Question

a used car dealer says that the mean price of a 2008 honda cr - v is at least $20,500. you suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $19,850 and a standard deviation of $1084. is there enough evidence to reject the dealers claim at $\alpha = 0.05$? assume the population is normally distributed. find the t - score for the sample.

Explanation:

Step1: Set up hypotheses

Let $\mu$ be the population mean price. The dealer's claim is $\mu\geq20500$. The alternative hypothesis (what we suspect) is $\mu < 20500$.

Step2: Calculate the t - score

The formula for the t - score is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$. Here, $\bar{x} = 19850$, $\mu=20500$, $s = 1084$, $n = 14$.

$$ t=\frac{19850 - 20500}{1084/\sqrt{14}}=\frac{- 650}{1084/3.742}\approx\frac{-650}{290}\approx - 2.24 $$

Step3: Determine the critical value and make a decision

For a one - tailed test with $\alpha = 0.05$ and $n-1=14 - 1=13$ degrees of freedom, the critical value from the t - distribution table is $t_{critical}=- 1.771$. Since the calculated t - score ($t=-2.24$) is less than the critical value ($-2.24<-1.771$), we reject the dealer's claim.

Answer:

t - score = - 2.244 (rounded). There is enough evidence to reject the dealer's claim at $\alpha = 0.05$.