Answer to Question #207314 in Statistics and Probability for Abel

Question

according to textile engineering, an important property of fiber is its water absorbency. the average percent absorbency of 25 randomly selected pieces of cotton fiber was found to be 20 with standard deviation of 1.5. a random sample of 25 pieces of acetate yielded an average percent of 12 with a standard deviation of 1.25. is there strong evidence that the population mean percent absorbency is significantly higher for cotton fiber than acetate? assume that the percent absorbency is approximately normally distributed and that the population variances in percent absorbency for the two fibers are the same. use a significant level of 0.05.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1>\mu_2$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df_{total}=df_1+df_2=24+24=48$$

Based on the information provided, the significance level is $\alpha=0.05,$the degrees of freedom are $df=48,$ and the critical value for a right-tailed test is $t_c=1.677224.$

The rejection region for this right-tailed test is $R=\{t: t>1.677224\}.$

The t-statistic is computed as follows:

$$z=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{20-12}{\sqrt{\dfrac{(25-1)1.5^2+(25-1)1.25^2}{25+25-2}(\dfrac{1}{25}+\dfrac{1}{25})}}$$

$$\approx 20.4859$$

Since it is observed that $t=20.4859>1.677224=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than the population mean $\mu_2$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for right-tailde $df=48, \alpha=0.05,$ $t=20.4859$ is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is greater than the population mean $\mu_2$ at the $\alpha=0.05$ significance level.

Therefore, there is enough evidence to claim that the percent absorbency is significantly higher for cotton fiber than acetate at $\alpha=0.05.$