Using the t-distribution, it is found that since the test statistic is more than the critical value for the right-tailed test, it can be conclude that the isolated compound gives more counts than those from background, and hence, we can be 95% confident that the compound is indeed radioactive.
At the null hypothesis, it is tested if isolated compounds do not give more counts than those from background, that is:
[tex]H_0: \mu_1 - \mu_2 \leq 0[/tex]
At the alternative hypothesis, it is tested if the give more counts, that is:
[tex]H_1: \mu_1 - \mu_2 > 0[/tex]
Using a calculator, the mean and the standard deviation for each sample are given by:
[tex]\mu_1 = 33.2, s_1 = 6.058[/tex]
[tex]\mu_2 = 24.25, s_2 = 4.35[/tex]
Considering the sample sizes of 5 and 4, respectively, the standard errors are given by:
[tex]s_1 = \frac{6.058}{\sqrt{5}} = 2.71[/tex]
[tex]s_2 = \frac{4.35}{\sqrt{4}} = 2.175[/tex]
For the distribution of differences, the mean and standard error are given by:
[tex]\overline{x} = \mu_1 - \mu_2 = 33.2 - 24.25 = 8.95[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.71^2 + 2.175^2} = 3.4749[/tex]
The standard error was found from the standard deviation for each sample, hence, the t-distribution is used.
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{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{8.95 - 0}{3.4749}[/tex]
[tex]t = 2.58[/tex]
The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with 5 + 4 - 2 = 7 df and a 0.05 significance level is of [tex]t^{\ast} = 1.895[/tex]
Since the test statistic is more than the critical value for the right-tailed test, it can be conclude that the isolated compound gives more counts than those from background, and hence, we can be 95% confident that the compound is indeed radioactive.
A similar problem is given at brainly.com/question/24826023
