1) A bicycle safety organization claims that fatal bicycle accidents are uniformly distributed throughout the week. The table shows the day of the week for which 782 randomly selected fatal bicycle accidents occurred. At 𝛼=0.10, can you reject the claim that the distribution is uniform? Sun. Mon. Tues. Wed. Thurs. Fri. Sat. 108 112 105 111 123 118

1) A bicycle safety organization claims that fatal bicycle accidents are uniformly distributed throughout the week. The table shows the day of the week for which 782 randomly selected fatal bicycle accidents occurred. At 𝛼=0.10, can you reject the claim that the distribution is uniform? Sun. Mon. Tues. Wed. Thurs. Fri. Sat. 108 112 105 111 123 118 SOLUTION: 𝐻 0 : The distribution of fatal bicycle deaths by day of the week is uniform. 𝐻 𝑎 : The distribution of fatal bicycle deaths by day of the week is not uniform. STAT - Edit L1 holds the claimed percentages Since the claim is that the distribution is uniform, ALL the percentages will be equal to each other. There are 7 categories, so divide 1 by 7 to get the claimed percentage. Type “1/7” into L1 7 times (one for each category).

1) A bicycle safety organization claims that fatal bicycle accidents are uniformly distributed throughout the week. The table shows the day of the week for which 782 randomly selected fatal bicycle accidents occurred. At 𝛼=0.10, can you reject the claim that the distribution is uniform? Sun. Mon. Tues. Wed. Thurs. Fri. Sat. 108 112 105 111 123 118 STAT - Edit L2 holds the observed frequencies Enter the numbers shown in the table into L2 L3 holds the expected values Highlight L3 and type "L1*782", since n = 782. L3 should have 111.714 in all 7 rows.

1) A bicycle safety organization claims that fatal bicycle accidents are uniformly istributed throughout the week. The table shows the day of the week for which 782 randomly selected fatal bicycle accidents occurred. At 𝛼=0.10, can you reject the claim that the distribution is uniform? Sun. Mon. Tues. Wed. Thurs. Fri. Sat. 108 112 105 111 123 118 SOLUTION: STAT - TEST - D Observed is in L2, Expected is in L3, Degrees of Freedom (df) = 6 (k – 1, where k is the number of categories) 𝑋 2 =2.43 (Standardized test statistic) 𝑝=.876 Since 𝑝>𝛼, we fail to reject the null. This means that at the 10% significance level, there is not enough evidence to reject the claim that the distribution of fatal bicycle accidents throughout the week is uniform.

Section 10-2 – Independence Contingency Tables An r x c contingency table shows the observed frequencies for two variables. The observed frequencies are arranged in r rows and c columns. The intersection of a row and a column is called a cell. Assuming that the variables are independent, the formula for calculating the expected value for any given cell is 𝐸 𝑟,𝑐 = 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑤 (𝑠𝑢𝑚 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛) 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 . The sums of the rows and columns are called marginal frequencies.

Section 10-2 – Independence Contingency Tables Chi-Square Independence Test Used to test the independence of two variables. You can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable. Two conditions that must be met in order to conduct this test: 1) Samples must be random 2) Expected frequency (E) for each cell must be ≥ 5.

Section 10-2 – Independence Contingency Tables The hypotheses are written as follows: H0: The variables are independent Ha: The variables are dependent This is ALWAYS a right tail test!! As the Chi-Square value gets larger, it becomes more likely that the variables are dependent. For this reason, this is a right-tail test. The calculator won’t even ask you when you run the test; it knows that this test is a right-tail test.

Section 10-2 – Independence Contingency Tables Steps for conducting the test 1) State the hypotheses and identify the claim. 2) Identify α. 3) Access the Matrix menu, select Edit. Enter Observed values into Matrix A 4) STAT–TEST–C 5) Make a decision to reject H0 or Fail to Reject H0. If 𝑝≤𝛼, reject the null. If 𝑝>𝛼, fail to reject the null.

𝐻 0 : The CEOs’ ages are independent of the company size. Example 2 (Page 568) The contingency table shows the results of Example 1. At 𝛼=0.01, can you conclude that the CEOs’ ages are related to company size? 𝐻 0 : The CEOs’ ages are independent of the company size. 𝐻 𝑎 : The CEOs’ ages are dependent on the company size. Enter the data into the calculator using matrices. The Matrix menu is accessed by pressing 2nd and the 𝑥 −1 key, immediately below the MATH key. Arrow over to Edit, then press 1. Your table has 2 rows and 5 columns, so make your matrix a 2 x 5 matrix. Enter the numbers from the first table into this matrix Company Size 39 & Under 40-49 50-59 60-69 70 & Over Total Small / Midsize 42 69 108 60 21 300 Large 5 18 85 120 22 250 47 87 193 180 43 550

Example 2 (Page 568) Once you have the data inserted into Matrix A, go to STAT–TEST–C. Make sure that the test is being run on Matrix A for Observed and Matrix B for Expected. Highlight Calculate and press Enter. The REALLY good news is that if you enter the values into Matrix A correctly, the calculator will take care of finding the expected values for you. Run the test, and then access the Matrix menu, select Edit and [B]. Matrix B will be filled in with the expected values for you. Check to make sure that ALL expected values are at least 5. 𝛸 2 =77.887 𝑝=4.882 𝐸 − 16. Since 𝑝≤𝛼, reject the null. This means that there is enough evidence at the 1% significance level to conclude that the CEO’s ages and the company size are dependent.

Days per week spent exercising (Observed) Example 3 (Page 570) A health club manager wants to determine whether the number of days per week that college students spend exercising is related to gender. A random sample of 275 college students is selected and the results are classified as shown in the table. At 𝛼=0.05, is there enough evidence to conclude that the number of days spent exercising per week is related to gender? 𝐻 0 : The number of days spent exercising per week is independent of gender. 𝐻 𝑎 : The number of days spent exercising per week depends on gender. (Claim) Days per week spent exercising (Observed) Gender 0-1 2-3 4-5 6-7 Total Male 40 53 26 6 125 Female 34 68 37 11 150 74 121 63 17 275

Days per week spent exercising (Observed) Example 3 (Page 570) Step 1: Enter the Observed values into Matrix A. Days per week spent exercising (Observed) Gender 0-1 2-3 4-5 6-7 Total Male 40 53 26 6 125 Female 34 68 37 11 150 74 121 63 17 275

Days per week spent exercising (Expected) Example 3 (Page 570) Step 2: STAT – TEST – C Step 3: Check Matrix B to be sure that all expected values are > 5. Since this is in fact the case, we may proceed. Step 4: Find 𝛸 2 and p and make your decision. 𝛸 2 = 3.493 (this is your standardized test statistic). p = .322 Since 𝑝>𝛼, fail to reject the null This means that there is not enough evidence at the 5% significance level to conclude that the number of days spent exercising per week is related to gender. Days per week spent exercising (Expected) Gender 0-1 2-3 4-5 6-7 Total Male 33.636 55 28.636 7.727 125 Female 40.364 66 34.364 9.273 150 74 121 63 17 275

Classwork: Page 571-572 #1-12 All Homework: Pages 572-575 #13-24 All For part b), find p instead of the critical value and compare to 𝜶 instead of using rejection regions.