Coefficient of Linear thermal expansion (CTE) is a material property that describes the unit expansion of a material per one degree of temperature increase. The higher it is, the higher does the size of the material becomes under heat. A steel manufacturer claims that their newly-developed alloy has lesser thermal unit expansion than the common steel used in the market. 20 samples of the newly-developed alloy were tested, while 17 samples of common steel were tested under the same conditions. The average coefficient of the newly-developed alloy is 11.13 with a standard deviation of 3.2, while the average coefficient of common steel is 12.98 with a standard deviation of 4.1. Using a 0.05 level of significance, is the CTE of the newly-developed alloy significantly different from the common steel? Assume equal variances.
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Expert's answer
2021-10-21T14:54:12-0400
Summary of the above information.
n1=20,xˉ1=11.13,S1=3.2
n2=17,xˉ2=12.98,s2=4.1
The hypothesis to be tested are,
H0:μ1=μ2
Against
H1:μ1=μ2
Since the study assumes that the population variances are equal, we shall apply the student's t distribution as described below.
The test statistic is given as,
t∗=(xˉ1−xˉ2)/Sp2∗(1/n1+1/n2) where Sp2 is the pooled sample variance given by the formula,
Sp2=((n1−1)∗S12+(n2−1)∗S22)/(n1+n2−2)
Sp2=((19∗3.22)+(16∗4.12))/(20+17−2)
Sp2=(194.56+268.96)/35=463.52/35=13.2434286
Now,
t∗=(11.13−12.98)/13.24∗(1/20+1/17)
t∗=−1.85/1.441197
t∗=−1.541027
We compare t∗ with the t-table value (t) with at α level of significance with (n1+n2−2)=(20+17−2)=35 degrees of freedom and reject the null hypothesis if ∣t∗∣>tα/2,35 .
The table value(t) is, tα/2,df=t0.05/2,35=t0.025,35=2.030108.
Since ∣t∗∣=1.541027<t0.025,35=2.030108, we fail to reject the null hypothesis and we conclude that there is no sufficient evidence to show that the newly-developed alloy is significantly different from the common steel.
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