Question

a researcher wants to determine whether a 2-week low-sodium diet reduces systolic blood pressure in patients with mild hypertension. the systolic blood pressure (mmhg) of 8 patients is measured before starting the diet and after 2 weeks on the diet.

\

$$\begin{tabular}{ccc} patient & before & after \\\\ \\hline 1 & 142 & 135 \\\\ 2 & 150 & 145 \\\\ 3 & 138 & 130 \\\\ 4 & 145 & 140 \\\\ 5 & 155 & 148 \\\\ 6 & 148 & 142 \\\\ 7 & 152 & 149 \\\\ 8 & 140 & 134 \\\\ \\end{tabular}$$

a. state the null hypothesis and alternative hypothesis for testing whether the diet reduces the mean systolic blood pressure.

b. use the t.test() function in r to perform a hypothesis test. report the value of the test statistic and the p-value.

c. at the significance level \\(\alpha = 0.01\\), state the decision and conclusion in the context of the study.

d. use r to construct a 99% confidence interval for the mean difference. interpret the interval in the context of the study.

Explanation:

⚡ Using what you learned: Hypothesis Testing · 🆕 New: Paired t-test in R
*Comparing paired measurements before and after a treatment*

Step 1: State the Hypotheses

Because the measurements are taken on the same 8 patients before and after the diet, this is a paired design. Let \( d_i = \text{Before}_i - \text{After}_i \) be the difference in systolic blood pressure for patient \( i \).

Let \( \mu_d \) be the mean difference in the population (\( \mu_d = \mu_{\text{Before}} - \mu_{\text{After}} \)).

If the diet reduces systolic blood pressure, the "Before" values should be higher than the "After" values on average, meaning the mean difference \( \mu_d \) is positive (\( \mu_d > 0 \)).

- **Null Hypothesis (\( H_0 \)):** The diet does not reduce mean systolic blood pressure.
\[ H_0: \mu_d = 0 \]
- **Alternative Hypothesis (\( H_a \)):** The diet reduces mean systolic blood pressure.
\[ H_a: \mu_d > 0 \]

---

Step 2: Perform the Paired t-test in R

First, we calculate the differences \( d = \text{Before} - \text{After} \) for each patient:

| Patient | Before | After | Difference (\( d = \text{Before} - \text{After} \)) |
| :---: | :---: | :---: | :---: |
| 1 | 142 | 135 | \( 142 - 135 = 7 \) |
| 2 | 150 | 145 | \( 150 - 145 = 5 \) |
| 3 | 138 | 130 | \( 138 - 130 = 8 \) |
| 4 | 145 | 140 | \( 145 - 140 = 5 \) |
| 5 | 155 | 148 | \( 155 - 148 = 7 \) |
| 6 | 148 | 142 | \( 148 - 142 = 6 \) |
| 7 | 152 | 149 | \( 152 - 149 = 3 \) |
| 8 | 140 | 134 | \( 140 - 134 = 6 \) |

Summary Statistics of Differences:

- Sample size: \( n = 8 \)
- Mean difference:
\[ \bar{d} = \frac{7 + 5 + 8 + 5 + 7 + 6 + 3 + 6}{8} = \frac{47}{8} = 5.875 \]
- Sample standard deviation of differences (\( s_d \)):
\[ s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \approx 1.5526 \]

R Code:

```R
before <- c(142, 150, 138, 145, 155, 148, 152, 140)
after <- c(135, 145, 130, 140, 148, 142, 149, 134)

Perform paired t-test (one-sided, alternative is 'greater' since we test if Before > After)

t.test(before, after, paired = TRUE, alternative = "greater")
```

Test Statistic and p-value Calculation:

- **Standard Error (\( SE \)):**
\[ SE = \frac{s_d}{\sqrt{n}} = \frac{1.5526}{\sqrt{8}} \approx 0.5489 \]
- **t-statistic:**
\[ t = \frac{\bar{d} - 0}{SE} = \frac{5.875}{0.5489} \approx 10.703 \]
- **Degrees of Freedom:**
\[ df = n - 1 = 7 \]
- **p-value:**
Using a \( t \)-distribution with 7 degrees of freedom for a one-tailed test:
\[ p\text{-value} = P(T_{7} \ge 10.703) \approx 0.0000064 \text{ (or } 6.4 \times 10^{-6}\text{)} \]

---

Step 3: Decision and Conclusion (\( \alpha = 0.01 \))

- **Decision:** Since the \( p\text{-value} \approx 0.0000064 \) is much smaller than the significance level \( \alpha = 0.01 \), we **reject the null hypothesis (\( H_0 \))**.
- **Conclusion:** There is extremely strong evidence to conclude that the 2-week low-sodium diet significantly reduces the mean systolic blood pressure in patients with mild hypertension.

---

Step 4: Construct and Interpret the 99% Confidence Interval

To find the two-sided 99% confidence interval for the mean difference \( \mu_d \):

R Code:

```R

Two-sided 99% confidence interval for the paired difference

t.test(before, after, paired = TRUE, conf.level = 0.99)
```

Calculation:

- Critical value \( t^* \) for \( df = 7 \) at a 99% confidence level (two-tailed):
\[ t^*_{0.005, 7} \approx 3.4995 \]
- **Margin of Error (\( ME \)):**
\[ ME = t^* \times SE = 3.4995 \times 0.5489 \approx 1.921 \]
- **Confidence Interval:**
\[ \bar{d} \pm ME = 5.875 \pm 1.921 \implies (3.954, 7.796) \]

Interpretation:

We are 99% confident that the true mean reduction in systolic blood pressure after 2 weeks on the low-sodium diet lies between \( 3.95 \text{ mmHg} \) and \( 7.80 \text{ mmHg} \).

Answer:

a. Hypotheses

• \( H_0: \mu_d = 0 \)
• \( H_a: \mu_d > 0 \)

(where \( \mu_d = \mu_{\text{Before}} - \mu_{\text{After}} \))

b. R Test Results

• Test Statistic (\( t \)): \( 10.703 \)
• p-value: \( 6.4 \times 10^{-6} \) (or \( 0.0000064 \))

c. Decision and Conclusion (\( \alpha = 0.01 \))

• Decision: Reject \( H_0 \) because \( p\text{-value} < 0.01 \).
• Conclusion: There is statistically significant evidence that the 2-week low-sodium diet reduces mean systolic blood pressure in patients with mild hypertension.

d. 99% Confidence Interval

• 99% CI: \( (3.95, 7.80) \text{ mmHg} \)
• Interpretation: We are 99% confident that the true mean reduction in systolic blood pressure due to the diet is between \( 3.95 \text{ mmHg} \) and \( 7.80 \text{ mmHg} \).