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:
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:
Two groups of 22 patients, hence, the standard errors are:
The distribution of the differences is has:
The test statistic is given by:
In which 
is the value tested at the null hypothesis.
Hence:
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
