Using the data below from a simple random paired sample from two (2) populations under consideration, **at the 10% significance level, use the paired t-test to determine if the mean of population one is not equal to that of population 2**

**Pair**

**Population 1**

**Population 2**

1

10

12

2

8

7

3

13

11

4

13

16

5

17

15

6

12

9

7

12

12

8

11

7

Question 7 options:

t=2.513, critical values=+/-1.895, reject the null hypothesis

t=1.024, critical values=+/-1.895, do not reject the null hypothesis

t=0.875, critical value=+/-0.895, do not reject the null hypothesis

t=2.513, critical value=+/-0.895, reject the null hypothesis

To perform a **paired t-test** and determine if the mean of population one is significantly different from the mean of population two, we follow these steps:

### Step 1: Calculate the differences
For each pair, subtract the value of Population 2 from Population 1:

\[
\text{Difference} = \text{Population 1} - \text{Population 2}
\]

| Pair | Population 1 | Population 2 | Difference (d) |
|------|--------------|--------------|----------------|
| 1    | 10           | 12           | -2             |
| 2    | 8            | 7            | 1              |
| 3    | 13           | 11           | 2              |
| 4    | 13           | 16           | -3             |
| 5    | 17           | 15           | 2              |
| 6    | 12           | 9            | 3              |
| 7    | 12           | 12           | 0              |
| 8    | 11           | 7            | 4              |

### Step 2: Calculate the mean difference ( \(\bar{d}\) )
\[
\bar{d} = \frac{\sum d}{n} = \frac{-2 + 1 + 2 - 3 + 2 + 3 + 0 + 4}{8} = \frac{7}{8} = 0.875
\]

### Step 3: Calculate the standard deviation of the differences ( \(s_d\) )
\[
s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}
\]
First, calculate each \( (d_i - \bar{d})^2 \):

| Pair | Difference (d) | \( (d - \bar{d})^2 \) |
|------|----------------|-----------------------|
| 1    | -2             | (-2 - 0.875)^2 = 8.39 |
| 2    | 1              | (1 - 0.875)^2 = 0.02  |
| 3    | 2              | (2 - 0.875)^2 = 1.27  |
| 4    | -3             | (-3 - 0.875)^2 = 15.14 |
| 5    | 2              | (2 - 0.875)^2 = 1.27  |
| 6    | 3              | (3 - 0.875)^2 = 4.54  |
| 7    | 0              | (0 - 0.875)^2 = 0.77  |
| 8    | 4              | (4 - 0.875)^2 = 9.77  |

\[
\sum (d - \bar{d})^2 = 41.17
\]

\[
s_d = \sqrt{\frac{41.17}{8-1}} = \sqrt{\frac{41.17}{7}} \approx \sqrt{5.88} \approx 2.425
\]

### Step 4: Calculate the t-statistic
The t-statistic is calculated as:
\[
t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{0.875}{2.425 / \sqrt{8}} = \frac{0.875}{0.857} \approx 1.021
\]

### Step 5: Determine the critical value
For a two-tailed test at the 10% significance level (α = 0.10) and degrees of freedom \(df = n - 1 = 8 - 1 = 7\), the critical value from the t-distribution table is approximately \( \pm 1.895 \).

### Step 6: Compare the t-statistic with the critical value
Our calculated t-statistic is \( t = 1.021 \), which falls within the range of \( -1.895 \) and \( 1.895 \). Therefore, we **do not reject the null hypothesis**.

### Conclusion:
At the 10% significance level, we **do not reject the null hypothesis**, meaning there is **insufficient evidence** to conclude that the means of Population 1 and Population 2 are different.

### The correct answer is:
**t = 1.024, critical values = ±1.895, do not reject the null hypothesis**

---

Would you like a more detailed breakdown or any additional help? Here are some related questions:

1. What is the difference between paired and unpaired t-tests?
2. How is the t-statistic different from the z-statistic?
3. How do you interpret the results of a hypothesis test in simple terms?
4. What is the significance level in hypothesis testing, and why is it important?
5. How can you reduce the standard error in a paired t-test?

**Tip**: Always check the degrees of freedom when using critical values from statistical tables!

Using the data below from a simple random paired sample from two (2) populations under consideration, at the 10% significance level, use the paired t-test to determine if the mean of population one is not equal to that of population 2. The paired Population 1 = [10, 8, 13, 13, 17, 12, 12, 11], Population 2 = [12, 7, 11, 16, 15, 9, 12, 7].

Perform a paired t-test to determine if the means of Population 1 and Population 2 are significantly different at the 10% significance level. This statistical problem involves calculating the mean difference, standard deviation, and t-statistic using a random paired sample. Suitable for undergraduate statistics students.

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