Using the t-distribution, as we have the standard deviation for the samples, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that the treatment increases the survival time of patients.
At the null hypothesis, we test if the treatment did not increase the survival time, that is, the subtraction of means is 0, hence:
[tex]H_0: \mu_T - \mu_W = 0[/tex]
At the alternative hypothesis, we test if the treatment has increased the survival time, hence:
[tex]H_1: \mu_T - \mu_W > 0[/tex]
For each sample, we have that:
[tex]\mu_W = 2.2625, s_W = \frac{0.75769858}{\sqrt{8}} = 0.2679[/tex]
[tex]\mu_T = 4.5875, s_T = \frac{0.50267143}{\sqrt{8}} = 0.1777[/tex]
Hence, for the distribution of differences, the mean and the standard error are given by:
[tex]\overline{x} = \mu_T - \mu_W = 4.5875 - 2.2625 = 2.325[/tex]
[tex]s = \sqrt{s_W^2 + s_T^2} = \sqrt{0.2679^2 + 0.1777^2} = 0.3215[/tex]
It 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{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{2.325 - 0}{0.3215}[/tex]
[tex]t = 7.23[/tex]
Considering a right-tailed test, as we are testing if the mean is greater than a value, with a standard significance level of 0.05 and 8 + 8 - 2 = 14 df, we have that the critical value is of [tex]t^{\ast} = 1.7613[/tex].
Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that the treatment increases the survival time of patients.
More can be learned about the t-distribution at https://brainly.com/question/13873630
