Question

Medical researchers conducted a study to determine whether treadmill exercise could improve the walking ability of patients suffering from claudication, which is pain caused by insufficient blood flow to the muscles of the legs. A sample of 51 patients walked on a treadmill for six minutes every day. After six months, the mean distance walked in six minutes was 362 meters, with a standard deviation of 80 meters. For a control group of 54 patients who did not walk on a treadmill, the mean distance was 374 meters with a standard deviation of 86 meters. Can you conclude that the mean distance walked for patients using a treadmill is less than the mean for the controls? Let uv1 denote the mean distance walked for patients who used a treadmill. Use the a=0.05 level of significance.

State the null and alternate hypothesis.
Is this a right/left or two tailed test?

Answer 1

Compute the P-value. Round the answer to at least four decimal places

Answer 2

Do we reject Hv0 at the a=0.05 level

Answer 3

To analyze whether treadmill exercise improves the walking ability of patients suffering from claudication, we can set up the hypotheses:

### Null and Alternative Hypothesis

- **Null Hypothesis (H0)**: The mean distance walked by patients using a treadmill is equal to or greater than the mean distance walked by the control group.
\[
H_0: \mu_1 \geq \mu_2
\]

- **Alternative Hypothesis (H1)**: The mean distance walked by patients using a treadmill is less than the mean distance walked by the control group.
\[
H_1: \mu_1 < \mu_2
\]

Where:
- \( \mu_1 \) = mean distance walked for the treadmill (treatment) group.
- \( \mu_2 \) = mean distance walked for the control group.

### Type of Test

This is a **left-tailed test** because we are specifically testing if the mean distance walked by the treadmill group is **less than** that of the control group.

### Summary
- **H0**: \( \mu_1 \geq \mu_2 \)
- **H1**: \( \mu_1 < \mu_2 \)
- **Test Type**: Left-tailed test

Answer 4

To compute the P-value for this hypothesis test, we will perform a two-sample z-test because we are comparing the means of two independent groups.

Here are the key statistics we have:

- Sample size for treadmill group (\(n_1\)): 51
- Sample mean for treadmill group (\(\bar{x}_1\)): 362 meters
- Standard deviation for treadmill group (\(s_1\)): 80 meters

- Sample size for control group (\(n_2\)): 54
- Sample mean for control group (\(\bar{x}_2\)): 374 meters
- Standard deviation for control group (\(s_2\)): 86 meters

### Step 1: Calculate the Z-score

The formula for the Z-score for two independent samples is:

\[
Z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

Substituting the values:

\[
Z = \frac{(362 - 374)}{\sqrt{\frac{80^2}{51} + \frac{86^2}{54}}}
\]

Calculating the variances:

\[
s_1^2 = 80^2 = 6400 \quad \Rightarrow \quad \frac{s_1^2}{n_1} = \frac{6400}{51} \approx 125.49
\]

\[
s_2^2 = 86^2 = 7396 \quad \Rightarrow \quad \frac{s_2^2}{n_2} = \frac{7396}{54} \approx 137.74
\]

Now, plug these variance components into the Z-score formula:

\[
Z = \frac{-12}{\sqrt{125.49 + 137.74}} = \frac{-12}{\sqrt{263.23}} \approx \frac{-12}{16.23} \approx -0.739
\]

### Step 2: Find the P-value

To find the P-value corresponding to \(Z = -0.739\), we look up this Z-value in the standard normal distribution table or use a calculator.

You can also use statistical software or a calculator with cumulative distribution functions to find the P-value.

Using a standard normal distribution table or an online calculator:

The P-value for \(Z = -0.739\) is approximately:

\[
P(Z < -0.739) \approx 0.2296
\]

### Conclusion

Rounding to at least four decimal places, the P-value is:

\[
\text{P-value} \approx 0.2296
\]

Answer 5

To determine whether to reject the null hypothesis \( H_0 \) at the \(\alpha = 0.05\) level, we compare the P-value we calculated to the significance level.

### Summary of Results:
- **Calculated P-value**: \( \approx 0.2296 \)
- **Significance Level (\(\alpha\))**: \(0.05\)

### Decision Rule:
- If the P-value is less than \(\alpha\) (i.e., \( \text{P-value} < \alpha \)), we reject the null hypothesis \( H_0 \).
- If the P-value is greater than or equal to \(\alpha\) (i.e., \( \text{P-value} \geq \alpha \)), we fail to reject the null hypothesis \( H_0 \).

### Comparison:
- \( 0.2296 \) (P-value) \( > 0.05 \) (α)

### Conclusion:
Since the P-value is greater than \( \alpha \), we **fail to reject** the null hypothesis \( H_0 \). There is not enough statistical evidence at the \(0.05\) level of significance to conclude that the mean distance walked by patients using a treadmill is less than that of the control group.