Answer in Statistics and Probability for blossomqt #248889
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? = 0.01?
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_dnot=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(4space decimalspace 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(4space decimalspace 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|=3.4805lt 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.
