Answer to Question #254410 in Statistics and Probability for blossom

Question

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.

Answer 1

Summary of the above information.

$n_1=20,\space \bar{x}_1=11.13,\space S_1=3.2$

$n_2=17,\space \bar{x}_2=12.98,\space s_2=4.1$

The hypothesis to be tested are,

$H_0:\mu_1=\mu_2$

$Against$

$H_1:\mu_1\not=\mu_2$

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,

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

$Sp^2=((n_1-1)*S_1^2+(n_2-1)*S_2^2)/(n_1+n_2-2)$

$Sp^2=((19*3.2^2)+(16*4.1^2))/(20+17-2)$

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

Now,

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

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

$t^*=-1.541027$

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

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

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