Answer to Question #226891 in Statistics and Probability for Joseph

Question

1)  The strength limit of a specific type of a cable is random variable with mean value of 1500 kg and standard deviation 175 kg. The factory that manufactures this type of cables claims that they improved the materials that they used and the new resistance limit of the cable has increased. We randomly chose a sample of 50 cables and the mean strength limit was 1570 kg. If the strength limit in the specific type of cable is normally distributed with a significance level of 0.05, can we state that the claim of the manufacture factory is true? Justify your answer.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq1500$

$H_1:\mu>1500$

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a right-tailed test is $z_c=1.6449.$

The rejection region for this right-tailed test is $R=\{z:z>1.6449\}$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{1570-1500}{175/\sqrt{50}}\approx2.8284$$

Since it is observed that $z=2.8284>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z>2.8284)=0.002339,$ and since $p=0.002339<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population $\mu$ is greater than $1500,$ at the $\alpha=0.05$ significance level.

Therefore, there is enough evidence to state that the claim of the manufacture factory is true, at the $\alpha=0.05$ significance level.