When the two independent small samples of sizes n_1 and n_2 with means x_1 and x_2 respectively are drawn from populations with identical population variances, the test-statistic is computed as Option 1: t = (x_1 - x_2)/(S_p√(1/n_1 + 1/n_2)), where S_p is pooled standard deviation. Option 2: t = (x_1 - x_2)/(S_p√(1/n_1 - 1/n_2)), where S_p is pooled standard deviation. Option 3: t = (x_1 + x_2)/(S_p√(1/n_1 - 1/n_2)), where S_p is pooled standard deviation. Option 4: t = (x_1 + x_2)/(√(1/n_1 - 1/n_2)), where S_p is pooled standard deviation.

Answer: t = (x_1 - x_2)/(S_p√(1/n_1 + 1/n_2)), where S_p is pooled standard deviation. Solution: When two independent small samples are drawn from populations with identical population variances, the two-sample t-test compares the means of the two groups. The numerator represents the difference between the two sample means: x_1 - x_2 This eliminates Options 3 and 4, which incorrectly use addition in the numerator. For the denominator, the variance of the difference between two independent sample means is: Variance(x_1 - x_2) = (^2)/n_1 + (^2)/n_2 = ^2(1/n_1 + 1/n_2) When dealing with independent samples, the variances add (not subtract). The standard error is obtained by taking the square root: Standard Error = S_p√(1/n_1 + 1/n_2) where S_p is the pooled standard deviation. The test statistic is: t = (x_1 - x_2)/(S_p√(1/n_1 + 1/n_2)) This matches Option 1. Option 2 incorrectly uses a minus sign in the denominator, while Option 4 is missing S_p entirely. Therefore, the correct test statistic is: t = (x_1 - x_2)/(S_p√(1/n_1 + 1/n_2))

Statistics & Applications

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When the two independent small samples of sizes $n_1$ and $n_2$ with means $\bar{x}_1$ and $\bar{x}_2$ respectively are drawn from populations with identical population variances, the test-statistic is computed as

$t = \frac{\bar{x}_1 - \bar{x}_2}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$ , where $S_p$ is pooled standard deviation.

$t = \frac{\bar{x}_1 - \bar{x}_2}{S_p\sqrt{\frac{1}{n_1} - \frac{1}{n_2}}}$ , where $S_p$ is pooled standard deviation.

$t = \frac{\bar{x}_1 + \bar{x}_2}{S_p\sqrt{\frac{1}{n_1} - \frac{1}{n_2}}}$ , where $S_p$ is pooled standard deviation.

$t = \frac{\bar{x}_1 + \bar{x}_2}{\sqrt{\frac{1}{n_1} - \frac{1}{n_2}}}$ , where $S_p$ is pooled standard deviation.

✅ Correct Option: 1

When two independent small samples are drawn from populations with identical population variances, the two-sample t-test compares the means of the two groups.

The numerator represents the difference between the two sample means:

$\bar{x}_1 - \bar{x}_2$

This eliminates Options 3 and 4, which incorrectly use addition in the numerator.

For the denominator, the variance of the difference between two independent sample means is:

$\text{Variance}(\bar{x}_1 - \bar{x}_2) = \frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}$

$= \sigma^2\left(\frac{1}{n_1} + \frac{1}{n_2}\right)$

When dealing with independent samples, the variances add (not subtract). The standard error is obtained by taking the square root:

$\text{Standard Error} = S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

where $S_p$ is the pooled standard deviation.

The test statistic is:

$t = \frac{\bar{x}_1 - \bar{x}_2}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

This matches Option 1. Option 2 incorrectly uses a minus sign in the denominator, while Option 4 is missing $S_p$ entirely.

Therefore, the correct test statistic is:

$t = \frac{\bar{x}_1 - \bar{x}_2}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

2026: 31 May Shift 2

Consider the following hypothesis test :

$H_0: \mu \le 35$

$H_a: \mu > 35$

A sample of 49 provided a sample mean of 38.2 The population standard deviation is 8. The value of test-statistic is,

2025: 27 May Shift 1

Consider the following hypothesis test.

$H_0: \mu \leq 12$

$H_a: \mu > 12$

If a sample of 25 is taken with sample mean 15 and a sample standard deviation of 6, then the value of t-test statistic is:

2026: 19 May Shift 1

Match List-I with List-II

List-I	List-II
(A) It describes the accuracy of a sampling method.	(I) Confidence interval
(B) It describes the uncertainity of a sampling method.	(II) Hypothesis
(C) It is used to express the precision and uncertainity of a sampling process.	(III) Margin of error
(D) An educated guess which needs to be tested.	(IV) Confidence level

Choose the correct answer from the options given below:

2026: 15 May Shift 1

A simple random sample of 49 items from a population with standard deviation, σ = 6 resulted in a sample mean of 32. At 95% confidence level, the margin of error is (Given $Z_{0.025} = 1.96$ )

2025: 22 May Shift 2

Rajesh calculated 95% confidence level. What does he mean by that?

(A) He can be 95% confident that his sample will include the population parameter.

(B) He can be 95% confused that his sample will include the population parameter.

(D) He can be 5% confident that his sample will include the population parameter.

Choose the correct answer from the options given below:

2026: 29 May Shift 1

A simple random sample of 64 items from a population with standard deviation $\sigma=4$ resulted in a sample mean of 35. What is the margin of error for the population mean at a 95% confidence level? (Given $Z_{0.025}=1.96$ )

When the two independent small samples of sizes n1n_1n1​ and n2n_2n2​ with means xˉ1\bar{x}_1xˉ1​ and xˉ2\bar{x}_2xˉ2​ respectively are drawn from populations with identical population variances, the test-statistic is computed as

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CUET Mathematics Past Year Questions (Topic-Wise)

When the two independent small samples of sizes $n_1$ and $n_2$ with means $\bar{x}_1$ and $\bar{x}_2$ respectively are drawn from populations with identical population variances, the test-statistic is computed as