To test the efficacy of a certain energy drink a random sample of individuals is asked to run a certain distance rest consume the energy drink, and run the distance a second time. The time it takes the individual to run the distance is recorded each time. The data (in minutes) were recorded as: Subject: $$ \begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\end{array} $$ Pro-drink time: $$ 43544755 \quad 51445853 $$ Post-drink time 3956485449435951 A paired t test is conducted on the data, with the difference Pre-drink - Post-drink palred by subject and $$ \mathrm{H}_{\mathrm{a}} $$ uniff $$ 0 $$. Assume that the paired differences come from a normally distributed population. The p-value for this paired t test is = 0.075 0.160 0.198

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To test the efficacy of a certain energy drink a random sample of individuals is asked to run a certain distance rest consume the energy drink, and run the distance a second time. The time it takes the individual to run the distance is recorded each time. The data (in minutes) were recorded as: Subject: $$ \begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\end{array} $$ Pro-drink time: $$ 43544755 \quad 51445853 $$ Post-drink time 3956485449435951 A paired t test is conducted on the data, with the difference Pre-drink - Post-drink palred by subject and $$ \mathrm{H}_{\mathrm{a}} $$ uniff $$ >0 $$. Assume that the paired differences come from a normally distributed population. The p-value for this paired t test is = 0.075 0.160 0.198

UpStudy AI · Accepted Answer

To determine the p-value for the paired $$ t $$-test, let's carefully analyze the provided data and perform the necessary calculations.

### **Data Interpretation**

Based on the formatting, we interpret the data as follows:

- **Subjects:** 1, 2, 3, 4, 5, 6, 7, 8
- **Pre-drink (Pro-drink) times (in minutes):** 43, 54, 47, 55, 51, 44, 58, 53
- **Post-drink times (in minutes):** 39, 56, 48, 54, 49, 43, 59, 51

### **Calculating Differences**

Compute the difference $$ D = \text{Pre} - \text{Post} $$ for each subject:

| Subject | Pre (Pro-drink) | Post (Post-drink) | Difference $$ D $$ |
|---------|------------------|--------------------|---------------------|
| 1       | 43               | 39                 | 4                   |
| 2       | 54               | 56                 | -2                  |
| 3       | 47               | 48                 | -1                  |
| 4       | 55               | 54                 | 1                   |
| 5       | 51               | 49                 | 2                   |
| 6       | 44               | 43                 | 1                   |
| 7       | 58               | 59                 | -1                  |
| 8       | 53               | 51                 | 2                   |

### **Statistical Calculations**

1. **Mean Difference ($$ \bar{D} $$):**
   $$
   \bar{D} = \frac{4 + (-2) + (-1) + 1 + 2 + 1 + (-1) + 2}{8} = \frac{6}{8} = 0.75
   $$

2. **Sample Standard Deviation ($$ s_D $$):**
   - Calculate each squared deviation from the mean:
     $$
     \begin{align*}
     (4 - 0.75)^2 &= 10.5625 \
     (-2 - 0.75)^2 &= 7.5625 \
     (-1 - 0.75)^2 &= 3.0625 \
     (1 - 0.75)^2 &= 0.0625 \
     (2 - 0.75)^2 &= 1.5625 \
     (1 - 0.75)^2 &= 0.0625 \
     (-1 - 0.75)^2 &= 3.0625 \
     (2 - 0.75)^2 &= 1.5625 \
     \end{align*}
     $$
   - Sum of squared deviations:
     $$
     SS = 27.5
     $$
   - Sample variance:
     $$
     s^2 = \frac{SS}{n-1} = \frac{27.5}{7} \approx 3.9286
     $$
   - Sample standard deviation:
     $$
     s \approx \sqrt{3.9286} \approx 1.981
     $$

3. **Test Statistic ($$ t $$):**
   $$
   t = \frac{\bar{D}}{s / \sqrt{n}} = \frac{0.75}{1.981 / \sqrt{8}} \approx \frac{0.75}{0.700} \approx 1.071
   $$
   - **Degrees of Freedom (df):** $$ n - 1 = 7 $$

### **Determining the p-value**

Using a $$ t $$-distribution table or appropriate statistical software:

- For $$ df = 7 $$ and $$ t \approx 1.071 $$, the one-tailed p-value falls between 0.15 and 0.20.
- Interpolating more precisely, the p-value is approximately **0.160**.

### **Conclusion**

The p-value for the paired $$ t $$-test is **0.160**.

**Answer:** 0.160
The p-value for the paired t-test is 0.160.

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