Answer to Question #173599 in Statistics and Probability for ANJU JAYACHANDRAN

Question

9. a) 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

In this question it is important to test both the null and alternative hypothesis.

$H_o​\::μ=5.11\:kg/mm^2$

$H_1​\::μ \neq 5.11\:kg/mm^2$

This is a two-tailed test in which the sample standard deviation would be used in a t-test with one mean with unknown population standard deviation. This is a two-tailed test in which the sample standard deviation would be used in a t-test with one mean with unknown population standard deviation.

The significance level is based on the information given as $α=0.05$ and $df=n−1=30−1=29\:\:$degrees of freedom.

Two-tailed test $t_c$ critical value is $2.04523.$

Region of rejection of this two-tailed test is $R={t:∣t∣>2.04523}$

$t$ - statistic is computed as follows:

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}=\frac{8.12−5.11}{\frac{2.0}{\sqrt{30}}}=8.24322\approx 8.243$

Therefore, it is inferred that $\:|t|=8.24322>2.04523=|t_c|,$ hence a conclusion that the null hypothesis is rejected.

There is a concrete proof to claim that the population mean $\mu$  is different than $5.11,$ at the $\alpha=0.05$ significance level.

By utilizing the P value strategy, the significance level for the p-value for two-tailed is:

$\alpha=0.05, t=8.24322$ and $df=29$ degrees of freedom is $p<0.00001,$

and since p<0.00001<0.05=α it is concluded that the null hypothesis is rejected.

As a result, there is sufficient evidence to conclude that the population mean $\mu$ is distinct from  $5.11,$ at the $\alpha=0.05$ significance level.