Answer to Question #350859 in Statistics and Probability for lala

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.