(a) identify the expected distribution and state H0 and Ha,...

Key Concepts

Decision Making and Interpretation

Once the test statistic is calculated and compared against the critical value, a decision is made: if the test statistic exceeds the critical value, the null hypothesis is rejected; otherwise, it is not rejected. The final step involves interpreting this statistical decision in context, explaining what it means in terms of the original claim or research question.

Critical Value and Rejection Region

The critical value is determined from the chi-square distribution table using the specified significance level (alpha) and degrees of freedom (typically the number of categories minus one). The rejection region consists of all values of the test statistic that are greater than the critical value, indicating that if the test statistic falls into this region, the null hypothesis can be rejected.

Chi-Square Test Statistic

The test statistic is calculated using the formula ?² = ?((observed - expected)² / expected), summing over all categories. This statistic quantifies the deviation of the observed frequencies from the expected frequencies and serves as the basis for comparing against a reference chi-square distribution.

Null and Alternative Hypotheses

In hypothesis testing, the null hypothesis (H?) typically posits that there is no difference between the observed distribution and the expected distribution, while the alternative hypothesis (H?) asserts that a difference does exist. Formulating these correctly is critical, as they define the basis for testing claims and drawing conclusions from the data.

Chi-Square Goodness-of-Fit Test

This is a nonparametric statistical test used to determine whether an observed frequency distribution differs from a theoretical or expected distribution. It compares the observed counts in various categories to the counts expected under a specified hypothesis, which in many cases is based on historical or known proportions.

Expected Frequency Distribution

The expected distribution comprises the counts anticipated in each category if the null hypothesis were true. These are derived by multiplying the total sample size by the theoretical proportion for each category. This step is crucial because it provides the baseline against which the actual observations are compared.

Transcript

00:01 Okay, so in this question we have a pie chart that shows the distribution of the opinions of us parents on whether a college education is worth the expense.

00:09 Okay, so we have the probabilities.

00:14 Now, there is an economist that claims that the distribution of the opinions of the us teenagers is different from the distribution of us parents.

00:24 So what do we do? as always, in order to test the claim, we are very skeptical.

00:29 So in order to test the claim, we randomly select 200 us.

00:32 Teenagers and ask them whether a college education is worth the expense.

00:38 So we have our survey results with us.

00:42 The first column is the response.

00:46 In order to solve these kinds of questions, always the first thing that we do is draw the table.

00:53 Response.

00:54 The first one is strongly agree.

01:02 Strongly agree.

01:04 I fall in this category by the way.

01:14 What is happening? all right, we need to okay, so the first category is strongly agree.

01:25 The second one is somewhat agree.

01:35 This is somewhat agree.

01:39 Then it is neither agree nor disagree.

01:49 There is neither agree nor disagree.

01:54 Then there is somewhat disagree.

02:02 Somewhat disagree.

02:03 Then in the end, there is strongly disagree.

02:10 I think these are the people who dropped out of college.

02:14 And then went on to become billionaires.

02:17 Yeah, who knows? like, maybe bill gates or zuckerberg or these kind of people.

02:27 Okay.

02:31 Then we have the observed values.

02:36 The observed values.

02:42 The frequencies that we observed in our sample were 56, oh, sorry, 86, 86, 62, 34, 14, and 4.

02:56 All right now what are the probabilities now we have been given a pi chart which has the probability so the probability for strongly agree is 55 so i will write this as 0 .55 then we have somewhat agree which is 0 .3 oh just a moment this would be wrong yeah there should be 0 .55 there should be 0 .30 then neither agree or disagrees 5%.

03:29 So this should be 0 .55.

03:29 This is so this should be 0 .55.

03:30 0 .05 then we have somewhat disagree is 6%.

03:41 What is happening? this should be 0 .05.

03:49 Then this should be 0 .06.

03:53 And strongly disagrees 4%.

03:56 0 .4.

04:05 Now, what would be our null and alternative hypothesis? now, this is very important.

04:10 You frame these correctly.

04:13 You get half of your question right.

04:14 You get a good understanding of what you are going to do.

04:17 The null hypothesis will be.

04:18 Be that the distribution of opinion of the parents of the parents and teenagers about the education or about the education about the college education about let's write it like this about the college education is similar is similar okay the distribution that whether they think that the college education is what the expense are not, this distribution for parents as well as the teenagers is similar.

05:28 And what will be the alternative hypothesis? the alternative hypothesis will be that the distribution of opinions of the parents and teenagers about the college education, about the college education, differ.

06:14 All right.

06:17 Now we are going to use the kai square goodness of fit test.

06:20 We have an expected distribution and we have the observed frequencies.

06:25 Now we want to see if the expected distribution fits the observed frequencies.

06:30 What is the first step? the first step in this test is always finding the expected values for the categories.

06:36 So expected value ei for a given category will be e i is equal to the sample size the sample size multiplied by the probability for that category the probability for that category probability of the category i what is the sample size the sample size is 200 okay so just a moment just one yeah the sample size is 200 so this overall addition is 200 and over here this is going to be the column of the expected values okay so 200 into 0 .55 this should be 110 then 200 into 0 .30 this should be 60 200 into 0 .05 this should be 10 200 into point 6 this should be 12 and 200 into point 4 this should be 8 is this adding up.

08:10 Let us just check.

08:11 I think we have missed something.

08:13 Strongly agree is 55.

08:15 Make sense.

08:16 Somewhat agree is 30.

08:18 Neither agree nor disagree is five.

08:23 Somewhat disagree is six.

08:25 And strongly disagree is four.

08:31 Oh, so this is where we made a mistake.

08:33 This should be 100 in 110.

08:37 Okay.

08:38 Now let's see.

08:39 Look at the addition.

08:40 80.

08:41 This is 192.

08:41 This 200.

08:42 Yes.

08:42 So these are our expected values.

08:44 Now what is the next step? now we have the expected values and the next step is calculating the kai square statistic.

08:49 How do we do that? for every category we are going to apply the formula o minus e that is a difference between the observed and the expected values.

08:56 We square them.

08:58 We divide the value by the expected value and in the end we sum them all up.

09:03 So this will give us the overall chi square statistic for our table.

09:07 Now let us do that.

09:08 Let us find these values...

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