Answer to Question #248889 in Statistics and Probability for blossomqt

Question

One company produces premium grade carbon steel for Katana Sword manufacturers. Two different analytical methods were used to detect the contamination level of the carbon steel in 8 random specimens

  SPECIMEN

Subjects 12345678

         Method 1 1.2 1.3 1.5 1.4 1.7 1.8 1.4 1.3 Method 2 1.4 1.7 1.5 1.3 2.0 2.1 1.7 1.6

Is there sufficient evidence to conclude that tests differ in the mean contamination level use? 𝛼 = 0.01?

Answer 1

There are two samples in which observations in sample1(contamination level by method 1) can be paired with observations in sample 2(contamination level by method 2)

The hypotheses tested are,

$H_0:\mu_d=0$

$Against$

$H_1:\mu_d\not=0$, where $d$ is the difference between each pair of observations in method 1 and method 2.

To test these hypotheses, we first determine the mean of the differences$(\bar{d})$ given by,

$\bar{d}=\sum (d)/n$ ,where $n=8$ and use the students' t distribution to make a decision.

Now,

$\sum(d)=-1.7$

Therefore,

$\bar{d}=-1.7/8=-0.2125$

Variance of the differences is given by,

$V(d)=\sum(d-\bar{d})^2/(n-1)$

$V(d)=0.20875/7=0.02982143$

The standard deviation $SD(d)$ is given by,

$SD(d)=\sqrt{V(d)}=\sqrt{0.02982143}=0.1727(4\space decimal\space places)$

The t-test statistic is given by,

$t=\bar{d}/(SD(d)/\sqrt{n})$

$t=-0.2125/(0.1727/\sqrt{8})$

$t=-0.2125/0.061055=-3.4805(4\space decimal\space places)$

The t-test statistic is compared with the t-table value with $\alpha=0.01$ and $(n-1)=8-1=7$ degrees of freedom.

Table value is,

$t_{\alpha/2,7}=t_{0.01/2,7}=t_{0.005,7}=3.499$

To make comparison, we use the absolute value of the t-test statistic, that is,

$|t|=|-3.4805|=3.4805$ and reject the null hypothesis if $|t|\gt t_{0.005,7}.$

Since $|t|=3.4805\lt t_{0.005,7}=3.499,$ we fail to reject the null hypothesis and conclude that, there is no sufficient evidence to show that the tests differ in the mean contamination level at $1\%$ level of significance.