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A crossover trial is a type of experiment used to compare two drugs. Subjects take one drug for a period of time, then switch to the other. The responses of the subjects are then compared using matched-pair methods. In an experiment to compare two pain relievers, seven subjects took one pain reliever for two weeks, then switched to the other. They rated their pain level from 1 to 10, with larger numbers representing higher levels of pain. The results are listed below. Can you conclude that the mean pain level is more with drug B? Let uv1 represent the mean pain level with drug A and uvd-uv1-uv2. Use the a=0.05 level and the P-value

Drug A:
Subject 1- 1
Subject 2- 1
Subject 3 - 5
Subject 4- 2
Subject 5- 5
Subject 6- 6
Subject 7- 5
Drug B:
Subject 1- 3
Subject 2- 7
Subject 3- 3
Subject 4- 2
Subject 5- 6
Subject 6- 6
Subject 7- 7

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

5 answers

In the context of this crossover trial comparing pain levels for two drugs, we need to state the null and alternate hypotheses clearly.

Null Hypothesis (H0):

The mean pain level with Drug A is equal to the mean pain level with Drug B. \[ H_0: \mu_A - \mu_B = 0 \]

Alternate Hypothesis (H1):

The mean pain level with Drug B is greater than the mean pain level with Drug A. \[ H_1: \mu_A - \mu_B < 0 \] or equivalently, \[ H_1: \mu_B > \mu_A \]

Type of Test:

Since we are testing whether the mean pain level is specifically greater for Drug B compared to Drug A, this is a one-tailed test (specifically a right-tailed test since we are looking for evidence that the pain level with Drug B is higher).

Summary:

  • Null Hypothesis (H0): \(\mu_A - \mu_B = 0\)
  • Alternate Hypothesis (H1): \(\mu_B > \mu_A\)
  • Type of Test: One-tailed test (right-tailed).

Next, to conduct the statistical test, you'd typically calculate the differences in pain levels for each subject (Drug B - Drug A), find the mean and standard deviation of those differences, and perform a paired t-test to obtain the p-value and test the hypotheses. Let me know if you would like to proceed with that calculation!

(b) Compute the p-value. Round the answer to four decimal places.

To compute the p-value for this crossover trial, we will first calculate the differences in pain levels for each subject when switching from Drug A to Drug B. Then, we'll conduct a paired t-test.

Step 1: Calculate the Differences

We'll find the difference in pain levels for each subject:

  • Difference = Pain Level with Drug B - Pain Level with Drug A

| Subject | Pain Level (Drug A) | Pain Level (Drug B) | Difference (B - A) | |----------|----------------------|----------------------|---------------------| | 1 | 1 | 3 | 2 | | 2 | 1 | 7 | 6 | | 3 | 5 | 3 | -2 | | 4 | 2 | 2 | 0 | | 5 | 5 | 6 | 1 | | 6 | 6 | 6 | 0 | | 7 | 5 | 7 | 2 |

Step 2: Calculate the Mean and Standard Deviation of Differences

Now we compute the mean of the differences and the standard deviation.

Differences: 2, 6, -2, 0, 1, 0, 2

Mean of Differences (\( \bar{D} \)): \[ \bar{D} = \frac{(2 + 6 - 2 + 0 + 1 + 0 + 2)}{7} = \frac{9}{7} \approx 1.2857 \]

Standard Deviation of Differences (D):

  1. Calculate the variance: \[ D = \sqrt{\frac{\sum (D_i - \bar{D})^2}{n-1}} \] Calculating each squared deviation:
  • \( (2 - 1.2857)^2 \approx 0.5102 \)
  • \( (6 - 1.2857)^2 \approx 21.5102 \)
  • \( (-2 - 1.2857)^2 \approx 10.8787 \)
  • \( (0 - 1.2857)^2 \approx 1.6531 \)
  • \( (1 - 1.2857)^2 \approx 0.0817 \)
  • \( (0 - 1.2857)^2 \approx 1.6531 \)
  • \( (2 - 1.2857)^2 \approx 0.5102 \)

Sum of Squared Deviations: \[ 0.5102 + 21.5102 + 10.8787 + 1.6531 + 0.0817 + 1.6531 + 0.5102 \approx 36.8932 \]

Variance: \[ s^2 = \frac{36.8932}{7 - 1} = \frac{36.8932}{6} \approx 6.1489 \]

Standard Deviation: \[ s = \sqrt{6.1489} \approx 2.4772 \]

Step 3: Calculate the t-statistic

Using the mean difference, standard deviation, and number of pairs: \[ t = \frac{\bar{D} - 0}{s / \sqrt{n}} = \frac{1.2857}{2.4772 / \sqrt{7}} \]

First, calculate the denominator: \[ s / \sqrt{n} = 2.4772 / \sqrt{7} \approx 0.935 \]

Thus, \[ t = \frac{1.2857}{0.935} \approx 1.376 \]

Step 4: Determine the degrees of freedom and find the p-value

Degrees of freedom \( df = n - 1 = 7 - 1 = 6 \).

Using a t-table or calculator for \( t = 1.376 \) and \( df = 6 \), we can find the p-value for a one-tailed test.

Using a t-distribution calculator:

  • p-value for \( t = 1.376 \) with \( df = 6 \) is approximately \( 0.1143 \).

Final Result

The p-value rounded to four decimal places is: \[ \text{p-value} \approx 0.1143 \]

This p-value indicates that there is not sufficient evidence to reject the null hypothesis.

Is there enough evidence to conclude that the mean pain level is more with drug B?

To determine whether there is sufficient evidence to conclude that the mean pain level is more with Drug B than Drug A, we compare the calculated p-value to the significance level (\(\alpha\)) that we set earlier.

Given:

  • Significance Level (\(\alpha\)): 0.05
  • Calculated p-value: 0.1143

Decision Rule:

  • If the p-value is less than \(\alpha\), we reject the null hypothesis.
  • If the p-value is greater than or equal to \(\alpha\), we fail to reject the null hypothesis.

Comparison:

  • p-value: 0.1143
  • \(\alpha\): 0.05

Since \(0.1143 > 0.05\), we fail to reject the null hypothesis.

Conclusion:

There is not enough evidence to conclude that the mean pain level is significantly higher with Drug B compared to Drug A at the 0.05 significance level. In other words, based on this data, we cannot assert that Drug B leads to higher pain levels than Drug A.