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Problem 32 Hard Difficulty

According to the National Automobile Dealers Association, the mean price for used cars is $\$ 10,192 .$ A manager of a Kansas City used car dealership reviewed a sample of 50 recent used car sales at the dealership in an attempt to determine whether the population mean price for used cars at this particular dealership differed from the national mean. The prices for the sample of 50 cars are shown in the file named UsedCars.
a. Fomulate the hypotheses that can be used to determine whether a difference exists in the mean price for used cars at the dealership.
b. What is the $p$ -value?
c. At $\alpha=.05,$ what is your conclusion?

Answer

a. $H_{0} : \mu=10192, H_{a} : \mu \neq 10192$
b. $0.01<P<0.025$
c. There is sufficient evidence to support that a difference exists in the mean
price for used cars at the dealership.

Discussion

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Yuktaben S.

December 8, 2020

32. According to the National Automobile Dealers Association, the mean price for used cars is $10,192. A manager of a Kansas City used car dealership reviewed a sample of 50 recent used car sales at the dealership in an attempt to determine whether the po

Video Transcript

our mu is equal to 10,192 and our alternative hypothesis would be the opposite of that. So it would say our population means is not equal to 10,192. And also given in this problem is at our sample size, our end equals 50. Now, here's an interesting situation in this problem. We're not given an ex bar or an s. Um, and this problem, we're not going to standard deviation either. So how do we compute these? So for calculating X bar, we follow the this formula. Export equals this sum of all the individual data points divided by the sample size. So we're using the Web file that the book provides us. We can calculate this sum, which is 491,000 500 divided by 50. Now, if you want to use excel for this, you can use the following formula equals in all caps, some and then the range of the values. So for me, it was a one a 12 a 51 okay. And that evaluates to 9750 now to calculate the standard deviation we use the following formula s is equal to the square root of this son of the difference between each individual data point and the sample mean squared over and minus one, and that is equal to 1399.99 If you wanted to do that in excel, you would do S t d e v dot es And then your date arrange a 12 a 51. So now that we have our our S and R x bar, we can compute a T test statistic to find our P value. So to compute a T test statistic that is equal to our sample mean minus our population mean divided by our sample, um, standard deviation divided by the square root of our sample size, which is equal to forth out or 9750 minus 10,192 over our sample standard deviation of 1399.99 divided by the square root of 50. And this evaluates to roughly negative 2.3 are native to 0.23 Okay, so if we draw our t distribution, T equals zero in the middle. This value would lie to the left of T equals zero somewhere here. And we're interested in finding the area to the left of this T equals negative 2.23 This represents that this area represents the probability that tea is less than or equal to negative 2.23 Now, in order to find this probability value, we will need to compute a degrees of freedom. So we take our end. So a degree of freedom is equal to and minus one, which is equal to 49. Now we go to our table, we find where and is equal to or degrees of freedom is equal to 49. We've scanned across that row and we figure out where our test statistic lies. So our test statistic was equal to ah, where is it is equal to negative 2.23 Now, remember that because this this charge asymmetrical from tea from T equals 02 um, the left and it's symmetrical from T equals zero to the right. Uh, this value over here is also equal to the probability that he is greater than or equal to 2.23 It's equal to this value so we can use this chart to find where to 0.23 lies that lies be in between 2.1 and 2.405 So it lies. So our probability lies between 0.25 and 0.1 So I will write this on a new page Probability that tea is less than or equal to negative. Two point 23 is between. Is there a 0.1 and zero point 0 to 5? Make sure those are the values. Yes. Okay. But this is not our p value because we have a Our alternative hypothesis is that mu is not equal to something. So this requires a two tailed test and to tail test. So in order to find a p value, we would multiply this times two. So 0.2 is less than or equal to two times the probability that tea is less than or equal to negative 2.23 which is a sinner equal to 0.5 So our P value is a member of the set. You're a 0.2 You're a point verified R p value lies in this range somewhere. Now, in the in part, C were asked if compare our P value to an Alfa of 0.5 So our P value lies in this range and because the two is less than R P value, which is less than or equal to 0.5 which is less than or equal to 0.5 and this is our Alfa over here, we can reject the no hypothesis. So what does this mean for us? That means that there is sufficient evidence sufficient evidence to claim that the mean used car price at this dealership is different, then the national mean of $10,192.