Question

Five samples of a ferrous-type substance are to be used to determine if there is a difference between a laboratory chemical analysis and an X-ray fluorescence analysis of the iron content. Each sample was split into two sub-samples and the two types of analysis were applied. Following are the coded data showing the iron content analysis:

Sample
Analysis 1 2 3 4 5
X-rays 2.0 2.0 2.3 2.1 2.4
Chemical 2.2 1.9 2.5 2.3 2.4

Assuming that the populations are normal, test at the 0.05 level of significance whether the two methods of analysis give, on the average, the same result.

Answer 1

Step-by-step explanation:

Step 1

From the given information,

number of ferrous-type substance, n = 5

Let $\mu_1$ and $\mu_2$ are the true population average for laboratory chemical and X-ray flourescence analysis

Level of significance, $\alpha = 0.05$

state the null and alternative hypotheses

$H_0 : \mu_1 - \mu_2 = 0\\\\H_1 : \mu_1 - \mu_2 \neq 0$

Attached are steps 2 to 7 of the remaining solution

Answer 2

Yes, the two methods of analysis give, on average, the same result and this can be determined by using the given data.

Given :

• Five samples of a ferrous-type substance are to be used.
• Each sample was split into two sub-samples and the two types of analysis were applied.
• 0.05 level of significance.

The mean of the first sample is:

$\bar{X_1} = \dfrac{\sum X_1}{n_1}$

$\bar{X_1} = \dfrac{ 10.8}{5}=2.16$

Now, the standard deviation of the first sample is:

$S_1 =\sqrt{\dfrac{\sum(X_1-\bar{X_1})^2}{n_1-1}}$

$S_1 =\sqrt{\dfrac{0.132}{4}}=0.182$

The mean of the second sample is:

$\bar{X_2} = \dfrac{\sum X_2}{n_2}$

$\bar{X_2} = \dfrac{ 11.3}{5}=2.26$

Now, the standard deviation of the second sample is:

$S_2 =\sqrt{\dfrac{\sum(X_2-\bar{X_2})^2}{n_2-1}}$

$S_2 =\sqrt{\dfrac{0.212}{4}}=0.230$

Now, the hypothesis test is given below:

Null Hypothesis  --  $H_0:\mu_1=\mu_2$

Alternative Hypothesis  --  $H_a:\mu_1\neq \mu_2$

The degree of freedom is calculated as:

$df=n_1+n_2-2\\df=8$

Now, the formula of the test statistics is given below:

$t = \dfrac{(\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2)}{\sqrt{\dfrac{S^2_1}{n_1}+\dfrac{S^2_2}{n_2}} }$

$t = \dfrac{2.16-2.26}{\sqrt{\dfrac{(0.182)^2}{5}+\dfrac{(0.230)^2}{5}} }$

$t = -0.76238688$

Now, according to the t-value, the p-value is 0.46771. Therefore, the null hypothesis is not rejected.

So, yes the two methods of analysis give, on average, the same result.

For more information, refer to the link given below:

https://brainly.com/question/2695653