Question #254410
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.
1
Expert's answer
2021-10-21T14:54:12-0400

Summary of the above information.

n1=20, xˉ1=11.13, S1=3.2n_1=20,\space \bar{x}_1=11.13,\space S_1=3.2

n2=17, xˉ2=12.98, s2=4.1n_2=17,\space \bar{x}_2=12.98,\space s_2=4.1

The hypothesis to be tested are,

H0:μ1=μ2H_0:\mu_1=\mu_2

AgainstAgainst

H1:μ1μ2H_1:\mu_1\not=\mu_2

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=(xˉ1xˉ2)/Sp2(1/n1+1/n2)t^*=(\bar{x}_1-\bar{x}_2)/\sqrt{Sp^2*(1/n_1+1/n_2)} where Sp2Sp^2 is the pooled sample variance given by the formula,

Sp2=((n11)S12+(n21)S22)/(n1+n22)Sp^2=((n_1-1)*S_1^2+(n_2-1)*S_2^2)/(n_1+n_2-2)

Sp2=((193.22)+(164.12))/(20+172)Sp^2=((19*3.2^2)+(16*4.1^2))/(20+17-2)

Sp2=(194.56+268.96)/35=463.52/35=13.2434286Sp^2=(194.56+268.96)/35=463.52/35=13.2434286

Now,

t=(11.1312.98)/13.24(1/20+1/17)t^*=(11.13-12.98)/\sqrt{13.24*(1/20+1/17)}

t=1.85/1.441197t^*=-1.85/\sqrt{1.441197}

t=1.541027t^*=-1.541027

We compare tt^* with the t-table value (t)(t) with at α\alpha level of significance with (n1+n22)=(20+172)=35(n_1+n_2-2)=(20+17-2)=35 degrees of freedom and reject the null hypothesis if t>tα/2,35|t^*|\gt t_{\alpha/2,35} .

The table value(t)(t) is, tα/2,df=t0.05/2,35=t0.025,35=2.030108t_{\alpha/2,df}=t_{0.05/2,35}=t_{0.025,35}=2.030108.

Since t=1.541027<t0.025,35=2.030108|t^*|=1.541027\lt t_{0.025,35}=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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