Question #345387

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.


1
Expert's answer
2022-05-30T03:54:12-0400

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

H0:μ20000H_0:\mu\ge20000

H1:μ<20000H_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 α=0.01,\alpha = 0.01, df=n1=99df=n-1=99 and the critical value for a left-tailed test is tc=2.364606.t_c =-2.364606.

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

The t-statistic is computed as follows:


t=xˉμs/n=23500200003900/100=8.9744t=\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=tc,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=99df=99 degrees of freedom, t=8.9744t=8.9744 is p=1,p=1, and since p=1>0.01=α,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 μ\mu

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


2. Mean  

110(10.2+9.7+10.1+10.3+10.1\dfrac{1}{10}(10.2+9.7+10.1+10.3+10.1




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

Variance



s2=Σ(xixˉ)2n1=1101(0.0196+0.1296s^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+0.0016+0.0576+0.0016+0.0676+0.0256

+0.1156+0.0576+0.0676)=0.5449+0.1156+0.0576+0.0676)=\dfrac{0.544}{9}


s=s2=0.54490.246s=\sqrt{s^2}=\sqrt{\dfrac{0.544}{9}}\approx0.246


The following null and alternative hypotheses need to be tested:

H0:μ=10H_0:\mu=10

H1:μ10H_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 α=0.01,\alpha = 0.01, df=n1=9df=n-1=9 and the critical value for a two-tailed test is tc=3.249823.t_c =3.249823.

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

The t-statistic is computed as follows:


t=xˉμs/n=10.06100.246/10=0.7713t=\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=tc,|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=9df=9 degrees of freedom, t=0.7713t=0.7713 is p=0.460299,p=0.460299, and since p=0.460299>0.01=α,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 μ\mu

is different than 10, at the α=0.01\alpha = 0.01 significance level.



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