Question

A manufacturer claims that the average tensile

strength of thread A exceeds the average tensile

strength of thread B. To test this claim, 50 pieces of

each type of thread are tested under similar

conditions. Type A thread had an average of tensile

strength of 86.7 kg with a standard deviation of 6.28

kg, while type B had an average tensile strength of

77.8 with a standard deviation of 5.61 kg. Test the

manufacturer’s

claim

using

a

0.05

level

of

significance.

Answer 1

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{6.28^2}{5.61^2}=1.2531$$

The critical values for $\alpha=0.05,$ $df_1=n_1-1=49$ degrees of freedom, $df_2=n_2-1=49$ degrees of freedom are $F_L = 0.5675$ and $F_U = 1.7622,$ and since $F = 1.2531$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\le\mu_2$

$H_1:\mu_1>\mu_2$

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:

$$df=df_1+df_2=n_1+n_2-2=98$$

Based on the information provided, the significance level is $\alpha = 0.05,$ and the degrees of freedom are $df = 98.$

The critical value for this right-tailed test, $\alpha = 0.05, df=98$ degrees of freedom is $t_c =1.660551.$

The rejection region for this right-tailed test is $R = \{t: t > 1.660551\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

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

$$=\dfrac{86.7-77.8}{\sqrt{\dfrac{(50-1)(6.28)^2+(50-1)(5.61)^2}{50+50-2}(\dfrac{1}{50}+\dfrac{1}{50})}}$$

$$=7.473426$$

Since it is observed that $t = 7.473426>1.660551= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, $df=98$ degrees of freedom, $t=7.473426$ is $p\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothes is rejected.

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