /\sqrt {n}} \sim N(0,1) \] From normal tables: \[ P(|Z| \leq 1.96) = 0.95 \] \[ P\left ( \left | \frac {\bar {X} - \mu }{\sigma /\sqrt {n}} \right | \leq 1.96 \right ) = 0.95 \] \[ P(|\bar {X} - \mu | \leq 1.96 \times \frac {\sigma }{\sqrt {n}}) = 0.95 \] Given that \( P(|\bar {X} - \mu | < E) > 0.95 \), we have: \[ E > 1.96 \times \frac {\sigma }{\sqrt {n}} \Rightarrow n > \frac {(1.96)^2 \sigma ^2}{E^2} \] Hence, the minimum sample size is: \[ n = \frac {3.84 \sigma ^2}{E^2} \]

Example 4.13.7 (Example 12.18). The mean muscular endurance score of a random sample of 60 subjects was found to be 145 with a s.d. of 40. Construct a \( 95\% \) confidence interval for the true mean. Assume the sample size to be large enough for normal approximation. What size of sample is required to estimate the mean within 5 of the true mean with a \( 95\% \) confidence?

Solution. We are given: \[ n = 60, \quad \bar {x} = 145, \quad s = 40 \] 95% Confidence Limits: \[ \bar {x} \pm 1.96 \times \frac {s}{\sqrt {n}} = 145 \pm 1.96 \times \frac {40}{\sqrt {60}} = 145 \pm 10.12 = (134.88, 155.12) \] Sample Size Required: \[ n = \left ( \frac {z_{\alpha } \sigma }{E} \right )^2 = \left ( \frac {1.96 \times 40}{5} \right )^2 = (15.68)^2 = 245.86 \approx 246 \]

Example 4.13.8 (Example 12.19). The standard deviation of a population is 2.70 inches. Find the probability that in a random sample of size 66 (i) the sample mean will differ from the population mean by 0.75 inch or more and (ii) the sample mean will exceed the population mean by 0.75 inch or more (given that the value of the standard normal probability integral from 0 to 2.25 is 0.4877).

Solution. Here \( n = 66 \), \( \sigma = 2.70 \). Since \( n \) is large: \[ Z = \frac {\bar {x} - \mu }{\sigma /\sqrt {n}} \sim N(0,1) \] (i) \begin {align*} P(|\bar {x} - \mu | \geq 0.75) &= 1 - P(|\bar {x} - \mu | < 0.75) \\ &= 1 - P\left ( |Z| < \frac {0.75\sqrt {66}}{2.70} \right ) \\ &= 1 - P(|Z| < 2.25) \\ &= 1 - 2(0.4877) = 0.0246 \end {align*} (ii) \[ P(\bar {x} - \mu > 0.75) = P(Z > 2.25) = 0.5 - 0.4877 = 0.0123 \]

Example 4.13.9 (Example 12.20). A normal population has a mean of 0.1 and standard deviation of 2.1. Find the probability that mean of a sample of size 900 will be negative.

Solution. Here \( \mu = 0.1 \), \( \sigma = 2.1 \), \( n = 900 \). \[ Z = \frac {\bar {x} - \mu }{\sigma /\sqrt {n}} = \frac {\bar {x} - 0.1}{2.1/30} = \frac {\bar {x} - 0.1}{0.07} \] \[ \bar {x} = 0.1 + 0.07Z \] Required probability: \[ P(\bar {x} < 0) = P(0.1 + 0.07Z < 0) = P(Z < -1.43) = 0.5 - 0.4236 = 0.0764 \]

Example 4.13.10 (Example 12.21). The guaranteed average life of a certain type of electric light bulbs is 1000 hours with a standard deviation of 125 hours. It is decided to sample the output so as to ensure that 90 per cent of the bulbs do not fall short of the guaranteed average by more than 2.5 per cent. What must be the minimum size of the sample?

Solution. Here \( \mu = 1000 \), \( \sigma = 125 \). We want \( \bar {x} > 1000 - 2.5\% \text { of } 1000 = 1000 - 25 = 975 \). \[ Z = \frac {\bar {x} - \mu }{\sigma /\sqrt {n}} > \frac {975 - 1000}{125/\sqrt {n}} = -\frac {\sqrt {n}}{5} \] We need: \[ P\left ( Z > -\frac {\sqrt {n}}{5} \right ) = 0.90 \Rightarrow P(0 < Z < \frac {\sqrt {n}}{5}) = 0.40 \] From normal tables, \( \frac {\sqrt {n}}{5} = 1.28 \Rightarrow n = 25 \times (1.28)^2 = 41 \)

Example 4.13.11 (Example 12.22). A survey is proposed to be conducted to know the annual earnings of the old Statistics graduates of Delhi University. How large should the sample be taken in order to estimate the mean annual earnings within plus and minus Rs. 1,000 at \( 95\% \) confidence level? The standard deviation of the annual earnings of the entire population is known to be Rs. 3,000.

Solution. We are given: \[ \sigma = 3000, \quad E = 1000, \quad z_{\alpha } = 1.96 \] \[ n = \left ( \frac {z_{\alpha } \sigma }{E} \right )^2 = \left ( \frac {1.96 \times 3000}{1000} \right )^2 = (5.88)^2 = 34.56 \approx 35 \]

4.14 Test of Significance for Difference of Means

Theorem 4.14.1 (Test for Difference of Means). Let \( \bar {x}_1 \) and \( \bar {x}_2 \) be the means of two independent random samples of sizes \( n_1 \) and \( n_2 \) from populations with means \( \mu _1, \mu _2 \) and variances \( \sigma _1^2, \sigma _2^2 \). For large samples: \[ \bar {x}_1 \sim N\left ( \mu _1, \frac {\sigma _1^2}{n_1} \right ), \quad \bar {x}_2 \sim N\left ( \mu _2, \frac {\sigma _2^2}{n_2} \right ) \] Under \( H_0: \mu _1 = \mu _2 \), the test statistic is: \[ Z = \frac {\bar {x}_1 - \bar {x}_2}{\sqrt {\frac {\sigma _1^2}{n_1} + \frac {\sigma _2^2}{n_2}}} \sim N(0,1) \] If \( \sigma _1^2 \) and \( \sigma _2^2 \) are unknown, for large samples we use: \[ \hat {\sigma }_1^2 = s_1^2, \quad \hat {\sigma }_2^2 = s_2^2 \] Then: \[ Z = \frac {\bar {x}_1 - \bar {x}_2}{\sqrt {\frac {s_1^2}{n_1} + \frac {s_2^2}{n_2}}} \sim N(0,1) \]

Remark 4.14.2. If \( \sigma _1^2 = \sigma _2^2 = \sigma ^2 \), then: \[ Z = \frac {\bar {x}_1 - \bar {x}_2}{\sigma \sqrt {\frac {1}{n_1} + \frac {1}{n_2}}} \sim N(0,1) \]

Example 4.14.3 (Example 12.23). The means of two single large samples of 1000 and 2000 members are 67.5 inches and 68.0 inches respectively. Can the samples be regarded as drawn from the same population of standard deviation 2.5 inches? (Test at 5% level of significance.)

Solution. We are given: \[ n_1 = 1000, \quad \bar {x}_1 = 67.5, \quad n_2 = 2000, \quad \bar {x}_2 = 68.0, \quad \sigma = 2.5 \] Null Hypothesis \( H_0: \mu _1 = \mu _2 \). Alternative Hypothesis \( H_1: \mu _1 \neq \mu _2 \) (Two-tailed). Test Statistic: \begin {align*} Z &= \frac {\bar {x}_1 - \bar {x}_2}{\sigma \sqrt {\frac {1}{n_1} + \frac {1}{n_2}}} \\ &= \frac {67.5 - 68.0}{2.5 \sqrt {\frac {1}{1000} + \frac {1}{2000}}} \\ &= \frac {-0.5}{2.5 \times \sqrt {0.0015}} = \frac {-0.5}{2.5 \times 0.0387} = \frac {-0.5}{0.0968} = -5.16 \end {align*} Conclusion: Since \( |Z| = 5.16 > 3 \), we reject \( H_0 \). The samples are not from the same population.

Example 4.14.4 (Example 12.24). In a survey of buying habits, 400 women shoppers are chosen at random in super market ’A’. Their average weekly food expenditure is Rs. 250 with a standard deviation of Rs. 40. For 400 women shoppers chosen at random in super market ’B’, the average weekly food expenditure is Rs. 220 with a standard deviation of Rs. 55. Test at 1% level of significance whether the average weekly food expenditure of the two populations of shoppers are equal.

Solution. We are given: \[ n_1 = 400, \quad \bar {x}_1 = 250, \quad s_1 = 40 \] \[ n_2 = 400, \quad \bar {x}_2 = 220, \quad s_2 = 55 \] Null Hypothesis \( H_0: \mu _1 = \mu _2 \). Alternative Hypothesis \( H_1: \mu _1 \neq \mu _2 \) (Two-tailed). Test Statistic: \begin {align*} Z &= \frac {\bar {x}_1 - \bar {x}_2}{\sqrt {\frac {s_1^2}{n_1} + \frac {s_2^2}{n_2}}} \\ &= \frac {250 - 220}{\sqrt {\frac {1600}{400} + \frac {3025}{400}}} \\ &= \frac {30}{\sqrt {4 + 7.5625}} = \frac {30}{\sqrt {11.5625}} = \frac {30}{3.40} = 8.82 \end {align*} Conclusion: Since \( |Z| = 8.82 > 2.58 \), we reject \( H_0 \) at 1% level. The average expenditures differ significantly.

Example 4.14.5 (Example 12.25). The average hourly wage of a sample of 150 workers in a plant ’A’ was Rs. 2.56 with a standard deviation of Rs. 1.08. The average wage of a sample of 200 workers in plant ’B’ was Rs. 2.87 with a standard deviation of Rs. 1.28. Can an applicant safely assume that the hourly wages paid by plant ’B’ are higher than those paid by plant ’A’?

Solution. We are given: \[ n_1 = 150, \quad \bar {x}_1 = 2.56, \quad s_1 = 1.08 \] \[ n_2 = 200, \quad \bar {x}_2 = 2.87, \quad s_2 = 1.28 \] Null Hypothesis \( H_0: \mu _1 = \mu _2 \). Alternative Hypothesis \( H_1: \mu _1 < \mu _2 \) (Left-tailed). Test Statistic: \begin {align*} Z &= \frac {\bar {x}_1 - \bar {x}_2}{\sqrt {\frac {s_1^2}{n_1} + \frac {s_2^2}{n_2}}} \\ &= \frac {2.56 - 2.87}{\sqrt {\frac {1.1664}{150} + \frac {1.6384}{200}}} \\ &= \frac {-0.31}{\sqrt {0.007776 + 0.008192}} = \frac {-0.31}{\sqrt {0.015968}} = \frac {-0.31}{0.1264} = -2.45 \end {align*} Conclusion: For a left-tailed test at 5%, critical value is -1.645. Since \( Z = -2.45 < -1.645 \), we reject \( H_0 \). Plant B pays significantly higher wages.

Example 4.14.6 (Example 12.26). In a certain factory there are two independent processes manufacturing the same item. The average weight in a sample of 250 items produced from one process is found to be 120 ozs. with a standard deviation of 12 ozs. while the corresponding figures in a sample of 400 items from the other process are 124 and 14. Obtain the standard error of difference between the two sample means. Is this difference significant? Also find the \( 99\% \) confidence limits for the difference in the average weights of items produced by the two processes respectively.

Solution. We are given: \[ n_1 = 250, \quad \bar {x}_1 = 120, \quad s_1 = 12 \] \[ n_2 = 400, \quad \bar {x}_2 = 124, \quad s_2 = 14 \] Standard Error: \begin {align*} \text {S.E.}(\bar {x}_1 - \bar {x}_2) &= \sqrt {\frac {s_1^2}{n_1} + \frac {s_2^2}{n_2}} \\ &= \sqrt {\frac {144}{250} + \frac {196}{400}} = \sqrt {0.576 + 0.49} = \sqrt {1.066} = 1.032 \end {align*} Test Statistic: \[ Z = \frac {\bar {x}_1 - \bar {x}_2}{\text {S.E.}} = \frac {120 - 124}{1.032} = -3.875 \] Conclusion: Since \( |Z| = 3.875 > 3 \), we reject \( H_0 \). The difference is significant. 99% Confidence Limits: \[ |\bar {x}_1 - \bar {x}_2| \pm 2.58 \times \text {S.E.} = 4 \pm 2.58 \times 1.032 = 4 \pm 2.66 = (1.34, 6.66) \]

Example 4.14.7 (Example 12.27). The mean height of 50 male students who showed above average participation in college athletics was 68.2 inches with a standard deviation of 2.5 inches; while 50 male students who showed no interest in such participation had a mean height of 67.5 inches with a standard deviation of 2.8 inches. (i) Test the hypothesis that male students who participate in college athletics are taller than other male students. (ii) By how much should the sample size of each of the two groups be increased in order that the observed difference of 0.7 inches in the mean heights be significant at the 5% level of significance.

Solution. We are given: \[ n_1 = 50, \quad \bar {x}_1 = 68.2, \quad s_1 = 2.5 \] \[ n_2 = 50, \quad \bar {x}_2 = 67.5, \quad s_2 = 2.8 \] (i) Null Hypothesis \( H_0: \mu _1 = \mu _2 \). Alternative Hypothesis \( H_1: \mu _1 > \mu _2 \) (Right-tailed). Test Statistic: \begin {align*} Z &= \frac {\bar {x}_1 - \bar {x}_2}{\sqrt {\frac {s_1^2}{n_1} + \frac {s_2^2}{n_2}}} \\ &= \frac {68.2 - 67.5}{\sqrt {\frac {6.25}{50} + \frac {7.84}{50}}} \\ &= \frac {0.7}{\sqrt {0.125 + 0.1568}} = \frac {0.7}{\sqrt {0.2818}} = \frac {0.7}{0.5308} = 1.32 \end {align*} Conclusion: For a right-tailed test at 5%, critical value is 1.645. Since \( Z = 1.32 < 1.645 \), we accept \( H_0 \). Athletes are not significantly taller. (ii) We need: \[ \frac {68.2 - 67.5}{\sqrt {\frac {(2.5)^2}{n} + \frac {(2.8)^2}{n}}} \geq 1.645 \] \begin {align*} \frac {0.7}{\sqrt {\frac {14.09}{n}}} &\geq 1.645 \\ \frac {0.7\sqrt {n}}{3.754} &\geq 1.645 \\ \sqrt {n} &\geq \frac {1.645 \times 3.754}{0.7} = 8.82 \\ n &\geq 77.8 \approx 78 \end {align*} Each sample size should be increased by \( 78 - 50 = 28 \).

4.15 Test of Significance for Difference of Standard Deviations

Theorem 4.15.1 (Test for Difference of Standard Deviations). For two independent samples, under \( H_0: \sigma _1 = \sigma _2 \), the test statistic for large samples is: \[ Z = \frac {s_1 - s_2}{\sqrt {\frac {s_1^2}{2n_1} + \frac {s_2^2}{2n_2}}} \sim N(0,1) \]

Example 4.15.2 (Example 12.28). Random samples drawn from two countries gave the following data relating to the heights of adult males: \[ \begin {array}{c|c|c|c} & \text {Mean height (in inches)} & \text {Standard deviation (in inches)} & \text {Number in samples} \\ \hline \text {Country A} & 67.42 & 2.58 & 1000 \\ \text {Country B} & 67.25 & 2.50 & 1200 \end {array} \] (i) Is the difference between the means significant? (ii) Is the difference between the standard deviations significant?

Solution. (i) Test for Means: \[ Z = \frac {67.42 - 67.25}{\sqrt {\frac {(2.58)^2}{1000} + \frac {(2.50)^2}{1200}}} = \frac {0.17}{\sqrt {0.00666 + 0.00521}} = \frac {0.17}{0.1089} = 1.56 \] Since \( |Z| = 1.56 < 1.96 \), the difference in means is not significant. (ii) Test for Standard Deviations: \begin {align*} Z &= \frac {2.58 - 2.50}{\sqrt {\frac {(2.58)^2}{2 \times 1000} + \frac {(2.50)^2}{2 \times 1200}}} \\ &= \frac {0.08}{\sqrt {\frac {6.6564}{2000} + \frac {6.25}{2400}}} = \frac {0.08}{\sqrt {0.003328 + 0.002604}} = \frac {0.08}{0.0770} = 1.04 \end {align*} Since \( |Z| = 1.04 < 1.96 \), the difference in standard deviations is not significant.

Example 4.15.3 (Example 12.29). Two populations have their means equal, but S.D. of one is twice the other. Show that in the samples of size 2000 from each drawn under simple sampling conditions, the difference of means will, in all probability, not exceed \(0.15\sigma \), where \(\sigma \) is the smaller S.D. What is the probability that the difference will exceed half this amount?

Solution. Let the standard deviations be \( \sigma \) and \( 2\sigma \), and \( n_1 = n_2 = 2000 \). \[ \text {S.E.}(\bar {x}_1 - \bar {x}_2) = \sqrt {\frac {\sigma ^2}{2000} + \frac {4\sigma ^2}{2000}} = \sigma \sqrt {\frac {5}{2000}} = \frac {\sigma }{20} = 0.05\sigma \] In all probability, \( |\bar {x}_1 - \bar {x}_2| < 3 \times \text {S.E.} = 0.15\sigma \). \[ P(|\bar {x}_1 - \bar {x}_2| > 0.075\sigma ) = P\left ( \left | \frac {\bar {x}_1 - \bar {x}_2}{0.05\sigma } \right | > 1.5 \right ) = P(|Z| > 1.5) = 0.1336 \]