Answer to Question #207320 in Statistics and Probability for Abel

Question

the following are the results of extension tests carried out on two types of yarn(percentage extension at break). yarn A: 14.1,14.7,15.1,14.3,15.6,14.8 and yarn B: 16.9,16.3,15.9,15.7,15.7. do these results suggest that one yarn is significantly more extensible than the other?

Answer 1

$\bar{X}_A=14.77$

$s_A=0.543$

$n_A=6$

$\bar{X}_B=16.1$

$sB​=0.51$

$n_B=5$

Since population standard deviations are unknown, t-distribution is used. similarly, since standard deviations are approximately equal, and the samples are independent, two independent samples t-test assuming equal variances is appropriate.

$H0:\mu_A=\mu_B$

$Ha:\mu_A<\mu_B$

$t = \frac{\overline{x}_{A} - \overline{x}_{B}}{s_{p}\sqrt{\frac{1}{n_{A}} + \frac{1}{n_{B}}}}$ which follows $t_{n_A+n_B-2}$ distribution

$s_{p} = \sqrt{\frac{(n_{A} - 1)s_{A}^{2} + (n_{B} - 1)s_{B}^{2}}{n_{A} + n_{B} - 2}}$

$s_{p} = \sqrt{\frac{5\times0.543^{2} + 4\times0.51^{2}}{6 + 5 - 2}}=0.52845$

$t = \frac{14.77 - 16.1}{0.52845\sqrt{\frac{1}{6} + \frac{1}{5}}}=-4.167$

let $\alpha=0.05$

$Cv=t_{0.05,9}=-1.833$

since the absolute value of the test statistic (4.167) is grater than the absolute value of the critical value (1.833) we reject the null hypothesis. there is sufficient evidence to support the suggestion that one yarn is significantly extensible than the other.