Answer to Question #194451 in Statistics and Probability for ANJU JAYACHANDRAN

Question

A manufacture of rayon wants to compare that the yield strength of 5 11⋅ kg/ 2 mm is 

met or not at 5% level of significance. The manufacturer draws a sample and 

calculates the mean to be 8 12⋅ kg/ 2 mm and the standard derivation is known to be 

2⋅0kg/ mm .

2

 Carry out the statistical test appropriate for this.

Answer 1

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu=5.11\ kg/mm^2$

$H_1: \mu\not=5.11\ kg/mm^2$

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 $\alpha=0.05$ and $df=n-1=30-1=29$ degrees of freedom.

The critical value for a two-tailed test is $t_c=2.04523.$  

The rejection region for this two-tailed test is $R=\{t:|t|>2.04523\}.$

The $t$ - statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{8.12-5.11}{2.0/\sqrt{30}}\approx8.24322$$

Since it is observed that $|t|=8.24322>2.04523=|t_c|,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$  is different than $5.11,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value for two-tailed, the significance level  $\alpha=0.05, t=8.24322$ and $df=29$ degrees of freedom is $p<0.00001,$

and since $p<0.00001<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$  is different than $5.11,$ at the $\alpha=0.05$ significance level.

Therefore, there is enough evidence to claim that the yield strength is different than $5.11,$ at the $\alpha=0.05$ significance level.