Question

"It is claimed that a car is driven on the average less than 28700 kilometers per year with a population standard deviation of 2100 kilometers. To test this claim, a random sample of 58 car owners are asked to keep a record of the kilometers they travelled. Would you agree with this claim if the random sample showed an average of 22500 kilometers? Use a 0.01 level of significance.

What are the given? Write only the number. :

population mean:Blank 1 km

population standard deviation:Blank 2 km

sample size:Blank 3

sample mean:Blank 4 km

level of significance:Blank 5

What is the critical value?

z:Blank 6

What is the value of the calculated z? Round your answer to the nearest hundredths.

z:Blank 7"

Answer 1

population mean: $28700$ km

population standard deviation: $2100$ km

sample size:$58$

sample mean: $22500$ km

level of significance: $0.01$

What is the critical value?

$z_c=-2.3263$

What is the value of the calculated z?

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{22500-28700}{2100/\sqrt{58}}\approx-22.48$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge28700$

$H_1:\mu<28700$

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

Since it is observed that $z=-22.48<-2.3263= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(z<-22.48)\approx0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is less than 28700, at the $\alpha = 0.01$ significance level.