Hypothesis Testing in Statistics 

Definition

Hypothesis Testing in Statistics is a method used to make decisions or inferences about population parameters based on sample data. It helps to determine whether there is enough statistical evidence to support a certain belief (hypothesis) about a population. In simple terms, it’s a technique to test assumptions using statistical data.

1.0Hypothesis Testing in Statistics Meaning

The meaning of Hypothesis Testing revolves around verifying claims or ideas regarding a population. This is done by collecting sample data, applying statistical techniques, and then deciding whether to accept or reject a hypothesis.

2.0Understanding Hypothesis Testing

To understand Hypothesis Testing, you must know these key terms:

• Null Hypothesis (H₀): The default assumption stating no effect or no difference.
• Alternative Hypothesis (H₁ or Ha): The statement we aim to test, suggesting an effect or difference exists.
• Significance Level (α): The probability of rejecting the null hypothesis when it is actually true, usually set at 0.05.
• Test Statistic: A value calculated from the sample data, which is compared against critical values to make decisions.
• p-value: The probability of obtaining test results at least as extreme as the observed results, under the assumption that the null hypothesis is correct.

3.0Hypothesis Testing Steps

Here are the steps of a Hypothesis Test:

1. State the Hypotheses:
2. • Null Hypothesis (H₀)
   • Alternative Hypothesis (H₁)
2. Set the Significance Level (α):

Commonly used values are 0.05, 0.01, or 0.10.

3. Choose the Appropriate Test:

Based on sample size, type of data, and test objective.

4. Compute the Test Statistic:

Apply the formula based on the test selected.

5. Calculate the p-value or Critical Value:

Determine the likelihood of observing the sample result.

6. Make the Decision:

If p-value ≤ α, reject H₀; otherwise, fail to reject H₀.

7. State the Conclusion:

Clearly interpret the statistical decision in context.

4.0Hypothesis Testing in Statistics Formula

The Hypothesis Testing formula varies based on the test type. A common formula for a z-test is: 

$z=\frac{\bar{x}-\mu }{\sigma /\sqrt{n}}$

Where:

• = sample mean
• μ = population mean
• σ = population standard deviation
• n = sample size

For a t-test (when population standard deviation is unknown):

$t=\frac{\bar{x}-\mu }{s/\sqrt{n}}$

Where:

• s = sample standard deviation

5.0Hypothesis Testing Types

There are two main types of Hypothesis Testing in Statistics:

1. Parametric Tests (assume data follows a specific distribution)
2. • Z-Test
   • T-Test
   • ANOVA
   • Chi-Square Test (for variances)
2. Non-Parametric Tests (distribution-free)
3. • Mann-Whitney U Test
   • Wilcoxon Signed-Rank Test
   • Kruskal-Wallis Test

What are the Two Types of Hypothesis Testing?

1. One-tailed Test: Tests for effect in one direction.
2. Two-tailed Test: Tests for effect in both directions (more common).

6.0Solved Examples on Hypothesis Testing in Statistics

Example 1: A company claims its battery lasts 500 hours. A sample of 36 batteries has an average lifespan of 485 hours with a standard deviation of 60 hours. Test at the 5% significance level whether the claim is valid.

Solution:

• Null Hypothesis (H₀): μ = 500
• Alternative Hypothesis (H₁): μ ≠ 500

Given:

, , s = 60, n = 36

Using t-test formula:

Degrees of freedom = 36 - 1 = 35

Critical t-value (two-tailed, α = 0.05) ≈ ±2.030

Since -1.5 lies between -2.030 and 2.030, we fail to reject H₀.

Conclusion: There’s insufficient evidence to reject the company's claim.

Example 2: A company claims that the average weight of its product is 150 grams. A sample of 49 items gives a mean of 152 grams with a standard deviation of 5 grams. Test the claim at 5% significance level.

Solution:

• Null Hypothesis (H₀): μ = 150
• Alternative Hypothesis (H₁): μ ≠ 150 (Two-tailed test)

Given:

$\bar{x}=152,\mu =150,\sigma =5,n=49$

Test Statistic (Z-test):

$$\begin{gathered} z=\frac{\bar{x}-\mu }{\sigma /\sqrt{n}} \\ z=\frac{152-150}{5/\sqrt{49}} \\ z=\frac{2}{5/7}=\frac{2\times 7}{5}=2.8 \end{gathered}$$

Critical Z-values for α = 0.05 (two-tailed): ±1.96

Decision:
|2.8| > 1.96 → Reject H₀.

Conclusion: The claim is not valid; the average weight differs from 150 grams.

$t=\frac{485-500}{60/\sqrt{36}}=\frac{-15}{10}=-1.5$

Example 3: The average lifespan of bulbs produced by a factory is claimed to be 1200 hours. A sample of 16 bulbs has a mean lifespan of 1150 hours with a standard deviation of 100 hours. Test at 5% significance level.

Solution:

• H₀: μ = 1200
• H₁: μ ≠ 1200

Given:
$\bar{x}=1150,s=100,n=16$

Test Statistic (T-test):

$$\begin{gathered} t=\frac{\bar{x}-\mu }{s/\sqrt{n}} \\ t=\frac{1150-1200}{100/\sqrt{16}} \\ t=\frac{-50}{25}=-2 \end{gathered}$$

Degrees of freedom = 16 - 1 = 15

Critical t-value at α = 0.05 (two-tailed, df = 15) ≈ ±2.131

Decision:
|-2| < 2.131 → Fail to reject H₀.

Conclusion: No significant evidence to reject the factory's claim.

Example 4: In a city, it is believed that 40% of people prefer public transport. A survey of 500 people shows that 230 prefer public transport. Test the belief at 5% significance level.

Solution:

• H₀: p = 0.40
• H₁: p ≠ 0.40

Sample proportion

$\hat{p}=\frac{230}{500}=0.46$

Test Statistic:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \\ z=\frac{0.46-0.40}{\sqrt{\frac{0.40\times 0.60}{500}}} \\ z=\frac{0.06}{\sqrt{0.00048}} \\ z=\frac{0.06}{0.0219}\approx 2.74 \end{gathered}$$

Critical z-value at α = 0.05 (two-tailed): ±1.96

Decision:
|2.74| > 1.96 → Reject H₀.

Conclusion: There is significant evidence that the proportion differs from 40%.

Example 5: A researcher wants to test whether gender and preference for a product are independent. The observed data:

Test at 5% significance level.

Solution:

• H₀: Gender and product preference are independent.
• H₁: They are not independent.

Expected frequencies:

$$\begin{gathered} E_{\text{Male, Like}}=\frac{50\times 50}{100}=25 \\ {E}_{\text{Male, Dislike}}=\frac{50\times 50}{100}=25, \\ {E}_{\text{Female, Like}}=25,E_{\text{Female, Dislike}}=25 \end{gathered}$$

Chi-square statistic:

$$\begin{gathered} {\chi }^2=\sum \frac{(O-E{)}^2}{E} \\ {\chi }^2=\frac{(30-25{)}^2}{25}+\frac{(20-25{)}^2}{25}+\frac{(20-25{)}^2}{25}+\frac{(30-25{)}^2}{25} \\ {\chi }^2=\frac{25}{25}+\frac{25}{25}+\frac{25}{25}+\frac{25}{25}=4 \end{gathered}$$

Degrees of freedom = (2 - 1)(2 - 1) = 1

Critical value at α = 0.05, df = 1 → 3.841

Decision:
4 > 3.841 → Reject H₀.

Conclusion: Gender and product preference are not independent.

7.0Practice Questions on Hypothesis Testing in Statistics

Question 1: The mean score of students in a school is 68 with a standard deviation of 10. A sample of 36 students has a mean of 70. Test if the mean score is significantly different at 5% significance level.

Question 2: A shopkeeper claims that 30% of customers buy product A. A random sample of 400 customers showed that 140 customers bought product A. Test the claim at 1% significance level.

Question 3: A manufacturer claims the average lifetime of a machine part is 2000 hours. A sample of 25 parts gives a mean of 1950 hours with a standard deviation of 120 hours. Test the claim at 5% significance level.

Question 4: In a survey, 70% of people said they prefer online shopping. A sample of 150 people showed that 95 people prefer online shopping. Test if this result differs from the survey at 5% level.

Question 5: A school wants to know whether there is an association between class level (junior or senior) and participation in sports. The observed data is:

Test for independence at 5% significance level.