Basic Business Statistics 1B Study Notes

Fundamentals of Statistics: Populations, Samples, and Variables

• Random Variable: Any attribute or characteristic being measured or observed.
• Data: The actual values (numbers) or outcomes recorded on random variables.
• Sampling Unit: The item being measured, observed, or counted with respect to the random variable(s) under study.
• Population: Every possible item that contains a data value of the random variable under study.
• Sample: A subset of items drawn from a population.
• Census: The gathering of data on a random variable from every member of the target population.
• Survey: The gathering of data on a random variable from a sample.
• Parameter: The numeric characteristic of a population.
• Statistic: The numeric characteristic of a sample.
• Sampling: The process of coming up with the sample.

Comparison of Population Parameters and Sample Statistics

• Single Mean:
  • Population Parameter: $\mu$
  • Sample Statistic: $\bar{x}$
• Single Proportion:
  • Population Parameter: $\pi$
  • Sample Statistic: $p$
• Difference Between Two Means:
  • Population Parameter: $\mu_1 - \mu_2$
  • Sample Statistic: $\bar{x}_1 - \bar{x}_2$
• Difference Between Two Proportions:
  • Population Parameter: $\pi_1 - \pi_2$
  • Sample Statistic: $p_1 - p_2$

Sampling Techniques and Data Collection Errors

Reasons for Sampling

• Cost: It is generally less expensive to gather sample data than population data.
• Timeliness: Sample data can be gathered much quicker than population data.
• Destructive Testing: Data on certain random variables can only be generated by consuming or destroying the sampling unit.
• Accuracy: Better control can usually be exercised over the data collection process in a sample than in an entire population.

Classification of Sampling Methods

Sampling methods are broadly categorized into Probability Sampling Methods (random-based) and Non-Probability Sampling Methods.

Probability Sampling Methods

Probability sampling includes any selection method where sample members are chosen from the target population on a purely random basis.

• Advantages:

  • Reduces the likelihood of selection bias.
  • Makes it possible to quantitatively measure sampling error.

• Four Types of Probability Sampling:

  1. Simple Random Sampling: Every member in the target population has an equal chance of being selected. This method is used when the population is assumed to be relatively homogeneous with respect to the random variable under study.
  2. Systematic Random Sampling: Similar to simple random sampling, but differs in how units are selected. Sampling begins by randomly selecting the first sampling unit from the sampling frame, after which subsequent units are selected at a uniform interval relative to the first unit.
  3. Stratified Random Sampling: Used when the population is assumed to be heterogeneous with respect to the random variable under study. The population is divided into segments (strata) where members in each stratum are relatively homogeneous. Random samples are then independently selected from each stratum.
  4. Cluster Random Sampling: Used when the target population can be naturally divided into clusters where each cluster has a profile similar to every other cluster. A subset of clusters is randomly selected for full sampling. This method is typically used when the population is large and geographically dispersed.

Non-Probability Sampling Methods

• Four Types of Non-Probability Sampling:
  1. Convenience Sampling: Applied when sampling units are selected purely to suit the convenience of the researcher.
  2. Judgment Sampling: Samples drawn based on the researcher's personal knowledge and experience in selecting the most appropriate units.
  3. Quota Sampling: Proportions of various subgroups in the population are determined, and the sample is drawn to reflect those exact percentages. It is similar to stratified random sampling, except that selection within subgroups lacks randomization.
  4. Snowball Sampling: Used to reach target populations where sampling units are difficult to identify. Each identified member of the target population is asked to identify other units belonging to the same target population.

Errors in Statistical Analysis

• Sampling Errors: Errors that occur as a result of using a sample rather than considering the entire population. It is defined as the numerical difference between the actual population parameter value and its corresponding sample statistic.
• Non-Sampling Errors: Errors that occur during data collection processes that are not caused by using a sample. Examples include misrecording responses, interviewer errors, lack of interviewer knowledge, or flawed measurement instruments.

Parameter Estimation and Sample Size Determination

Key Factors in Determining Sample Size

1. Degree of Confidence: The probability associated with a measure of confidence that the required sample size will yield an accurate estimate. Higher confidence (e.g., $99\%$) implies greater accuracy and requires a larger sample size.
2. Maximum Allowable Error ($E$): The maximum sampling error tolerated at a specified level of confidence.
3. Variation of the Population: Measured by the population standard deviation ($\sigma$), which must be known or estimated.

Sample Size Determination for Estimating a Population Mean

• Formula:$n = \left(\frac{Z \times \sigma}{E}\right)^2$   Where:

  • $Z$ = Critical value from the standardized normal distribution corresponding to the specified confidence level.
  • $\sigma$ = Population standard deviation.
  • $E$ = Maximum allowable sampling error.

• Worked Example 1 (Plumbing Company):

  • Desired sampling error $E = \pm 5$
  • Confidence level = $95\%$ ($Z = 1.96$
  • Standard deviation $\sigma = \$25$
  • Calculation:     $n = \left(\frac{1.96 \times 25}{5}\right)^2 = (9.8)^2 = 96.04 \approx 97$

• Worked Example 2 (Home Heating Oil Usage):

  • Desired sampling error $E = \pm 50$
  • Confidence level = $90\%$ ($Z = 1.645$
  • Standard deviation $\sigma = 325$
  • Calculation:     $n = \left(\frac{1.645 \times 325}{50}\right)^2 = (10.6925)^2 = 114.33 \approx 115$

• Exercise Problem:

  • Desired confidence level = $99\%$ ($Z = 2.58$
  • Sampling error $E = \pm 20$
  • Standard deviation $\sigma = 100$
  • Calculation:     $n = \left(\frac{2.58 \times 100}{20}\right)^2 = (12.9)^2 = 166.41 \approx 167$

Sample Size Determination for Estimating a Population Proportion

• Formula:$n = \frac{Z^2 \times p(1-p)}{E^2}$   Where:

  • $p$ = Estimated proportion of success in the population.
  • Rule: If no prior information exists regarding the proportion of success, fix $p = 0.05$.

• Worked Example 1 (Sales Invoices Errors):

  • Confidence level = $95\%$ ($Z = 1.96$
  • Sampling error $E = \pm 0.07$
  • Past data maximum proportion $p = 0.15$
  • Calculation:     $n = \frac{(1.96)^2 \times 0.15 \times (1 - 0.15)}{(0.07)^2} = \frac{3.8416 \times 0.1275}{0.0049} = \frac{0.489804}{0.0049} = 99.96 \approx 100$

• Worked Example 2 (Non-Conforming Newspapers):

  • Confidence level = $90\%$ ($Z = 1.645$
  • Sampling error $E = \pm 0.05$
  • No past information available ($p = 0.05$
  • Calculation:     $n = \frac{(1.645)^2 \times 0.05 \times (1 - 0.05)}{(0.05)^2} = \frac{2.706025 \times 0.0475}{0.0025} = 51.41 \approx 52$

• Exercise Problem:

  • Confidence level = $90\%$ ($Z = 1.645$
  • Sampling error $E = \pm 0.04$
  • No prior information ($p = 0.05$
  • Calculation:     $n = \frac{(1.645)^2 \times 0.05 \times (1 - 0.05)}{(0.04)^2} = \frac{0.128536}{0.0016} = 80.345 \approx 81$

Sampling Distributions and the Central Limit Theorem

Definitions and Theoretical Properties

• Sampling Distribution: Shows the relationship between a sample statistic (e.g., sample mean $\bar{x}$) and its corresponding population parameter (e.g., population mean $\mu$). The distribution of all possible sample means and their associated probabilities is the sampling distribution of the means.
• Properties of the Sampling Distribution of the Sample Mean:
  1. Sample means are random variables because their values vary from sample to sample.
  2. The mean of all possible sample means ($\mu_{\bar{x}}$) equals the true population mean ($\mu$):      $\mu_{\bar{x}} = \bar{\bar{x}} = \mu$

The Central Limit Theorem (CLT)

If a random sample of $n$ observations is selected from any population, then when the sample size is sufficiently large ($n \ge 30$), the sampling distribution of the mean tends to approximate a normal distribution regardless of the underlying population shape.

• $Z$-Transformation Formula for Sample Means:$Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

Probability Calculations with Known Population Parameters

• Example 1 (Highway Truck Speeds):

  • Population parameter: speed limit $\mu = 60\text{ km/h}$, $\sigma = 12.5\text{ km/h}$
  • Sample details: $n = 120$, $\bar{x} = 62\text{ km/h}$
  • Problem: Find $P(\bar{x} \ge 62)$
  • $Z$-score calculation:     $Z = \frac{62 - 60}{\frac{12.5}{\sqrt{120}}} = \frac{2}{1.141} = 1.75$
  • Probability: $P(Z \ge 1.75) = 0.5 - 0.4599 = 0.0401$

• Example 2 (High School Boys IQ):

  • $\mu = 100$, $\sigma = 16$, $n = 36$
  • Problem: Find $P(\bar{x} \le 96)$
  • $Z$-score calculation:     $Z = \frac{96 - 100}{\frac{16}{\sqrt{36}}} = \frac{-4}{2.667} = -1.50$
  • Probability: $P(Z \le -1.50) = 0.5 - 0.4332 = 0.0668$

• Example 3 (Boxers Weight Distribution):

  • $\mu = 94\text{ kg}$, $\sigma = 4.2\text{ kg}$, $n = 81$
  • Problem: Find $P(93.5 \le \bar{x} \le 95.1)$
  • $Z$-score calculations:     $Z_1 = \frac{93.5 - 94}{\frac{4.2}{\sqrt{81}}} = \frac{-0.5}{0.4667} = -1.07$$Z_2 = \frac{95.1 - 94}{\frac{4.2}{\sqrt{81}}} = \frac{1.1}{0.4667} = 2.36$
  • Probability: $P(-1.07 \le Z \le 2.36) = 0.3577 + 0.4909 = 0.8486$

Probability Calculations with Unknown Population Means

Used to evaluate the probability that a single sample mean lies within a specific margin of error around an unknown population mean.

• Required Conditions:

  1. Normally distributed population.
  2. Population standard deviation ($\sigma$) is known.
  3. Large sample size ($n \ge 30$

• Example 1 (Typists Speed):

  • $n = 100$, $\sigma = 8\text{ wpm}$
  • Margin: Differs from $\mu$ by no more than $1\text{ wpm}$
  • $Z = \frac{1}{\frac{8}{\sqrt{100}}} = \frac{1}{0.8} = 1.25$
  • Probability: $P(-1.25 \le Z \le 1.25) = 2 \times 0.3944 = 0.7888$

• Example 2 (BBS1b Student Marks):

  • $n = 81$, $\sigma = 6\%$
  • Margin: Differs from $\mu$ by no more than $2\%$
  • $Z = \frac{2}{\frac{6}{\sqrt{81}}} = \frac{2}{0.667} = 3.00$
  • Probability: $P(-3.00 \le Z \le 3.00) = 2 \times 0.4987 = 0.9974$

• Exercise (Newborn Babies Weight):

  • $n = 225$, $\sigma = 0.25\text{ kg}$
  • Margin: Differs from $\mu$ by no more than $0.04\text{ kg}$
  • $Z = \frac{0.04}{\frac{0.25}{\sqrt{225}}} = \frac{0.04}{0.01667} = 2.40$
  • Probability: $P(-2.40 \le Z \le 2.40) = 2 \times 0.4918 = 0.9836$

Sampling Distribution of the Sample Proportion

• The sample proportion ($p$) measures central location when the random variable under study is qualitative.

• Properties:

  1. Sample proportions are random variables that vary across samples.
  2. The mean of all sample proportions equals the true population proportion ($\pi$).
  3. Standard error of the sample proportion:      $\sigma_p = \sqrt{\frac{\pi(1-\pi)}{n}}$
  4. The shape of the sampling distribution of sample proportions is normally distributed.

• $Z$-Transformation Formula:$Z = \frac{p - \pi}{\sqrt{\frac{\pi(1-\pi)}{n}}}$

• Example (Burglarized Homes):

  • $\pi = 0.55$, $n = 100$, observed burglaries = $64$ ($p = 0.64$
  • Calculation:     $Z = \frac{0.64 - 0.55}{\sqrt{\frac{0.55(1-0.55)}{100}}} = \frac{0.09}{\sqrt{0.002475}} = \frac{0.09}{0.04975} = 1.81$

Interval Estimation for Means and Proportions

Point Estimation vs. Confidence Interval Estimation

• Point Estimate: A single numeric value calculated from a sample statistic used to represent a true but unknown population parameter (e.g., $\bar{x}$ is a point estimate of $\mu$
• Point Estimator: The formula or rule producing the point estimate.
• Disadvantage: Gives no information regarding reliability or closeness to the actual parameter.
• Confidence Interval (CI): A range of values constructed around a sample statistic within which the population parameter is expected to fall at a specified confidence level.
  • Level of Confidence: The probability that the constructed interval includes the true population parameter.
  • Margin of Error: The maximum distance added to and subtracted from the statistic.

CI Estimation for Single Mean ($\sigma$ Known)

• Formula:$\bar{x} \pm Z \left(\frac{\sigma}{\sqrt{n}}\right)$

• Assumptions:

  1. Population standard deviation ($\sigma$) is known.
  2. Sample size is large ($n \ge 30$
  3. Population is normally distributed.

• Worked Example 1 (Windhoek Car Commuters):

  • $n = 100$, $\bar{x} = 35.8\text{ minutes}$, $\sigma = 20\text{ minutes}$, Level = $95\%$ ($Z = 1.96$
  • Calculation:     $\text{CI} = 35.8 \pm 1.96 \left(\frac{20}{\sqrt{100}}\right) = 35.8 \pm 1.96(2) = 35.8 \pm 3.92 = [31.88, 39.72]\text{ minutes}$
  • Interpretation: We are $95\%$ confident that the true population mean daily travel time for all car commuters in Windhoek is between $31.88$ and $39.72$ minutes.

• Worked Example 2 (Radial Tires Tread Life):

  • $n = 49$, $\bar{x} = 52345\text{ km}$, $\sigma = 12943\text{ km}$, Level = $99\%$ ($Z = 2.58$
  • Calculation:     $\text{CI} = 52345 \pm 2.58 \left(\frac{12943}{\sqrt{49}}\right) = 52345 \pm 2.58(1849) = 52345 \pm 4770.42 = [47574.58, 57115.42]\text{ km}$

• Homework Problem (Mores Desserts):

  • $n = 49$, $\bar{x} = \text{R}3140$, $\sigma = \text{R}345$, Level = $98\%$ ($Z = 2.33$
  • Calculation:     $\text{CI} = 3140 \pm 2.33 \left(\frac{345}{\sqrt{49}}\right) = 3140 \pm 2.33(49.286) = 3140 \pm 114.84 = [\text{R}3025.16, \text{R}3254.84]$

Distribution Selection Rules ($Z$ vs. $t$)

• When population is normally distributed and $\sigma$ is known: Use $Z$-distribution regardless of sample size.
• When population is normally distributed and $\sigma$ is unknown:
  • Large samples ($n \ge 30$): Use $Z$-distribution.
  • Small samples ($n < 30$): Use Student $t$-distribution.

Properties of Student $t$-Distribution

• Bell-shaped and symmetrical about its mean of zero.
• Standard deviation is not constant; it changes based on sample size.
• Standard deviation depends on sample standard deviation ($s$) and degrees of freedom ($df = n - 1$

CI Estimation for Single Mean ($\sigma$ Unknown, $n < 30$)

• Formula:$\bar{x} \pm t_{\frac{\alpha}{2}, n-1} \left(\frac{s}{\sqrt{n}}\right)$

• Worked Example 1 (Credit Card Purchases):

  • $n = 25$, \bar{x} = \text{N}\\$165.45, s = \text{N}\\$38.60, Level = $95\%$
  • Degrees of freedom $df = 25 - 1 = 24$; Critical value $t_{0.025, 24} = 2.064$
  • Calculation:     $\text{CI} = 165.45 \pm 2.064 \left(\frac{38.60}{\sqrt{25}}\right) = 165.45 \pm 2.064(7.72) = 165.45 \pm 15.93 = [\text{N}\\$149.52, \text{N}\\$181.38]$

• Worked Example 2 (Taxi Travel Times):

  • $n = 49$, $\bar{x} = 20\text{ mins}$, $s = 6\text{ mins}$
  • For $98\%$ Level ($Z = 2.33$ as $n \ge 30$$\text{CI} = 20 \pm 2.33 \left(\frac{6}{\sqrt{49}}\right) = 20 \pm 2.33(0.857) = 20 \pm 2.00 = [18.00, 22.00]\text{ mins}$
  • For $90\%$ Level ($Z = 1.645$$\text{CI} = 20 \pm 1.645 \left(\frac{6}{\sqrt{49}}\right) = 20 \pm 1.645(0.857) = 20 \pm 1.41 = [18.59, 21.41]\text{ mins}$

Precision of Confidence Intervals

• Precision is determined by interval width:
  • Narrower CI = Higher precision.
  • Wider CI = Lower precision.
• Width is influenced by: confidence level, sample size ($n$), and population variation ($\sigma$

CI Estimation for Single Population Proportion

• Formula for Large Samples ($n \ge 30$):$p \pm Z \sqrt{\frac{p(1-p)}{n}}$

• Formula for Small Samples ($n < 30$):$p \pm t \sqrt{\frac{p(1-p)}{n}}$

• Worked Example 1 (Johannesburg Street Vendors):

  • $n = 120$, $x = 42$ vendors ($p = \frac{42}{120} = 0.35$
  • Confidence level = $95\%$ ($Z = 1.96$
  • Calculation:     $\text{CI} = 0.35 \pm 1.96 \sqrt{\frac{0.35(0.65)}{120}} = 0.35 \pm 1.96(0.0435) = 0.35 \pm 0.0853 = [0.2647, 0.4353]$

• Worked Example 2 (Overdrawn Cheque Accounts):

  • $n = 29$, $x = 6$ overdrawn ($p = \frac{6}{29} = 0.2069$
  • Level = $95\%$, $df = 28$, $t_{0.025, 28} = 2.048$
  • Calculation:     $\text{CI} = 0.2069 \pm 2.048 \sqrt{\frac{0.2069(0.7931)}{29}} = 0.2069 \pm 2.048(0.0752) = 0.2069 \pm 0.1540 = [0.0529, 0.3609]$

Principles and Applications of Hypothesis Testing

Definition and Purpose

Hypothesis testing is a statistical process used to test the validity of a claim regarding the true value of a population parameter. It determines whether sample analysis provides sufficient evidence to conclude that the population possesses a specific property.

Five-Step Procedure

1. Formulate Hypotheses: Define Null Hypothesis ($H_0$) and Alternative Hypothesis ($H_1$
2. Decision Rule: Determine critical value limits ($Z$ or $t$) based on significance level ($\alpha$) and directionality.
3. Compute Sample Test Statistic: Calculate $Z_{cal}$ or $t_{cal}$.
4. Comparison: Compare calculated statistic with critical limits.
5. Conclusion: State conclusion to reject or fail to reject $H_0$

Hypothesis Formulation Structures

• One-sided Upper-tailed Test:
  • $H_0: \mu \le a$
  • $H_1: \mu > a$
  • Reject $H_0$ if $Z_{cal} > Z_{\alpha}$
• One-sided Lower-tailed Test:
  • $H_0: \mu \ge a$
  • $H_1: \mu < a$
  • Reject $H_0$ if $Z_{cal} < -Z_{\alpha}$
• Two-sided Test:
  • $H_0: \mu = a$
  • $H_1: \mu \neq a$
  • Reject $H_0$ if $|Z_{cal}| > Z_{\frac{\alpha}{2}}$

Test Statistic Formulas

• For Single Sample Mean:$Z_{cal} = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \quad \text{or} \quad t_{cal} = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$
• For Single Sample Proportion:$Z_{cal} = \frac{p - \pi}{\sqrt{\frac{\pi(1-\pi)}{n}}}$

Standard $Z$-Limits for Critical Values

• $99\%$ Confidence ($\alpha = 0.01$):
  • Two-sided: $\pm 2.58$
  • Lower-tailed: $-2.33$
  • Upper-tailed: $+2.33$
• $95\%$ Confidence ($\alpha = 0.05$):
  • Two-sided: $\pm 1.96$
  • Lower-tailed: $-1.645$
  • Upper-tailed: $+1.645$
• $90\%$ Confidence ($\alpha = 0.10$):
  • Two-sided: $\pm 1.645$
  • Lower-tailed: $-1.28$
  • Upper-tailed: $+1.28$

Decision Errors

• Type I Error ($\alpha$): Rejecting the null hypothesis $H_0$ when it is actually true.
• Type II Error ($\beta$): Failing to reject the null hypothesis $H_0$ when it is actually false.

Worked Examples on Single Mean

• Example 1 (Domestic Cleaners Hourly Wage):

  • Claim: \mu = \text{N}\\$2.85, $n = 250$, \bar{x} = \text{N}\\$3.03, \sigma = \text{N}\\$1.00, $\alpha = 0.05$
  • Hypotheses: $H_0: \mu = 2.85 \quad \text{vs} \quad H_1: \mu \neq 2.85$
  • Decision Rule: Reject $H_0$ if $|Z_{cal}| > 1.96$
  • Test Statistic:     $Z_{cal} = \frac{3.03 - 2.85}{\frac{1.00}{\sqrt{250}}} = \frac{0.18}{0.0632} = 2.85$
  • Decision: Since $2.85 > 1.96$, reject $H_0$. Average hourly rate differs significantly from N\$2.85.

• Example 2 (Typing Machine Battery Life):

  • Claim: Stop buying if $\mu < 24\text{ months}$, $n = 9$, $\bar{x} = 22$, $s = 3$, $\alpha = 0.10$
  • Hypotheses: $H_0: \mu \ge 24 \quad \text{vs} \quad H_1: \mu < 24$
  • Critical value: $df = 8$, $t_{0.10, 8} = -1.397$
  • Test Statistic:     $t_{cal} = \frac{22 - 24}{\frac{3}{\sqrt{9}}} = \frac{-2}{1} = -2.00$
  • Decision: Since $-2.00 < -1.397$, reject $H_0$. School should stop buying these batteries.

• Example 3 (Namibian CEOs Work Hours):

  • Claim: $\mu < 60\text{ hours}$, $n = 90$, $\bar{x} = 61.3$, $\sigma = 8.8$, $\alpha = 0.01$
  • Hypotheses: $H_0: \mu \ge 60 \quad \text{vs} \quad H_1: \mu < 60$
  • Decision Rule: Reject $H_0$ if $Z_{cal} < -2.33$
  • Test Statistic:     $Z_{cal} = \frac{61.3 - 60}{\frac{8.8}{\sqrt{90}}} = \frac{1.3}{0.9276} = 1.40$
  • Decision: Since $1.40 > -2.33$, fail to reject $H_0$

• Example 4 (Battery Life Exceeds 50 Hours):

  • Claim: $\mu > 50$, $n = 36$, $\bar{x} = 52$, $\sigma = 6$, $\alpha = 0.05$
  • Hypotheses: $H_0: \mu \le 50 \quad \text{vs} \quad H_1: \mu > 50$
  • Decision Rule: Reject $H_0$ if $Z_{cal} > 1.645$
  • Test Statistic:     $Z_{cal} = \frac{52 - 50}{\frac{6}{\sqrt{36}}} = \frac{2}{1} = 2.00$
  • Decision: Since $2.00 > 1.645$, reject $H_0$. Mean battery life is significantly greater than 50 hours.

Worked Examples on Single Proportion

• Example 1 (Polytechnic Students Newspaper Reading):

  • Claim: $\pi = 0.25$, $n = 200$, $x = 45$ readers ($p = \frac{45}{200} = 0.225$
  • Hypotheses: $H_0: \pi = 0.25 \quad \text{vs} \quad H_1: \pi \neq 0.25$
  • Decision Rule: Reject $H_0$ if $|Z_{cal}| > 1.96$
  • Test Statistic:     $Z_{cal} = \frac{0.225 - 0.25}{\sqrt{\frac{0.25(0.75)}{200}}} = \frac{-0.025}{0.0306} = -0.817$
  • Decision: Fail to reject $H_0$. Statement is accurate.

• Example 2 (Auro Cosmetics Market Share):

  • Claim: $\pi > 0.35$, $n = 360$, $x = 108$ ($p = 0.30$
  • Hypotheses: $H_0: \pi \le 0.35 \quad \text{vs} \quad H_1: \pi > 0.35$
  • Decision Rule: Reject $H_0$ if $Z_{cal} > 2.33$
  • Test Statistic:     $Z_{cal} = \frac{0.30 - 0.35}{\sqrt{\frac{0.35(0.65)}{360}}} = \frac{-0.05}{0.0251} = -1.99$
  • Decision: Fail to reject $H_0$

• Example 3 (Manufacturer Market Share):

  • Claim: $\pi < 0.60$, $n = 500$, $x = 275$ ($p = 0.55$
  • Hypotheses: $H_0: \pi \ge 0.60 \quad \text{vs} \quad H_1: \pi < 0.60$
  • Decision Rule: Reject $H_0$ if $Z_{cal} < -1.645$
  • Test Statistic:     $Z_{cal} = \frac{0.55 - 0.60}{\sqrt{\frac{0.60(0.40)}{500}}} = \frac{-0.05}{0.0219} = -2.28$
  • Decision: Reject $H_0$

Statistical Inferences on Population Variance and Standard Deviation

Sample Variance

• Formula:$S^2 = \frac{\sum x_i^2 - n \bar{x}^2}{n - 1}$

Confidence Interval Estimation for Single Variance ($\sigma^2$)

To construct a CI for a single unknown population variance, the Chi-Square ($\chi^2$) distribution is used.

• Interval Definition:$LCL \le \sigma^2 \le UCL$
• Limits Formulas:$LCL = \frac{(n-1) \times S^2}{\chi^2_{\frac{\alpha}{2}, n-1}}, \quad UCL = \frac{(n-1) \times S^2}{\chi^2_{1-\frac{\alpha}{2}, n-1}}$
• Limits Formulas for Standard Deviation ($\sigma$):$LCL = \sqrt{\frac{(n-1) \times S^2}{\chi^2_{\frac{\alpha}{2}, n-1}}}, \quad UCL = \sqrt{\frac{(n-1) \times S^2}{\chi^2_{1-\frac{\alpha}{2}, n-1}}}$

Characteristics of the Chi-Square Distribution

1. Non-symmetric shape (positively skewed).
2. Values are always non-negative (positive).
3. Applicable under conditions with small sample sizes.

Hypothesis Testing for Single Population Variance

• Hypotheses (Two-sided):

  • $H_0: \sigma^2 = a$
  • $H_1: \sigma^2 \neq a$

• Decision Rule: Reject $H_0$ if $\chi^2_{stat} > \chi^2_{1-\frac{\alpha}{2}, n-1}$ or $\chi^2_{stat} < \chi^2_{\frac{\alpha}{2}, n-1}$

• Test Statistic Formula:$\chi^2_{cal} = \frac{(n-1)S^2}{\sigma_0^2}$

• Worked Example 1 (Post Office Waiting Times):

  • Historical standard deviation $\sigma = 7.2\text{ mins}$ ($\sigma^2 = 51.84$
  • Sample $n = 25$, $s = 3.5\text{ mins}$ ($S^2 = 12.25$
  • Level $\alpha = 0.10$, $df = 24$
  • Hypotheses: $H_0: \sigma^2 = 51.84 \quad \text{vs} \quad H_1: \sigma^2 \neq 51.84$
  • Test Statistic:     $\chi^2_{cal} = \frac{(25-1)(12.25)}{51.84} = \frac{294}{51.84} = 5.67$
  • Critical values ($df=24$): $\chi^2_{0.05, 24} = 36.415$, $\chi^2_{0.95, 24} = 13.848$
  • Decision: Since $5.67 < 13.848$, reject $H_0$

• Worked Example 2 (Process Control Variance):

  • Process in control if variance $\sigma^2 < 0.05$
  • Sample $n = 26$, $S^2 = 0.06$
  • Level $\alpha = 0.10$, $df = 25$
  • Hypotheses: $H_0: \sigma^2 \le 0.05 \quad \text{vs} \quad H_1: \sigma^2 > 0.05$
  • Test Statistic:     $\chi^2_{cal} = \frac{(26-1)(0.06)}{0.05} = \frac{1.5}{0.05} = 30.00$
  • Critical value: $\chi^2_{0.10, 25} = 34.382$
  • Decision: Since $30.00 < 34.382$, fail to reject $H_0$. Process is in control.

Chi-Square Tests for Independence and Contingency Tables

Overview and Classification

A Chi-square test is a test of significance based on bivariate contingency tables, measuring differences between observed frequencies ($f_o$) and expected frequencies ($f_e$).

• Three Basic Chi-Square Tests:
  1. Test for independence of association.
  2. Test for equality (homogeneity) in more than two population proportions.
  3. Chi-square goodness-of-fit test.

Test for Independence of Association

• Hypotheses:
  • $H_0$: The variables are independent (no association).
  • $H_1$: The variables are dependent (there is an association).
• Decision Rule: Reject $H_0$ if $\chi^2_{cal} > \chi^2_{(\alpha, df)}$
  • Degrees of Freedom: $df = (r-1)(c-1)$, where $r$ = number of rows, $c$ = number of columns.
• Test Statistic Formula:$\chi^2_{cal} = \sum \frac{(f_o - f_e)^2}{f_e}$
• Expected Frequency Formula:$f_e = \frac{\text{row total} \times \text{column total}}{\text{grand total}} = \frac{RT \times CT}{GT}$

Practical Contingency Table Problems

• Example 1 (Gender vs. Magazine Preference):

  • Significance Level $\alpha = 0.05$
  • Data Table:
  • Girls: Beat = $18$, Youth = $12$, Grow = $20$, Live = $28$ (Row Total = $78$
  • Boys: Beat = $38$, Youth = $26$, Grow = $34$, Live = $24$ (Row Total = $122$
  • Column Totals: Beat = $56$, Youth = $38$, Grow = $54$, Live = $52$ (Grand Total = $200$
  • $df = (2-1)(4-1) = 3$
  • Critical Value: $\chi^2_{0.05, 3} = 7.815$

• Example 2 (Buyer's Age vs. Car Size Bought):

  • Significance Level $\alpha = 0.01$
  • Data Table:
  • Under 30: Small = $10$, Medium = $22$, Large = $34$ (Row Total = $66$
  • 30-45: Small = $24$, Medium = $42$, Large = $48$ (Row Total = $114$
  • Over 45: Small = $52$, Medium = $32$, Large = $36$ (Row Total = $120$
  • Column Totals: Small = $86$, Medium = $96$, Large = $118$ (Grand Total = $300$
  • $df = (3-1)(3-1) = 4$
  • Critical Value: $\chi^2_{0.01, 4} = 13.277$

• Example 3 (Warehouse Advertisement vs. Response Type):

  • Significance Level $\alpha = 0.01$
  • Data Table:
  • Response: TV = $200$, Radio = $150$, Newspaper = $175$ (Row Total = $525$
  • Non-response: TV = $300$, Radio = $250$, Newspaper = $125$ (Row Total = $675$
  • Column Totals: TV = $500$, Radio = $400$, Newspaper = $300$ (Grand Total = $1200$
  • $df = (2-1)(3-1) = 2$
  • Critical Value: $\chi^2_{0.01, 2} = 9.210$

Time-Series Analysis, Trend Estimation, and Forecasting

Fundamental Concepts

• Time Series: A set of quantitative observations recorded over regular intervals of time (hourly, daily, weekly, monthly, quarterly, annually).
• Core Assumption: Continuation of past patterns into the future (stable operational environment).
• Purpose: To identify recurring patterns for forecasting future values and to isolate environmental influences.
• Cross-Sectional vs. Time-Series Data:
  • Cross-Sectional Data: Data gathered from a sample survey at a single point in time.
  • Time Series Data: Data gathered over time across sequential periods.

Decomposition of Time Series: Multiplicative Model

• Formula:$y = T \times C \times S \times I$   Where:
  • $y$ = Actual observed value.
  • $T$ = Long-term Trend component.
  • $C$ = Cyclical fluctuation component.
  • $S$ = Seasonal pattern component.
  • $I$ = Irregular/random variation component.
• Note: Statistical methods effectively isolate Trend ($T$) and Seasonal ($S$) components; Cyclical ($C$) and Irregular ($I$) components are difficult to quantify precisely.

Trend Fitting Methods

1. Linear Regression Analysis (Least Squares Method)

Fits a straight trend line equation $\hat{y} = a + bx$ where time ($x$) is coded.

• Sequential Coding Method:   Assign sequential numbers $0, 1, 2, 3, \dots$ or $1, 2, 3, 4, \dots$ to consecutive time periods.

  • Formulas:     $b = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$$a = \bar{y} - b \bar{x}$

• Zero-Sum Coding Method:   Codes $x$ such that $\sum x = 0$

  • Odd Number of Periods ($n$): Set center period $x = 0$, first period $x = -\frac{n-1}{2}$, increment by $1$
  • Even Number of Periods ($n$): First period $x = -(n-1)$, increment by $2$
  • Simplified Coefficient Formulas:     $b = \frac{\sum xy}{\sum x^2}$$a = \frac{\sum y}{n}$

• Worked Example 1 (Namibia Electricity Demand 1993–2003):

  • Dataset ($n = 11$ years): 1993 (3.3), 1994 (3.8), 1995 (3.5), 1996 (3.3), 1997 (3.1), 1998 (3.4), 1999 (3.7), 2000 (3.9), 2001 (4.1), 2002 (4.5), 2003 (4.9).
  • Sequential coding starting at $x = 1$ (1993) to $x = 11$ (2003).
  • Evaluation yields equation coefficients to forecast for 1990 ($x = -2$) and 2007 ($x = 15$

• Worked Example 2 (Rideon Motor Vehicle Sales):

  • Data ($n = 7$ quarters): 1990 Q1 (117), Q2 (144), Q3 (191), Q4 (198); 1991 Q1 (270), Q2 (243), Q3 (300).
  • Estimation of 1991 Q4 ($x = 8$) and target value prediction for $y = 400$

2. Moving Average Method

Used when a specific mathematical trend line equation is not required. Smooths short-term fluctuations by taking averages of consecutive groups of observations.

Economic Index Numbers: Construction and Interpretation

Definitions and Interpretation

An index number is a single summary statistic that measures overall relative changes in price, quantity, or value from a base period to a current period.

• Base period index value is defined as $100$
• Interpretation:
  • $\text{Index} - 100$ = Percentage increase over base period.
  • $100 - \text{Index}$ = Percentage decrease from base period.
  • Example: An index of $104$ indicates a $4\%$ increase over the base period.

Classification of Index Numbers

• Price Indexes: Single Price Index, Composite Price Index.
• Quantity Indexes: Single Quantity Index, Composite Quantity Index.

Mathematical Notation

• $p_0$ = Base period price.
• $p_1$ = Current period price.
• $q_0$ = Base period quantity.
• $q_1$ = Current period quantity.

Single Price Index (Price Relative)

$\text{Price Relative} = \frac{p_1}{p_0} \times 100$

• Worked Example (Cape Town Fuel Prices, Base 2004 = R3.87):
  • 2004: $p_0 = 3.87 \implies \text{Index} = 100.0$
  • 2005: $p_1 = 4.22 \implies \frac{4.22}{3.87} \times 100 = 109.04$
  • 2006: $p_1 = 4.68 \implies \frac{4.68}{3.87} \times 100 = 120.93$
  • 2007: $p_1 = 5.29 \implies \frac{5.29}{3.87} \times 100 = 136.69$

Composite Price Index Methods

1. Laspeyres Price Index (LPI)

Uses base period quantities ($q_0$) as weights. $\text{LPI} = \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100$

2. Paasche Price Index (PPI)

Uses current period quantities ($q_1$) as weights. $\text{PPI} = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100$

• Worked Calculation Example (Car Dealer Sales 1995 vs. 1997):
  • 1995 Data (Base 0):
  • Model A: $p_0 = 40$, $q_0 = 10$
  • Model B: $p_0 = 25$, $q_0 = 50$
  • Model C: $p_0 = 20$, $q_0 = 25$
  • 1997 Data (Current 1):
  • Model A: $p_1 = 50$, $q_1 = 15$
  • Model B: $p_1 = 30$, $q_1 = 60$
  • Model C: $p_1 = 18$, $q_1 = 20$
  • Summations:
  • $\sum p_0 q_0 = (40 \times 10) + (25 \times 50) + (20 \times 25) = 400 + 1250 + 500 = 2150$
  • $\sum p_1 q_0 = (50 \times 10) + (30 \times 50) + (18 \times 25) = 500 + 1500 + 450 = 2450$
  • $\sum p_0 q_1 = (40 \times 15) + (25 \times 60) + (20 \times 20) = 600 + 1500 + 400 = 2500$
  • $\sum p_1 q_1 = (50 \times 15) + (30 \times 60) + (18 \times 20) = 750 + 1800 + 360 = 2910$
  • Index Calculations:     $\text{LPI} = \frac{2450}{2150} \times 100 = 113.95$$\text{PPI} = \frac{2910}{2500} \times 100 = 116.40$

Single Quantity Index (Quantity Relative)

$\text{Quantity Relative} = \frac{q_1}{q_0} \times 100$