Take this test as you would take a test in class.For each exer...

Question

Take this test as you would take a test in class.For each exercise, perform the steps below.

c.Find the critical value(s) and identify the rejection region(s).

A real estate agency says that the mean home sales price in Olathe, Kansas, is greater than in Rolla, Missouri. The mean home sales price for 39 homes in Olathe is $392,453. Assume the population standard deviation is $224,902. The mean home sales price for 38 homes in Rolla is $285,787. Assume the population standard deviation is $330,578. At α=0.05, is there enough evidence to support the agency’s claim? (Adapted from Realtor.com)

Accepted Answer

Hello. In this video, we are told that a university administrator wants to know if the average number of credits taken by freshmen is less than that taken by sophomores. The sample of 30 freshmen has a mean of 13.2 credits with a population standard deviation of 2.5 credits. A sample of 35 sophomores has a mean of 14 credits with a population standard deviation of 2.1 credits at a significance level of 0.05. What is the critical value and rejection region for the hypothesis test? So the first thing we want to do is we want to go ahead and create our null and alternate hypothesis. Now the cling From the university administrator is that the average number of credits taken by freshmen is less than that taken by sophomores. So let's go ahead and allow a new one to equal to the credits of the freshman. And U2 is going to be the credits taken by sophomores. So, the claim here is that they are saying that the units taken by freshmen is less than that taken by sophomore, so the original claim is that new 1 is less than U2. Now in this case here, because we are given known population standard deviations between the two samples, our claim is going to be the null or the alternate hypothesis. So the alternate hypothesis is going to be that new one is less than U2. And the null hypothesis is going to be the complement of this claim, which is that new1 is greater than equal to m2. Now, what's important here is that because our alternate hypothesis uses the less than symbol, we are going to be using a left tailed test. Furthermore, because we are working with small sample populations, we are going to use a left-tailed Z test. Now, we are also working with a significance level of 0.05. So, let's go ahead and start by determining the critical value. Now, we are given the critical value of zero or the significance level of 0.05. So if we use a Z table, and for the left tail test, the Z critical value with the significance level of 0.05 is going to approximate to be -1.645. This is going to be our critical value for the following claim. But now what we want to do is we want to go ahead and identify the rejection region. Now, whenever we are determining the rejection region, we want to always compare a Z test statistic to a critical value. But because we are using a left tail test, whenever the Z test statistic is less than the Z critical value, That tells us that the rejection is going to be any value of z that is less than -1.645 for a left tail test. So this is going to be our area of rejection. And the option that matches this solution is going to be option B. So option B has a critical value of -1.645, and the area of rejection is going to be all values that are less than -1.645. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Hypothesis Testing

Critical Value and Rejection Region

Two-Sample Z-Test for Means with Known Population Standard Deviations

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