Respuesta :
Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is less than the standard significance level of 0.05, 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 tested if the consumption is not different, that is, if the subtraction of the means is 0, hence:
[tex]H_0: \mu_1 - \mu_2 = 0[/tex]
At the alternative hypothesis, it is tested if the consumption is different, that is, if the subtraction of the means is not 0, hence:
[tex]H_1: \mu_1 - \mu_2 \neq 0[/tex]
Two groups of 22 patients, hence, the standard errors are:
[tex]s_1 = \frac{45.1}{\sqrt{22}} = 9.6154[/tex]
[tex]s_2 = \frac{26.4}{\sqrt{22}} = 5.6285[/tex]
The distribution of the differences is has:
[tex]\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14[/tex]
The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence:
[tex]t = \frac{25 - 0}{11.14}[/tex]
[tex]t = 2.2438[/tex]
The p-value of the test is found using a two-tailed test, as we are testing if the mean is different of a value, with t = 2.2438 and 22 + 22 - 2 = 42 df.
- 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 less than the standard significance level of 0.05, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.
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