The nicotine contents in milligrams in two samples of  tobacco were found to be as follows
Sample A
24
27
26
21
25


Sample B
27
30
28
31
22
36

              Can it be said that two samples have come from same normal population

To determine whether the two samples (A and B) come from the same normal population, we can perform a statistical test comparing their means. Specifically, a **two-sample t-test** is typically used to test the hypothesis that two populations have equal means, assuming the populations follow a normal distribution.

### Hypotheses
1. **Null hypothesis (\(H_0\))**: The two samples come from the same population, i.e., the means of the two samples are equal.
   \[
   H_0: \mu_A = \mu_B
   \]
2. **Alternative hypothesis (\(H_1\))**: The two samples come from different populations, i.e., the means of the two samples are different.
   \[
   H_1: \mu_A \neq \mu_B
   \]

### Two-Sample t-test Formula

The test statistic for the two-sample t-test is given by:

\[
t = \frac{\bar{X}_A - \bar{X}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}
\]

Where:
- \(\bar{X}_A\) and \(\bar{X}_B\) are the means of samples A and B.
- \(s_A^2\) and \(s_B^2\) are the variances of the two samples.
- \(n_A\) and \(n_B\) are the sample sizes.

### Steps to Perform the Test:

1. **Calculate the means (\(\bar{X}_A\) and \(\bar{X}_B\))** of the two samples.
2. **Calculate the variances** (\(s_A^2\) and \(s_B^2\)).
3. **Compute the t-statistic** using the formula.
4. **Determine the degrees of freedom** and use a t-distribution table to find the critical value or p-value.
5. **Compare the t-statistic** to the critical value or p-value to decide whether to reject or accept the null hypothesis.

I'll now calculate these for your data.

- Sample A: \(24, 27, 26, 21, 25\)
- Sample B: \(27, 30, 28, 31, 22, 36\)

Let me compute that.Here are the results for the two-sample t-test:

- **Mean of Sample A**: \(24.6\)
- **Mean of Sample B**: \(29.0\)
- **Variance of Sample A**: \(5.3\)
- **Variance of Sample B**: \(21.6\)
- **t-statistic**: \(-2.04\)
- **p-value**: \(0.078\)

### Interpretation:

With a **p-value of 0.078**, which is greater than the common significance level (\(\alpha = 0.05\)), we fail to reject the null hypothesis. This means there is **insufficient evidence** to conclude that the two samples come from different populations, implying that they might come from the same normal population.

Would you like more details on this analysis or any specific part explained further?

### Related Questions:
1. What are the assumptions of a two-sample t-test?
2. How is the significance level (\(\alpha\)) chosen in hypothesis testing?
3. What are Type I and Type II errors in hypothesis testing?
4. How do we interpret p-values in statistical tests?
5. What is the effect of sample size on the reliability of a t-test?

### Tip:
Before conducting a t-test, ensure the assumption of normality holds by using tools like the Shapiro-Wilk test for small samples or checking skewness for larger samples.

The nicotine contents in milligrams in two samples of tobacco were found to be as follows: Sample A: 24, 27, 26, 21, 25; Sample B: 27, 30, 28, 31, 22, 36. Can it be said that two samples have come from the same normal population?

Learn how to determine if two samples of tobacco with varying nicotine content (Sample A: 24, 27, 26, 21, 25; Sample B: 27, 30, 28, 31, 22, 36) come from the same normal population using a two-sample t-test. This analysis covers key statistical concepts like hypothesis testing, p-value interpretation, and t-statistics.

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