Lecture 24 Slides

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University of Michigan *

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250

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Statistics

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Feb 20, 2024

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24

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Welcome to our last week of Stats 250! stats250F23@umich.edu v Due Wednesday, December 6, by 7:00 pm v HW 8 v Updated optional Lab schedule and Office hours are posted on Canvas v Extra credit opportunities v Exam 2 Playbook v ECoach End of Term Survey v Stats 250 Student Experience Survey
This test is used for assessing if a particular discrete model is a good fitting model for a discrete characteristic. Example 1: Is the color distribution (red, green, etc.) in a bag of M&Ms the same as the one stated on the Mars Candy website? Example 2: Has the model for the method of transportation (drive, bike, walk, other) used by students to get the class changed from that 5 years ago? Chi-Square Goodness of Fit Test
This test helps assess if two discrete (categorical) variables are independent for a population, or if there is an association between the two categorical variables. Example 1 : Is there a relationship between academic grade level (first year, second year, etc.) and living situation (dorm, apartment, house, etc.) for the population of UM students? Example 2 : Is there an association between satisfaction with quality of public schools (unsatisfied, neutral, satisfied) and political party (Republican, Democrat, etc.) for the population of all adults? Note: Chi-square Test of Independence is similar to regression but the two variables are categorical (not quantitative). Chi-Square Test of Independence
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Chi-Square Distributions Both tests based on a Chi-Square test statistic that, under the null hypothesis, follows a Chi-Square distribution with some degrees of freedom, ࠵? ࠵? (࠵?࠵?) Shape of a Chi-Square distribution changes based on the degrees of freedom
BIG IDEA Behind a Chi-Square Test 1. Sample data will consist of observed counts 2. Compute expected counts assuming the null is true 3. Check assumptions (based on expected counts) 4. Compare observed counts to expected using a ࠵? ! test statistic The larger the difference in counts à the. _____________ the ࠵? " test statistic, à the ____________ evidence against ࠵? 0 . ࠵? " = ( ࠵?࠵?࠵? − ࠵?࠵?࠵? " ࠵?࠵?࠵?
Scenario: Many UM courses have adopted flexible course modalities: In-Person, Synchronous streaming, asynchronous videos. Question: Among all undergraduate students at UM, is there an association between preferred course modality and whether a student commutes to campus or not? One Population: All UM ugrad students Two Variables: Preferred Modality (In-person, Synchronous, Asynchronous) and Commuter Status (Non-Commuter, Commuter) Chi-Square Test of Independence
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Data: Survey of random sample of 1000 ugrad students at UM. Course Modality and Commuter Status
Course Modality and Commuter Status Question: Among all undergraduate students at UM, is there an association between preferred course modality and whether a student commutes to campus or not? Observed Counts: Null Hypothesis: H 0 : There is _______________ between _______________________________ and ____________________________________________ for the population of___________________________________.
Course Modality and Commuter Status In-Person: 73% of sampled students (730 out of 1000) preferred in-person. Assuming no association between preferred modality and commuter status, we’d expect 73% of the 810 non-commuters to prefer in-person … Expected count: ________________________ and we’d expect 73% of the 190 commuters to prefer in-person… Expected count: ________________________
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Course Modality and Commuter Status Synchronous: 20 % of sampled students (200 out of 1000) preferred synchron. Assuming no association between preferred modality and commuter status, we’d expect 20% of the 810 non-commuters to prefer synchronous… Expected count: ________________________ and we’d expect 20% of the 190 commuters to prefer synchronous … Expected count: ________________________
Test of Independence Expected Counts A Closer Look at Expected Counts: Synchronous: expect 20% of the 190 commuters to prefer synchronous … Expected count: 38 = 0.20 ∗ 190 = !"" #""" ∗ 190 = !""∗#%" #""" Cross-Product Rule ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? = ࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?
Course Modality and Commuter Status If there is no association between commuter status and preferred modality, how many non-commuters would you expect to prefer asynchronous? Try the cross-product rule:
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Test of Independence Assumptions Stating Assumptions 1. The responses are a random sample from population of interest. 2. At least 80% of the expected counts are greater than 5. 3. None of the expected counts are less than 1. Checking Assumptions 1. Told the 1000 undergraduate UM students are a random sample. 2. We have 6 expected counts. What % are greater than 5? ______ 3. We have 6 expected counts. What % are less than 1? ________ Expected Counts
Our test statistic measures of how close observed counts are to expected counts Course Modality and Commuter Status ࠵? ! = , ࠵?࠵?࠵? − ࠵?࠵?࠵? ! ࠵?࠵?࠵? = 600 − 591.3 ! 591.3 + 160 − 162 ! 162 + 50 − 56.7 ! 56.7 +
Do you think a ࠵? ! test statistic of __________ is large enough to provide some evidence against the null hypothesis? If ࠵? ! is true, this ࠵? ! test statistic has a ______________ distribution. For Tests of Independence df = ____________________ We can use this distribution to find the corresponding p -value. Course Modality and Commuter Status
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Course Modality and Commuter Status Use the R Shiny App to find and then evaluate the ࠵? -value… iClicker : How much evidence is there against the null hypothesis? A. Not enough evidence B. Some evidence C. Strong evidence D. Very strong evidence
Course Modality and Commuter Status Which of the following is an appropriate conclusion? A. Based on the data, we have __________evidence to suggest that there is an association between preferred course modality and commuter status for the population of all UM ugrad students. B. Based on the data, we have __________evidence to suggest that there is no association between preferred course modality and commuter status for the population of all UM ugrad students. ࠵? $ : There is no association between preferred course modality and commuter status for the population of all UM ugrad students. p-value = _____________
Carbonated Beverage Name and Region Scenario : People in the US have different words for carbonated beverages depending on which region of the US they are from. Question : Among all US adults, is there an association between the word used when referring to carbonated beverages and the region of the US they are from? a. Based on the results, estimate the population proportion of US adults residing in the West who use the word “Soda” when referring to a carbonated drink.
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Carbonated Beverage Name and Region Question : Among all US adults, is there an association between the word used when referring to carbonated beverages and the region of the US they are from? b. State the null hypothesis for the test of independence: H 0 : There is _______________________________ between _________________________ and __________________ for the population of ______________________________.
Carbonated Beverage Name and Region c. Test statistic value is 18.03. Calculate numerical contribution to this test statistic value from the US adults residing in South who use word “ Coke ”.
Carbonated Beverage Name and Region Test statistic value is 18.03. Evaluate the ࠵? -value. iClicker : How much evidence is there against the null hypothesis? A. Not enough evidence B. Some evidence C. Strong evidence D. Very strong evidence
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Carbonated Beverage Name and Region iClicker : Complete the conclusion statement… Based on the data, we have very strong evidence to suggest that there __________ an association between region of US adult is from and word choice for carbonated beverages for the population of all US adults. A. IS B. IS NOT
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Test of Independence
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It has been my pleasure working with you! Thank YOU for Taking Stats 250 I hope you have a wonderful and restful Winter Break! Good Luck with all your end-of-term tasks! Stats 250 Exam 2: Monday, December 11 from 7:30-9:30 PM Many office hours this week! YOU ARE A AZING!
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