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Answer the following to summarize the test of the hypothesis that there is no dependence between the two variables result of picking the Crush and bet placed the following week. For your test, use the 0.10 level of significance. (a) Determine the type of test statistic to use. Type of test statistic: (Choose one) (b) Find the value of the test statistic. (Round to two ar mare decimal places.) (c) Find the critical value for a test at the 0.10 level af significance. (Round to two or mare decimal places.) (d) Can we reject the hypothesis that there is no dependence between the variables result of picking the Crush and bet placed the following week? Use the 0.10 level of significance. O Yes O No

Answer the following to summarize the test of the hypothesis that there is no dependence between the two variables result of picking the Crush and bet placed the following week. For your test, use the 0.10 level of significance. (a) Determine the type of test statistic to use. Type of test statistic: (Choose one) (b) Find the value of the test statistic. (Round to two ar mare decimal places.) (c) Find the critical value for a test at the 0.10 level af significance. (Round to two or mare decimal places.) (d) Can we reject the hypothesis that there is no dependence between the variables result of picking the Crush and bet placed the following week? Use the 0.10 level of significance. O Yes O No

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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A group of researchers wanted to examine some hypotheses about the behavior of people who bet on professional football games. One hypothesis was that people tend to bet according to how their picks turned out the previous week: if the team a person picked to win one week actually won, the person might also pick that team the next week, thinking that the team is "on a hot streak," or the person might bet against that team the next week, thinking that the team is "due for a loss." Another hypothesis was that there is no such rhyme or reason to betting: there is no association between the bets that people make one week and the bets they make the next. In an experimental study, the researchers had each of their participants bet on each game of a professional football season. The contingency table below contains some information from a random sample of 100 bets placed on the Columbus Crush (picking them to win) during the last 14 games of the season (the Crush had 7 wins and 7 losses over that period). The contingency table indicates, for each bet placed on the Crush, whether or not the team won and how a participant who placed the bet went the following week. So, there are two variables represented in this table: result of picking the Crush ("Crush won" or "Crush lost") and bet placed the following week ("picked Crush to win" or "picked Crush to lose"). Each cell of the table contains three numbers: the first number is the observed cell frequency (Oᵢⱼ); the second number is the expected cell frequency (Eᵢⱼ) under the assumption that there is no dependence between the two variables result of picking the Crush and bet placed the following week; and the third number is the following value. \[ \frac{(O_{i j}-E_{i j})^2}{E_{i j}} \] \[ \text{(Observed cell frequency $-$ Expected cell frequency)}^2 / \text{Expected cell frequency} \] The numbers labeled "Total" are totals for observed frequency. --- ### Part 1 Fill in the missing values in the contingency table. Round your expected frequencies to two or more decimal places, and round your \(\frac{(O_{i j}-E_{i j})^2}{E_{i j}}\) values to three or more decimal places. | | Bet placed the following week | | | | |
Transcribed Image Text:A group of researchers wanted to examine some hypotheses about the behavior of people who bet on professional football games. One hypothesis was that people tend to bet according to how their picks turned out the previous week: if the team a person picked to win one week actually won, the person might also pick that team the next week, thinking that the team is "on a hot streak," or the person might bet against that team the next week, thinking that the team is "due for a loss." Another hypothesis was that there is no such rhyme or reason to betting: there is no association between the bets that people make one week and the bets they make the next. In an experimental study, the researchers had each of their participants bet on each game of a professional football season. The contingency table below contains some information from a random sample of 100 bets placed on the Columbus Crush (picking them to win) during the last 14 games of the season (the Crush had 7 wins and 7 losses over that period). The contingency table indicates, for each bet placed on the Crush, whether or not the team won and how a participant who placed the bet went the following week. So, there are two variables represented in this table: result of picking the Crush ("Crush won" or "Crush lost") and bet placed the following week ("picked Crush to win" or "picked Crush to lose"). Each cell of the table contains three numbers: the first number is the observed cell frequency (Oᵢⱼ); the second number is the expected cell frequency (Eᵢⱼ) under the assumption that there is no dependence between the two variables result of picking the Crush and bet placed the following week; and the third number is the following value. \[ \frac{(O_{i j}-E_{i j})^2}{E_{i j}} \] \[ \text{(Observed cell frequency $-$ Expected cell frequency)}^2 / \text{Expected cell frequency} \] The numbers labeled "Total" are totals for observed frequency. --- ### Part 1 Fill in the missing values in the contingency table. Round your expected frequencies to two or more decimal places, and round your \(\frac{(O_{i j}-E_{i j})^2}{E_{i j}}\) values to three or more decimal places. | | Bet placed the following week | | | | |
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