# Answer in Statistics and Probability for fiz #250687

A car is driven 20,000 km/year. To test the claim, a sample at 150 car owners are picked randomly and asked to state the distance that their car had traveled in a year. Do you agree to the claim of the following sample data: x = 22,500 km, σ = 3800 km. Use the significance level of 5%.

Show working on critical value approach and p-value approach.

The following null and alternative hypotheses need to be tested:

“H_0:mu=20000”

“H_1:munot=20000”

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 = 0.05, df=n-1”

“=150-1=149” ​degrees of freedom, and the critical value for a two-tailed test is“t_c = 1.976013.”

The rejection region for this two-tailed test is “R = {t: |t| > 1.976013}.”

The t-statistic is computed as follows:

“t=dfrac{bar{x}-mu}{s/sqrt{n}}=dfrac{22500-20000}{3800/sqrt{150}}=8.0575”

Since it is observed that “|t|= 8.0575>1.976013=t_c” it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailed, “alpha=0.05, df=149,” “t=8.0575” is “papprox0,” and since “p=0<0.05=alpha,” it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean “mu” is different than 20000, at the “alpha = 0.05” significance level.

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