FAD1015 L23-L24 — Hypothesis Testing About the Mean

Lectures 23–24 introducing hypothesis testing methodology for the population mean using a one-sample test. Source file: (L23 L24) FAD 1015 -Week 14 Hypothesis Testing About the Mean_.pdf

Summary

Comprehensive introduction to hypothesis testing for a single population mean: null and alternative hypotheses, significance level, test statistics (Z and t), three decision approaches (traditional/critical-value, confidence-interval, and P-value), and the six-step testing procedure.

Key Concepts

Hypothesis Testing — statistical decision making about a population parameter
Null Hypothesis (H₀) — statement assumed true until evidence contradicts it
Alternative Hypothesis (H₁ or Hₐ) — research claim; true if H₀ is false
Significance Level (α) — measure of confidence in rejecting H₀; commonly 0.10, 0.05, 0.01
Test Statistic — Z (large sample) or t (small sample, σ unknown)
P-value — observed level of significance; smallest α for which H₀ can be rejected
Critical Value / Rejection Region — region where H₀ is rejected
Confidence Interval Method — alternative approach using interval estimation

1. Hypothesis Testing Framework

1.1 Definitions

Hypothesis A claim or statement made about a population parameter that we want to test. The statement may or may not be true.

Null Hypothesis (H₀) A statement or claim about a population parameter that is assumed to be true until it is declared false.

Usually contains equality (=, ≤, or ≥)
Denoted H₀ (pronounced “H-null”, “H-zero”, or “H-nought”)

Alternative Hypothesis (H₁ or Hₐ) A claim about a population parameter that will be true if the null hypothesis is false.

Represents the research hypothesis
Can be two-tailed (≠), left-tailed (<), or right-tailed (>)

1.2 Types of Tests

Test Type	Null Hypothesis	Alternative Hypothesis
Two-tailed	H₀: μ = μ₀	H₁: μ ≠ μ₀
Left-tailed	H₀: μ ≥ μ₀	H₁: μ < μ₀
Right-tailed	H₀: μ ≤ μ₀	H₁: μ > μ₀

Note: In practice H₀ is often written with equality only (μ = μ₀) regardless of tail direction.

1.3 Significance Level (α)

The level of significance is a measure of how confident you can be about rejecting the null hypothesis.

Also called the alpha level
Common values: 0.10, 0.05, 0.01
The most common value is α = 0.05
α specifies the size of the rejection region (critical region)

2. Test Statistic Selection

The choice of test statistic depends on whether the population standard deviation σ is known and on sample size:

graph TD
    Q1["Do you know sigma?"] -->|Yes| Z1["Use Z-test<br/>Z = (Xbar - mu) / (sigma/sqrt(n))"]
    Q1 -->|No| Q2["What is the sample size n?"]
    Q2 -->|n >= 30| Z2["Use Z-test<br/>approximate sigma with s<br/>Z = (Xbar - mu) / (s/sqrt(n))"]
    Q2 -->|n < 30| T["Use t-test<br/>df = n - 1<br/>t = (Xbar - mu) / (s/sqrt(n))"]

    style Q1 fill:#fff4e6,stroke:#e67700
    style Q2 fill:#fff4e6,stroke:#e67700
    style Z1 fill:#d3f9d8,stroke:#2f9e44
    style Z2 fill:#d3f9d8,stroke:#2f9e44
    style T fill:#e5dbff,stroke:#5f3dc4

2.1 Large Sample (n ≥ 30)

When the sample is large, the sampling distribution of the sample mean is approximately normal by the Central Limit Theorem.

σ known: $$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim N(0,1)$$

σ unknown: $$Z = \frac{\bar{X} - \mu}{s / \sqrt{n}} \sim N(0,1)$$

2.2 Small Sample (n < 30)

When the sample is small and σ is unknown, use the t-distribution (requires approximate normality of the population).

σ unknown: $$T = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \sim t_{n-1}$$

3. Three Approaches to Hypothesis Testing

graph TD
    subgraph trad["Traditional Method"]
        T1["Calculate test statistic"]
        T2["Compare to critical value"]
        T3["Reject H0 if in<br/>rejection region"]
    end
    subgraph ci["Confidence Interval Method"]
        C1["Construct CI for mu"]
        C2["Reject H0 if mu0<br/>is outside CI"]
    end
    subgraph pval["P-value Method"]
        P1["Calculate P-value"]
        P2["Reject H0 if<br/>P-value <= alpha"]
    end

    T1 --> T2 --> T3
    C1 --> C2
    P1 --> P2

    style trad fill:#e7f5ff,stroke:#1971c2
    style ci fill:#e5dbff,stroke:#5f3dc4
    style pval fill:#ffe8cc,stroke:#d9480f

The lecture presents three methods for making the decision:

Traditional method — critical value / rejection region
Confidence Interval method
P-value method

4. Steps for Hypothesis Testing

graph LR
    S1["Step 1: State H0 and H1"] --> S2["Step 2: Select alpha"]
    S2 --> S3["Step 3: Identify test statistic<br/>Z or t"]
    S3 --> S4["Step 4: Identify rejection region"]
    S4 --> S5["Step 5: Make decision<br/>Reject or Do not reject H0"]
    S5 --> S6["Step 6: Conclusion<br/>in context of problem"]

    style S1 fill:#e7f5ff,stroke:#1971c2
    style S2 fill:#e7f5ff,stroke:#1971c2
    style S3 fill:#ffe8cc,stroke:#d9480f
    style S4 fill:#ffe8cc,stroke:#d9480f
    style S5 fill:#fff4e6,stroke:#e67700
    style S6 fill:#d3f9d8,stroke:#2f9e44

State the null hypothesis and alternative hypothesis
Select a level of significance (α)
Identify the test statistic
Identify the rejection region / critical region
Make a decision
Conclusion

5. Decision Rules

5.1 Traditional (Critical-Value) Approach

Compare the calculated test statistic to the critical value(s):

Test statistic falls in the rejection region → Reject H₀
Test statistic falls in the non-rejection region → Do not reject H₀

5.2 P-Value Approach

The P-value is also called the observed level of significance. It is the smallest value of α for which H₀ can be rejected.

P-value ≤ α → Reject H₀
P-value > α → Do not reject H₀

The P-value is the smallest level of significance that would lead to the rejection of the null hypothesis H₀ with the given data.

5.3 Confidence Interval Approach

Construct a confidence interval for μ. If the hypothesized value μ₀ falls outside the interval, reject H₀. If it falls inside, do not reject H₀.

6. Worked Examples

Example 1 — Two-Tailed Z-Test (Large Sample, σ Unknown)

The processors of XYZ Ketchup indicate on the label that the bottle contains 16 ounces of ketchup. A sample of 36 bottles from last hour’s production revealed a mean weight of 16.12 ounces per bottle and a standard deviation of 0.5 ounces. At the 0.05 significance level, is the process out of control? That is, can we conclude that the mean amount per bottle is different from 16 ounces?

Solution outline:

H₀: μ = 16 vs H₁: μ ≠ 16
α = 0.05 (two-tailed)
n = 36 ≥ 30, σ unknown → use Z with s
$Z = \frac{16.12 - 16}{0.5 / \sqrt{36}} = \frac{0.12}{0.0833} = 1.44$
Critical values: ±1.96
Decision: |1.44| < 1.96 → Do not reject H₀
Conclusion: At the 5% significance level, there is not enough evidence to conclude that the mean amount per bottle differs from 16 ounces.

Example 2 — Right-Tailed Z-Test (Large Sample, σ Unknown)

A phone industry manager thinks that customers’ monthly cell phone bills have increased, and now average more than RM 52 per month. The company wishes to test this claim. Suppose a sample of 64 customers is randomly selected with a mean RM 53.1 and standard deviation RM10. Test the manager’s claim at 10% level of significance.

Solution outline:

H₀: μ ≤ 52 vs H₁: μ > 52
α = 0.10 (right-tailed)
n = 64 ≥ 30, σ unknown → use Z with s
$Z = \frac{53.1 - 52}{10 / \sqrt{64}} = \frac{1.1}{1.25} = 0.88$
Critical value: 1.28
Decision: 0.88 < 1.28 → Do not reject H₀
Conclusion: At the 10% significance level, there is not enough evidence to support the manager’s claim that the average monthly bill exceeds RM 52.

Example 3 — Left-Tailed Z-Test (Large Sample, σ Known)

A quality control engineer finds that a sample of 100 light bulbs had an average life-time of 470 hours. Assuming a population standard deviation of σ = 25 hours, test whether the population mean is 480 hours vs. the alternative hypothesis μ < 480 at a significance level of α = 0.05.

Solution outline:

H₀: μ = 480 vs H₁: μ < 480
α = 0.05 (left-tailed)
n = 100 ≥ 30, σ known → use Z
$Z = \frac{470 - 480}{25 / \sqrt{100}} = \frac{-10}{2.5} = -4.0$
Critical value: −1.645
Decision: −4.0 < −1.645 → Reject H₀
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean lifetime of the light bulbs is less than 480 hours.

Example 4 — Two-Tailed t-Test (Small Sample, σ Unknown)

The mean cost of a hotel room in KL is said to be RM168 per night. A random sample of 25 hotels resulted in $\bar{X}$ = RM172.50 and S = 15.40. Test at the α = 0.05 level. (A stem-and-leaf display and a normal probability plot indicate the data are approximately normally distributed.)

Solution outline:

H₀: μ = 168 vs H₁: μ ≠ 168
α = 0.05 (two-tailed)
n = 25 < 30, σ unknown → use t with df = 24
$t = \frac{172.50 - 168}{15.40 / \sqrt{25}} = \frac{4.50}{3.08} = 1.46$
Critical values: ±2.064
Decision: |1.46| < 2.064 → Do not reject H₀
Conclusion: At the 5% significance level, there is not enough evidence to conclude that the mean cost of a hotel room in KL differs from RM168.

This example is also used to illustrate the Confidence Interval Approach.

Example 5 — P-Value Approach (Same Data as Example 2)

A phone industry manager thinks that customers’ monthly cell phone bills have increased, and now average more than RM 52 per month. The company wishes to test this claim. Suppose a sample of 64 customers is randomly selected with a mean RM 53.1 and standard deviation RM10. Test the manager’s claim at 10% level of significance using the p-value approach.

Solution outline:

H₀: μ ≤ 52 vs H₁: μ > 52
α = 0.10
Test statistic: Z = 0.88 (as in Example 2)
P-value = P(Z > 0.88) = 1 − 0.8106 = 0.1894
Decision: 0.1894 > 0.10 → Do not reject H₀
Conclusion: At the 10% significance level, there is not enough evidence to support the manager’s claim.

Example 6 — P-Value Approach (Same Data as Example 1)

Solution outline:

H₀: μ = 16 vs H₁: μ ≠ 16
α = 0.05
Test statistic: Z = 1.44 (as in Example 1)
P-value = 2 × P(Z > 1.44) = 2 × (1 − 0.9251) = 0.1498
Decision: 0.1498 > 0.05 → Do not reject H₀
Conclusion: At the 5% significance level, there is not enough evidence to conclude that the mean amount differs from 16 ounces.

7. Exercises

Exercise 1

Effy’s Discount Store chain issues its own credit card. Bobby, the credit manager, wants to find out if the mean monthly unpaid balance is more than RM400. The level of significance is set at 0.05. A random check of 172 unpaid balances revealed the sample mean to be RM407 and the sample standard deviation to be RM38. Should Bobby conclude that the population mean is greater than RM400?

Exercise 2

The manager of a large factory believes that the average hourly wage of the employees is below RM 9.78 per hour. A sample of 40 employees has a mean hourly wage of RM9.60. The variance of all salaries is RM 2.20. Assume the variable is normally distributed. At α = 0.10, is there enough evidence to support the manager’s claim?

Exercise 3

The mean height of all female college basketball players is 69.5 inches. A random sample of 25 such players produced a mean height of 70.2 inches with a variance of 4.41 inches. Assuming that the heights of all female college basketball players are normally distributed, test at the 1% significance level if their mean height is different from 69.5 inches.

Exercise 4

The current rate for producing 5 amp fuses at Ah Seng Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. A sample of 10 randomly selected hours from last month revealed the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour. At the 0.05 significance level can Ah Seng conclude that the new machine is faster?

Exercise 5

We have a sample of 106 body temperatures having a mean of 98.20°F. Assume that the sample is a simple random sample and that the population standard deviation σ is known to be 0.62°F. Use a 0.05 significance level to test the common belief that the mean body temperature of healthy adults is equal to 98.6°F.

i) Traditional method (critical value / rejection region)
ii) Confidence Interval method
iii) P-value method

Related Course Page

FAD1015 - Mathematics III

FAD1015 L23-L24 — Hypothesis Testing About the Mean

Summary

Key Concepts

1. Hypothesis Testing Framework

1.1 Definitions

1.2 Types of Tests

1.3 Significance Level (α)

2. Test Statistic Selection

2.1 Large Sample (n ≥ 30)

2.2 Small Sample (n < 30)

3. Three Approaches to Hypothesis Testing

4. Steps for Hypothesis Testing

5. Decision Rules

5.1 Traditional (Critical-Value) Approach

5.2 P-Value Approach

5.3 Confidence Interval Approach

6. Worked Examples

Example 1 — Two-Tailed Z-Test (Large Sample, σ Unknown)

Example 2 — Right-Tailed Z-Test (Large Sample, σ Unknown)

Example 3 — Left-Tailed Z-Test (Large Sample, σ Known)

Example 4 — Two-Tailed t-Test (Small Sample, σ Unknown)

Example 5 — P-Value Approach (Same Data as Example 2)

Example 6 — P-Value Approach (Same Data as Example 1)

7. Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Related Topics

Related Course Page