Chapter 11 Chi-Square Distribution

Review So far, we have used several probability distributions for hypothesis testing and confidence intervals with normal distribution and Student’s t distribution. In this section, we will be using chi-squre.

What is Chi-Square?

Example:

Mode (high point)

Degrees of Freedom

Example: (The situation) Innovative Machines Incorporated has developed two new letter arrangements for computer keyboards. The company wishes to see if there is any relationship between the arrangement of letters on the keyboard and the number of hours it takes a new typing student to learn to type at 20 words per minute. Or, from another point of view, is the time it takes a student to learn to type independent of the arrangement of the letters on a keyboard? Use 5% level of significance

Example: (step 1)

Example: (chart) Step 2: Determine E

Answer for E (will show in class) Keyboard21-40 h41-60 h61-80 hRow Total AO:25 E:24 O:30 E:40 O:25 E:16 80 BO:30 E:36 O:71 E:60 O:19 E: StandardO:35 E:30 O:49 E:50 O:16 E: Column Total (sample size)

Cell

Example: (Degrees of freedom for test of independence)

Conclusion Look in the book with chi-square table. Since we have Chi-square as with d.f. 4 The corresponding P-value falls between and Since (.005< P-Value < 0.010) <.05, we reject null and accept alternate. Based on 5% level of significance, we are taking a chance to conclude that keyboard arrangement and learning time are not independent.

Group Work (the situation) Vending Machine is to install soda machines in elementary school and high school. The market analyst wish to know if flavor preference and school level are independent. A random sample of 200 students was taken. Their school level and soda preferences are given. Is independence indicated at the 1% level of significance?

Group Work (table) SodaHigh SchoolElementaryRow Total CokeO:33 E: O:57 E: 90 PepsiO:30 E: O:20 E: 50 Mountain DewO:5 E: O:35 E: 40 FantaO:12 E: O:8 E: 20 Column Total (sample size)

How to Test for independence of two statistical variables Look at Pg 582. Copy it and follow it!

Test of homogeneity The test claim that different populations share the sample proportions of specified characteristics.

Test of Homogeneity

Example: If you could own one pet, what kind would you choose? The possible responses were of the following. Does the same proportion of males same as females prefer each type of pet? Use 1 % level of significance GenderDogCatOther petNo Pet Female Male

Fill this out GenderDogCatOther petNo PetRow Total FemaleO:120 E: O:132 E: O:18 E: O:30 E: MaleO:135 E: O:70 E: O:20 E: O:25 E: Column Total

Answer GenderDogCatOther petNo PetRow Total FemaleO:120 E: O:132 E: O:18 E:20.73 O:30 E: MaleO:135 E: O:70 E:91.82 O:20 E:17.27 O:25 E: Column Total (sample size)

Fill this out Cell

Answer Cell

Final Answer Chi-square= d.f.=3 P-value=.001 Based on 1% level of significance, we are taking a chance to say that males and female students have different preferences when it comes to selecting a pet because we rejected the null saying preference is the same and accept the alternate saying the preference is different.

Homework Practice Pg 588 #1-15 even

CHI-SQUARE: GOODNESS OF FIT

Reason Behind Goodness of Fit Set up a test to investigate how well a sample distribution fits a given distribution Use observed and expected frequencies to compute the sample chi-square statistics Find or estimate the P-value and complete the test

Hypothesis Testing

Sample statistic

Question Does present distribution of favorable responses the same or different than last year? To test this hypothesis, a random sample of 500 employees was taken. The chart is on the next slide. Use 1% level of significance

Example CategoryPercentage of Favorable Responses Vacation time4% Salary65% Safety regulations13% Health and retirement benefits12% Overtime policy and pay6% CategoryObserved Vacation time30 Salary290 Safety regulations70 Health and retirement benefits 70 Overtime40

Answer CategoryOE Vacation time Salary Safety regulations Health and retirement benefits Overtime Total

Answer

Group Work The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community (Red Lake village) Use 5% level of significance to test the claim that the age distribution fits the age distribution of red lake village Age% populationObserved in Red Lake Village Under 57.2% % % %45

Answer

Homework Practice Pg 597 #1-18 even