🧪 t-Test

Last Updated: Jan 2026

A t-Test is a hypothesis test used to determine whether a sample mean (or means) is significantly different when:

• Sample size is small
• Population standard deviation is unknown

It uses the t-distribution, which is wider than the normal distribution.

🗣 Hinglish Tip: t-Test = jab sample chhota ho aur $\sigma$ pata na ho

When to Use t-Test?

Use t-Test when:

• Sample size: $n < 30$
• Population standard deviation unknown
• Data is approximately normal
• Samples are independent (except paired test)

Types of t-Test

1. One-Sample t-Test
2. Two-Sample (Independent) t-Test
3. Paired t-Test

👉 In this tutorial, we cover One-Sample t-Test

t-Test Notation (Math Standard)

• Population mean → $\mu$
• Sample mean → $\bar{x}$
• Sample standard deviation → $s$
• Sample size → $n$
• Degrees of freedom → $df = n - 1$
• Significance level → $\alpha$
• t statistic → $t$

t-Test Formula (One-Sample)

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

Example

A coaching institute claims that the average score of students is 70. A sample of 10 students is taken and their scores are:

$65, 68, 70, 72, 74, 69, 71, 73, 67, 66$

Test the claim at 5% significance level.

Step 1: State the Hypotheses

Two-tailed test (checking difference):

$H_0: \mu = 70
\\[0.5em]
H_1: \mu \ne 70$

Step 2: Identify Test Type

• $n = 10$ ($< 30$) $\checkmark$
• $\sigma$ unknown $\checkmark$

👉 Use One-Sample t-Test

Step 3: Calculate Sample Mean ($\bar{x}$)

$\sum x = 695
\\[0.5em]
\bar{x} = \frac{695}{10} = 69.5$

Step 4: Calculate Sample Standard Deviation ($s$)

$\sum (x - \bar{x})^2 = 82.5
\\[0.5em]
s^2 = \frac{82.5}{10 - 1} = 9.17
\\[0.5em]
s = \sqrt{9.17} \approx 3.03$

Step 5: Calculate t-Statistic

$t = \frac{69.5 - 70}{3.03 / \sqrt{10}}
\\[0.5em]
t = \frac{-0.5}{0.958}
\\[0.5em]
t \approx -0.52$

Step 6: Find Critical t-Value

• $\alpha = 0.05$
• $df = n - 1 = 9$

From t-table:

$t_{0.025,9} = \pm 2.262$

Step 7: Decision

• Calculated $t = -0.52$
• Critical $t = \pm 2.262$

Since:

$|t| = 0.52 < 2.262$

👉 Fail to Reject $H_0$

Step 8: Conclusion

There is no sufficient evidence to reject the institute's claim. The average score can be considered 70.

🗣 Hinglish Tip: $t$ limit ke andar hai → claim acceptable

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