Answer to Question #345387 in Statistics and Probability for Natalia

Question

Question:1

It is claimed that an automobile is driven in the average less than 20,000 km /year. To test this claim, a random sample of 100 auto-mobiles owners are asked to keep a record of the kilometers, they travel. Would you agree with this claim if the random sample showed an average of 23,500 kilometers and a standard deviation of 3900 kilometers. Use a 0.01 level of significance.

Question: 2

Test the hypothesis that the average content of containers of a particular lubricant is 10 liters if the contents of a random sample of 10 containers are 10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, and 9.8 litters.

Use a 0.01 level of significance and assume that the distribution is Normal.

Answer 1

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

$H_0:\mu\ge20000$

$H_1:\mu<20000$

This corresponds to a left-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 $\alpha = 0.01,$ $df=n-1=99$ and the critical value for a left-tailed test is $t_c =-2.364606.$

The rejection region for this left-tailed test is $R = \{t:t<-2.364606\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{23500-20000}{3900/\sqrt{100}}=8.9744$$

Since it is observed that $t=8.9744>-2.364606=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=99$ degrees of freedom, $t=8.9744$ is $p=1,$ and since $p=1>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is less than 20000, at the $\alpha = 0.01$ significance level.

2. Mean  

$$\dfrac{1}{10}(10.2+9.7+10.1+10.3+10.1$$

$$+9.8+9.9+10.4+10.3+9)=10.06$$

Variance

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=\dfrac{1}{10-1}(0.0196+0.1296$$

$$+0.0016+0.0576+0.0016+0.0676+0.0256$$

$$+0.1156+0.0576+0.0676)=\dfrac{0.544}{9}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{0.544}{9}}\approx0.246$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=10$

$H_1:\mu\not=10$

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 $\alpha = 0.01,$ $df=n-1=9$ and the critical value for a two-tailed test is $t_c =3.249823.$

The rejection region for this two-tailed test is $R = \{t:|t|>3.249823\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{10.06-10}{0.246/\sqrt{10}}=0.7713$$

Since it is observed that $|t|=0.7713<3.249823=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=9$ degrees of freedom, $t=0.7713$ is $p=0.460299,$ and since $p=0.460299>0.01=\alpha,$ 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 10, at the $\alpha = 0.01$ significance level.