Answer to Question #157981 in Statistics and Probability for Ratul Das

Question

Breaking strength (in kg) of the front part of a new vehicle is normally distributed. In 10

trials the breaking strengths were found to be 578, 572, 570, 568, 572, 570, 570, 572, 596,

and 584. Can we say that the mean breaking strength is significantly less than 570 at 1%

level of significance? Further test the hypothesis at 5% level of significance.

Answer 1

$$578+ 572+ 570+ 568+572+ 570+570$$

$$+ 572+ 596+584=5752$$

$$\bar{x}=\dfrac{5752}{10}=575.2$$

$$\displaystyle\sum_{i=1}^{10}(x_i-\bar{x})^2=681.6$$

$$s^2 =\dfrac{\displaystyle\sum_{i=1}^{10}(x_i-\bar{x})^2}{n-1}=\dfrac{681.6}{10-1}=\dfrac{227.3}{3}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{227.3}{3}}\approx8.7025$$

The provided sample mean is $\bar{x}=575.2$ and the sample standard deviation is $s=8.7025,$ and the size of the sample is $n=10.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq570$

$H_1:\mu<570$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

The number of degrees of freedom are $df=n-1=10-1=9,$ and the significance level is $\alpha=0.01.$ Based on the provided information, the critical t-value for $\alpha=0.01$ and $df=9$ degrees of freedom is $t_c=-2.8214.$

The rejection region for this two-tailed test is $R=\{t:t<-2.8214\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{575.2-570}{8.7025/\sqrt{10}}=1.89$$

Since it is observed that $t=1.89>-2.8214=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 570, at the 0.01 significance level.

Use the P-value approach.

The P-value $t=1.89, df=9, \alpha=0.01,$ left-tailed, is $p=1-0.045668=0.954332$ Since $p=0.954332>0.01=\alpha,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 570, at the 0.01 significance level.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq570$

$H_1:\mu<570$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

The number of degrees of freedom are $df=n-1=10-1=9,$ and the significance level is $\alpha=0.05.$ Based on the provided information, the critical t-value for $\alpha=0.05$ and $df=9$ degrees of freedom is $t_c=-1.8331$

The rejection region for this two-tailed test is $R=\{t:t<-1.8331\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{575.2-570}{8.7025/\sqrt{10}}=1.89$$

Since it is observed that $t=1.89>-1.8331=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 570, at the 0.01 significance level.

Use the P-value approach.

The P-value $t=1.89, df=9, \alpha=0.01,$ left-tailed, is $p=1-0.045668=0.954332$ Since $p=0.954332>0.05=\alpha,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 570, at the 0.05 significance level.