The meaning of 2 random sample size 9 and 7 are 196.42 and 198.82. the sum of the square of the deviation from the mean are 26.94 and 18.73 respectively. Then the sample can be considered as drawn from the normal population of equal means

To determine whether the two samples can be considered drawn from a normal population with equal means, we perform a **two-sample t-test**. This test compares the means of two independent samples to see if they are significantly different. Here’s the step-by-step process:

### 1. **Given Information:**

- Sample 1: 
  - Sample size \( n_1 = 9 \)
  - Mean \( \bar{X_1} = 196.42 \)
  - Sum of squares of deviations \( SS_1 = 26.94 \)

- Sample 2: 
  - Sample size \( n_2 = 7 \)
  - Mean \( \bar{X_2} = 198.82 \)
  - Sum of squares of deviations \( SS_2 = 18.73 \)

### 2. **Step 1: Calculate sample variances:**

The sample variance for each group can be found using the formula:
\[
s^2 = \frac{SS}{n - 1}
\]
Where \( SS \) is the sum of squares of deviations from the mean, and \( n \) is the sample size.

- For Sample 1:
\[
s_1^2 = \frac{26.94}{9 - 1} = \frac{26.94}{8} = 3.3675
\]

- For Sample 2:
\[
s_2^2 = \frac{18.73}{7 - 1} = \frac{18.73}{6} = 3.1217
\]

### 3. **Step 2: Check if population variances are equal:**

For a two-sample t-test, we usually assume equal variances if they are not significantly different. The ratio of variances can be used to check for equality, but it appears the variances are quite close.

Thus, we will proceed assuming equal variances.

### 4. **Step 3: Pooled variance:**

Since we assume equal variances, we calculate the **pooled variance** using:
\[
s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}
\]

Substituting the values:
\[
s_p^2 = \frac{(9 - 1)(3.3675) + (7 - 1)(3.1217)}{9 + 7 - 2}
\]
\[
s_p^2 = \frac{8(3.3675) + 6(3.1217)}{14} = \frac{26.94 + 18.73}{14} = \frac{45.67}{14} \approx 3.2621
\]

### 5. **Step 4: Compute the t-statistic:**

The t-statistic is given by:
\[
t = \frac{\bar{X_1} - \bar{X_2}}{\sqrt{s_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}
\]

Substituting the values:
\[
t = \frac{196.42 - 198.82}{\sqrt{3.2621 \left( \frac{1}{9} + \frac{1}{7} \right)}}
\]
\[
t = \frac{-2.40}{\sqrt{3.2621 \left( 0.1111 + 0.1429 \right)}} = \frac{-2.40}{\sqrt{3.2621(0.254)}}
\]
\[
t = \frac{-2.40}{\sqrt{0.828}} = \frac{-2.40}{0.91} \approx -2.64
\]

### 6. **Step 5: Degrees of freedom:**

The degrees of freedom (df) for the test is \( n_1 + n_2 - 2 = 9 + 7 - 2 = 14 \).

### 7. **Step 6: Determine the critical value or p-value:**

For a two-tailed test with 14 degrees of freedom, the critical value of \( t \) at a 5% significance level (0.05) is approximately \( \pm 2.145 \).

Since the calculated t-value (-2.64) is less than -2.145, we reject the null hypothesis.

### Conclusion:
The t-test indicates that the difference in means is statistically significant, meaning the two samples **cannot** be considered to be drawn from populations with equal means.

---

Do you want further details or have any questions? Here are five related questions:

1. How does the two-sample t-test work when variances are not equal?
2. What is the assumption of normality in the two-sample t-test?
3. How can I compute the p-value from the t-statistic?
4. What if I wanted to test the variance of two populations instead of their means?
5. Can I use another method instead of a t-test for comparing sample means?

**Tip**: Always check for the assumption of normality and equal variances before using a t-test!

The meaning of 2 random sample size 9 and 7 are 196.42 and 198.82. the sum of the square of the deviation from the mean are 26.94 and 18.73 respectively. Then the sample can be considered as drawn from the normal population of equal means.

Explore the process of determining whether two samples with sizes 9 and 7, means 196.42 and 198.82, and sums of square deviations 26.94 and 18.73, respectively, come from populations with equal means. This detailed solution includes calculations of sample variances, pooled variance, and the t-statistic using the two-sample t-test method. Ideal for undergraduate statistics students.

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