Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.
At the null hypothesis, it is <u>tested if the consumption is not different</u>, that is, if the subtraction of the means is 0, hence:
![H_0: \mu_1 - \mu_2 = 0](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu_1%20-%20%5Cmu_2%20%3D%200)
At the alternative hypothesis, it is <u>tested if the consumption is different</u>, that is, if the subtraction of the means is not 0, hence:
![H_1: \mu_1 - \mu_2 \neq 0](https://tex.z-dn.net/?f=H_1%3A%20%5Cmu_1%20-%20%5Cmu_2%20%5Cneq%200)
Two groups of 22 patients, hence, the standard errors are:
![s_1 = \frac{45.1}{\sqrt{22}} = 9.6154](https://tex.z-dn.net/?f=s_1%20%3D%20%5Cfrac%7B45.1%7D%7B%5Csqrt%7B22%7D%7D%20%3D%209.6154)
![s_2 = \frac{26.4}{\sqrt{22}} = 5.6285](https://tex.z-dn.net/?f=s_2%20%3D%20%5Cfrac%7B26.4%7D%7B%5Csqrt%7B22%7D%7D%20%3D%205.6285)
The distribution of the differences is has:
![\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%3D%20%5Cmu_1%20-%20%5Cmu_2%20%3D%2052.1%20-%2027.1%20%3D%2025)
![s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7Bs_1%5E2%20%2B%20s_2%5E2%7D%20%3D%20%5Csqrt%7B9.6154%5E2%20%2B%205.6285%5E2%7D%20%3D%2011.14)
The test statistic is given by:
![t = \frac{\overline{x} - \mu}{s}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Coverline%7Bx%7D%20-%20%5Cmu%7D%7Bs%7D)
In which
is the value tested at the null hypothesis.
Hence:
![t = \frac{25 - 0}{11.14}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B25%20-%200%7D%7B11.14%7D)
![t = 2.2438](https://tex.z-dn.net/?f=t%20%3D%202.2438)
The p-value of the test is found using a <u>two-tailed test</u>, as we are testing if the mean is different of a value, with <u>t = 2.2438</u> and 22 + 22 - 2 = <u>42 df.</u>
- Using a t-distribution calculator, this p-value is of 0.0302.
Since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.
A similar problem is given at brainly.com/question/25600813