Null Hypothesis

The Null Hypothesis (H₀) is a fundamental concept in statistics used to test assumptions or claims about a population. It represents a statement of no effect, no difference, or no relationship between variables. It is the default or starting assumption in hypothesis testing.

1.0Null Hypothesis Definition

In statistical terms, the null hypothesis is a hypothesis that states there is no statistical significance between the observed data and what is expected. It is tested to either be rejected or not rejected based on evidence from a sample.

2.0What Do You Mean by Null Hypothesis?

The null hypothesis assumes that any observed variation is due to chance or randomness. It’s the “innocent until proven guilty” principle of statistics. Only strong evidence can lead us to reject it in favor of the alternative hypothesis.

3.0Null Hypothesis and Alternative Hypothesis

Example:

• $H_0$: The average height of students is 170 cm.
• $H_1:$ The average height of students is not 170 cm.

4.0Null Hypothesis Formula

The most common formula is related to the Z-test or t-test.

Z-test formula:

$Z=\frac{\bar{X}-{\mu }_0}{\sigma /\sqrt{n}}$

Where:

$$\begin{gathered} \bar{X}=\text{sample mean} \\ {\mu }_0=\text{null hypothesis value (assumed population mean)} \\ \sigma \sigma =\text{standard deviation} \\ \text{n = sample size} \\ \text{The null hypothesis value}{\mu }_0\text{is what you’re testing your sample against.} \end{gathered}$$

5.0Understanding Null Hypothesis Testing

The process involves:

1. State the null and alternative hypotheses.
2. Choose a significance level (α, often 0.05).
3. Calculate the test statistic (Z or t).
4. Compare with critical value or use p-value.
5. Reject or fail to reject H_0.

If the p-value is less than α, reject the null hypothesis.

6.0Null Hypothesis Examples

Here are real-world examples:

1. Medical Research

• $H_0$: The new drug has no effect on blood pressure.
• $H_1:$ The new drug lowers blood pressure.

2. Education

• $H_0$: Online classes have no impact on student performance.
• $H_1:$ Online classes improve student performance.

3. Marketing

• $H_0:$The new advertisement does not increase sales.
• $H_1:$The new advertisement increases sales.

7.0What Is the Null Hypothesis Used for in Investing?

In investing and finance, the null hypothesis is used to test market theories, portfolio performance, and risk models. 

Example in investing:

• $H_0:$ A mutual fund does not outperform the market.
• $H_1:$ The mutual fund does outperform the market.

If statistical analysis shows no significant difference in returns, the fund manager's claims may be rejected.

8.0What Are Examples of Null Hypotheses?

• No change: “There’s no difference between product A and B.”
• No effect: “Caffeine does not affect reaction time.”
• No relationship: “Age and social media usage are unrelated.”

All these represent statements of no difference or no association—the hallmark of a null hypothesis.

9.0Null Hypothesis Value

The null hypothesis value is the specific number or assumption you're testing against.
For example, if testing whether the mean height is 170 cm, then:

• Null Hypothesis: → Here, 170 is the null hypothesis value.

10.0How Is the Null Hypothesis Used in Real Research?

• Science: To test experimental results
• Business: To measure effectiveness of strategies
• Social sciences: To explore behavioral trends
• Data science: To evaluate model assumptions and features

11.0Standard Solved Examples on Null Hypothesis

Example 1:  A battery manufacturer claims that its batteries last 100 hours. A random sample of 36 batteries showed an average life of 98 hours with a standard deviation of 6 hours. Test the claim at 5% significance.

Solution:

$$\begin{gathered} H_0:\mu =100\text{ (Null Hypothesis)} \\ {H}_1:\mu \ne 100\text{(Alternative Hypothesis)} \end{gathered}$$

Use Z-test because population standard deviation is known.

$Z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}=\frac{98-100}{6/\sqrt{36}}=\frac{-2}{1}=-2$

At 5% significance, critical Z-values are ±1.96.

Since -2 < -1.96, we reject the null hypothesis.

Conclusion: The data suggests battery life is significantly different from 100 hours.

Example 2: A researcher claims that a new teaching method improves student scores. The average score from a class using the new method is 78 (n=25), with a sample standard deviation of 5. Test if the mean is greater than 75 at 1% level.

Solution:

$$\begin{gathered} H_0:\mu =75 \\ {H}_1:\mu >75 \end{gathered}$$

Use one-tailed t-test:

$t=\frac{78-75}{5/\sqrt{25}}=\frac{3}{1}=3$

Degrees of freedom = 24.

At 1% significance,

$t_{0.01,24}\approx 2.492$

Since 3 > 2.492, reject $H_0$

Conclusion: Teaching method significantly improves performance.

Example 3: 60 out of 200 customers prefer a new product. Is this different from the company’s claim that 30% prefer it? Use 5% significance.

Solution:

• $H_0:p=0.30$
• $H_1:p\ne 0.30$

Sample proportion:

$$\begin{gathered} \hat{p}=\frac{60}{200}=0.30 \\ Z=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}=\frac{0.30-0.30}{\sqrt{0.30\cdot 0.70/200}}=\frac{0}{\sqrt{0.00105}}=0 \end{gathered}$$

Since Z = 0 lies within ±1.96 → fail to reject H_0

Conclusion: No significant difference in customer preference.

Example 4: An investor claims their portfolio return is more than the market return of 10%. From 10 observations, mean = 12%, s.d. = 3%.

Solution:

$H_0:\mu =10$

$H_1:mu>10$

$t=\frac{12-10}{3/\sqrt{10}}=\frac{2}{0.948}\approx 2.11$

$\text{Degrees of freedom = 9}$

At 5% significance $t_{0.05,9}\approx 1.833$

Since 2.11 > 1.833 → reject H_0

Conclusion: Portfolio outperforms the market.

12.0Practice Questions on Null Hypothesis

Q1. A diet program claims to reduce weight by 5 kg. A sample of 20 participants showed an average weight loss of 4.2 kg with s.d. = 1.5 kg. Test the claim at 0.05 significance.

Q2. A college asserts that 80% of its students pass the final exam. From a sample of 100 students, 76 passed. Is the claim valid at 1% significance?

Q3. A factory claims that machine A produces bulbs lasting 1200 hours. A sample of 30 bulbs shows an average of 1185 hours with s.d. of 40 hours. Test at 5%.

Q4. A sample of 50 packets weighs 202 g on average. Is it different from the labeled weight of 200 g (population s.d. = 5 g)? Use 1% significance.