Question

Rio's company is manufacturing steel wire with an average tensile strength of 50 kilos. The Laboratory test 16 pieces showed that the mean was 45kls with standard deviation of 5 kilos , is it correct to the company to claim that the tensile strength of steel wire is 50 kilos? Test the hypothesis at 0.01 level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=50$

$H_1:\mu\not=50$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=15$ and the critical value for a two-tailed test is $t_c =2.946712.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.946712\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{45-50}{5/\sqrt{16}}=4$$

Since it is observed that $|t|=4>2.946712=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=15$ degrees of freedom, $t=4$ is $p= 0.001159,$ and since $p= 0.001159<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 50, at the $\alpha = 0.01$ significance level.