Answer to Question #337545 in Statistics and Probability for che che

Question

1. in building an arena, steel bars with a mean ultimate tenslie strength of 400 Megapascal (MPa) with a variance of 81 MPa were delivered by the manufacturer. The project engineer tested 50 steel bars and found out that the mean ultimate tensile strength is 390 MPa. The decision for the extension of the contract with the manufacturer depends on the engineer. Test the hypothesis whether or not there is no significant difference between the two means using a two-tailed with u = 0.01.

a. What are the appropriate hypotheses for the two-tailed test?

b. What is the test statistic to be used and the reasons for its selection? G. What is the critical value c?

d. What is the value of the test statistic or the computed value?

e. Formulate a conclusion about the given situation.

Answer 1

a.The following null and alternative hypotheses need to be tested:

$H_0:\mu=400$

$H_a:\mu\not=400$

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

c. Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z|> 2.5758\}.$

d. The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{390-400}{81/\sqrt{50}}=-0.8730$$

e. Since it is observed that $z =0.8730<2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-0.8730)=0.382663,$ and since $p=0.382663>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 400, at the $\alpha = 0.01$ significance level.