Using the t-distribution, it is found that since the <u>test statistic is more than the critical value</u> 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 <u>tested if isolated compounds do not give more counts than those from background</u>, that is:
![H_0: \mu_1 - \mu_2 \leq 0](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu_1%20-%20%5Cmu_2%20%5Cleq%200)
At the alternative hypothesis, it is <u>tested if the give more counts</u>, that is:
![H_1: \mu_1 - \mu_2 > 0](https://tex.z-dn.net/?f=H_1%3A%20%5Cmu_1%20-%20%5Cmu_2%20%3E%200)
Using a calculator, the mean and the standard deviation for each sample are given by:
![\mu_1 = 33.2, s_1 = 6.058](https://tex.z-dn.net/?f=%5Cmu_1%20%3D%2033.2%2C%20s_1%20%3D%206.058)
![\mu_2 = 24.25, s_2 = 4.35](https://tex.z-dn.net/?f=%5Cmu_2%20%3D%2024.25%2C%20s_2%20%3D%204.35)
Considering the sample sizes of 5 and 4, respectively, the standard errors are given by:
![s_1 = \frac{6.058}{\sqrt{5}} = 2.71](https://tex.z-dn.net/?f=s_1%20%3D%20%5Cfrac%7B6.058%7D%7B%5Csqrt%7B5%7D%7D%20%3D%202.71)
![s_2 = \frac{4.35}{\sqrt{4}} = 2.175](https://tex.z-dn.net/?f=s_2%20%3D%20%5Cfrac%7B4.35%7D%7B%5Csqrt%7B4%7D%7D%20%3D%202.175)
For the distribution of differences, the mean and standard error are given by:
![\overline{x} = \mu_1 - \mu_2 = 33.2 - 24.25 = 8.95](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%3D%20%5Cmu_1%20-%20%5Cmu_2%20%3D%2033.2%20-%2024.25%20%3D%208.95)
![s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.71^2 + 2.175^2} = 3.4749](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7Bs_1%5E2%20%2B%20s_2%5E2%7D%20%3D%20%5Csqrt%7B2.71%5E2%20%2B%202.175%5E2%7D%20%3D%203.4749)
The standard error was found from the <u>standard deviation for each sample</u>, hence, the t-distribution is used.
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{\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)
![t = \frac{8.95 - 0}{3.4749}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B8.95%20-%200%7D%7B3.4749%7D)
![t = 2.58](https://tex.z-dn.net/?f=t%20%3D%202.58)
The critical value for a <u>right-tailed test</u>, as we are testing if the mean is greater than a value, with 5 + 4 - 2 = <u>7 df and a 0.05 significance level</u> is of ![t^{\ast} = 1.895](https://tex.z-dn.net/?f=t%5E%7B%5Cast%7D%20%3D%201.895)
Since the <u>test statistic is more than the critical value</u> 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