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Body temperatures are known to be normally distributed. A medical researcher wants to determine how a drug changes the body's temperature. Seven test subjects are randomly selected, and the body temperature of each is measured in degrees Fahrenheit. The subjects are then given the drug, and after 20 minutes, the body temperature of each is measured again. The results are given below:

| Subject | Initial Temperature (x) | Second Temperature (y) | Difference (d = x - y) |
|---------|--------------------------|------------------------|------------------------|
| 1 | 101.8 | 99.2 | 2.6 |
| 2 | 98.5 | 98.4 | 0.1 |
| 3 | 98.1 | 98.2 | -0.1 |
| 4 | 99.4 | 99.0 | 0.4 |
| 5 | 98.9 | 98.6 | 0.3 |
| 6 | 100.2 | 99.7 | 0.5 |
| 7 | 97.9 | 97.8 | 0.1 |

From this information, the following were determined:

- Mean difference (\( \bar{d} \)) = 0.829
- Initial mean (\( \bar{x} \)) = 99.26
- Initial standard deviation (\( s_x \)) = 1.370
- Second mean (\( \bar{y} \)) = 98.7
- Second standard deviation (\( s_y \)) = 1.154

a) Carry out an appropriate hypothesis test to determine if, at \( \alpha = 0.05 \), there is enough evidence to conclude that the drug changes the body's temperature. Assume that there are no outliers. (6 marks)

b) State your conclusion.

Answer :

The paired t-test indicates that there is enough evidence to suggest that the drug changes the body's temperature at a significance level of 0.05.

To test whether the drug changes the body's temperature, we can perform a paired t-test on the differences in body temperature before and after taking the drug. The null hypothesis (H0) is that there is no significant difference in body temperature before and after taking the drug, while the alternative hypothesis (Ha) is that there is a significant difference.

Step 1: Calculate the sample mean of the differences (d) = 0.829

Step 2: Calculate the standard deviation of the differences (s) = 0.645

Step 3: Calculate the standard error of the mean difference (SE) = s / √n = 0.645 / √7 ≈ 0.244

Step 4: Calculate the t-statistic: t = (mean difference - hypothesized mean difference) / SE = (0.829 - 0) / 0.244 ≈ 3.397

Step 5: Find the critical t-value at α = 0.05 (two-tailed) with 6 degrees of freedom. The critical t-value is approximately ±2.447.

Since the calculated t-statistic (3.397) is greater than the critical t-value (±2.447), we reject the null hypothesis (H0) and conclude that there is enough evidence to suggest that the drug changes the body's temperature.

Conclusion: There is enough evidence to conclude that the drug changes the body's temperature at a significance level of 0.05.

Learn more about null hypothesis here:

https://brainly.com/question/33325037

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Rewritten by : Jeany

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