Answer to Question #104493 in Statistics and Probability for Aj

Question

The breaking strengths of cable

Answer 1

The breaking strengths of cables produced by manufacturer have a mean of 1800 Kg and a standard deviation of 100 Kg. By a new technique in the manufacturing process, it is claimed that breaking strength can be increased. To test this claim, a sample of 50 cables is tested and it is found that the mean breaking strength is 1850Kg. Can we support the claim at the 0.01 significance level? 

Consider Z-test for one population mean $(\mu),$ with known population standard deviation $(\sigma).$

The provided sample mean is $\bar{X}=1850$ and the known population standard deviation is $\sigma=100,$ and the sample size is $n=50.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu=1800$

$H_1: \mu>1800$

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 $\alpha=0.01,$ and the critical value for a right-tailed test is $z_c=2.33.$

The rejection region for this right-tailed test is $R=\{z:z>2.33\}$

The z-statistic is computed as follows:

$$z={\bar{X}-\mu \over \sigma/\sqrt{n}}={1850-1800 \over 100/\sqrt{50}}\approx3.535534$$

Since it is observed that $z=3.535534>z_c=2.33,$  it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $0.000204,$ and since $p=0.0002<0.01,$ it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis $H_0$ is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 1800, at the 0.01 significance level.

We conclude that a new technique can improve the break strength.