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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?

5 answers

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

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:

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

Do we reject Hv0 at the a=0.05 level

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.