(.neck the assumptions: 1. Enter the matrix of expected values. Round values to 2 decimal places: a Are all of the values in the matrix of expected values greater than or equal to 5? 2. No N is given, we will assume N N: Name the test The conditions are met to use a - T: Test Statistic The formula set up of the first and last term of the test statistic is as follows: Round initial values to 2 decimal places: X2=2(@) =([:-[:]) ' 2/[:]+...+([:].[:]) \"2/ :1 Final answer from technology rounded to 2 decimal places: 0: Obtain the P-value Report the final answer to 4 decimal places. It is possible when rounded that a p-value is 0.0000 M: Make a decision Goodness of Fit Chi-Squared hypothesis test (a = 0.05) for the claim that all 6 outcomes of rolling 1 dice are equally likely. The number of trials is n = 89 Enter the expected value for each possible outcome the table; round these expected values to four decimal . laces Observed Expected X Frequency Frequency (counts) (counts) Round the following answers accurate to three decimal places. But retain unrounded numbers for future calculations. What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.) 2 _ X _ What is the p-value for this sample? (Report answer accurate to 3 decimal places.) p-value = The p-value is... 0 Less than (or equal to) a 0 Greater than a This P-Value leads to a decision to... O Reject the null hypothesis. 0 Fail to reject the null hypothesis. As such, the final conclusion is that... Q There is sufficient evidence to warrant rejection of the claim that all 6 categories are equally likely to be selected. 0 There is not sufficient evidence to warrant rejection of the claim that all 6 categories are equally likely to be selected. Goodness of Fit Chi-Squared hypothesis test (a = 0.05) for the claim that all 6 outcomes of rolling 1 dice are equally likely. The number of trials is n = 81 Enter the expected value for each possible outcome the table; round these expected values to four decimal laces Observed Expected Frequency Frequency (counts) (counts) Round the following answers accurate to three decimal places. But retain unrounded numbers for future calculations. What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.) X2: What is the p-value for this sample? (Report answer accurate to 3 decimal places.) p-value = The p-value is... 0 Less than (or equal to) a 0 Greater than a: This P-Value leads to a decision to... O Reject the null hypothesis. 0 Fail to reject the null hypothesis. As such, the final conclusion is that... Q There is sufficient evidence to warrant rejection of the claim that all 6 categories are equally likely to be selected. 0 There is not sufficient evidence to warrant rejection of the claim that all 6 categories are equally likely to be selected. Benford's law states that the probability distribution of the first digits of many items (e.g. populations and expenses) is not uniform, but has the probabilities shown in this table. Business expenses tend to follow Benford's Law, because there are generally more small expenses than large expenses Perform a "Goodness of Fit" Chi-Squared hypothesis test (a = 0.05) to see if these values are consistent with Benford's Law. If they are not consistent, it there might be embezzelment. Complete this table. The sum of the observed frequencies is 107 Observed Benford's Expected X Frequency Law P(X) Frequency (Counts) (Counts) 1 23 301 2 22 176 3 20 125 4 12 097 5 5 079 6 11 067 7 6 058 8 3 051 9 5 046 Report all answers accurate to three decimal places. What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.) x2 = What is the P-value for this sample? (Report answer accurate to 3 decimal places.) P-value = The P-value is... O less than (or equal to) a O greater than a This P-Value leads to a decision to... O reject the null hypothesis O fail to reject the null hypothesisBenford's law states that the probability distribution of the first digits of many items (e.g. populations and expenses) is not uniform, but has the probabilities shown in this table. Business expenses tend to follow Benford's Law, because there are generally more small expenses than large expenses. Perform a "Goodness of Fit" Chi-Squared hypothesis test (a = 0.05) to see if these values are consistent with Benford's Law. If they are not consistent, it there might be embezzelment. Complete this table. The sum of the observed frequencies is 112 Observed Benford's Expected X Frequency Law P(X) Frequency (Counts) (Counts) 26 301 2 19 176 3 21 125 7 097 5 12 079 6 11 067 7 4 .058 8 5 051 9 7 .046 Report all answers accurate to three decimal places. What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.) X = What is the P-value for this sample? (Report answer accurate to 3 decimal places.) P-value = The P-value is... O less than (or equal to) a greater than a This P-Value leads to a decision to... O reject the null hypothesis fail to reject the null hypothesisQuestion 1 You might think that if you looked at the first digit in randomly selected numbers that the distribution would be uniform. Actually, it is not! Simon Newcomb and later Frank Benford both discovered that the digits occur according to the following distribution: (digit, probability) (1, 0.301), (2, 0.176), (3, 0.125), (4, 0.097), (5, 0.079), (6, 0.067), (7, 0.058), (8, 0.051), (9, 0.046) The IRS currently uses Benford's Law to detect fraudulent tax data. Suppose you work for the IRS and are investigating an individual suspected of embezzling. The first digit of 148 checks to a supposed company are as follows: Digit 2 3 5 6 7 8 9 Observed 41 20 11 1 1 17 7 16 3 Frequency a. State the appropriate null and alternative hypotheses for this test. b. Explain why a = 0.01 is an appropriate choice for the level of significance in this situation. c. What is the P-Value? Report answer to 4 decimal places P-Value = d. What is your decision? O Fail to reject the Null Hypothesis O Reject the Null Hypothesis0 Question 1 v At the beginning of the semester we collected some data from you. Two of the questions on the survey were about gender and whether or not you have equal, more, or less energy in the afternoon compared to the morning. Below are the results. Assuming that gender and energy levels are not associated, what is the expected value for the number of males who have equal energy in the afternoon as the morning? 0 Question 3 v The following is data regarding gender and whether or not a person has equal, more, or less energy in the afternoon compared to the morning. Below are the partial results. The sample size is 126. Assuming that gender and energy levels are not associated, what is the expected value for the number of males who have equal energy in the afternoon as the morning? Round to 4 decimal places. Background: Morris Saldov conducted a study in Eastern and Central Newfoundland in 1988 to examine public attitudes towards social spending. In particular, the study tried to determine if knowing someone on public assistance (yes, no) affected one's views on social spending (too little, about right, too much). The data from the study is summarized in the table below. --m --n _--l nI-l- Source: Morris Saldov, Public Attitudes to Social Spending in Newfoundland," Canadian Review of Social Policy, 26, November 1990, pages 10-14. Directions: Conduct a chi-square test for independence to determine if the association between knowing someone on public assistance and views on social spending is statistically significant. 1. Choose the correct null and alternative h notheses. O HO: There is an association between knowing someone on public assistance and views on social spending. Ha There is no association between knowing someone on public assistance and views on social spending. H0: There is no association between knowing someone on public assistance and views on social spending. Ha There is an association between knowing someone on public assistance and views on social spending. \\/ 2. Compute the test statistic. Complete the following table of expected counts. (Round your answers to 3 decimal places). Too little About right Too much someone on public assistance and views on social spending is statistically significant. 1. Choose the correct null and alternative hypotheses. O HO: There is an association between knowing someone on public assistance and views on social spending. Ha There is no association between knowing someone on public assistance and views on social spending. (9 H0: There is no association between knowing someone on public assistance and views on social spending. Ha There is an association between knowing someone on public assistance and views on social spending. \\/ 2. Compute the test statistic. Complete the following table of expected counts. (Round your answers to 3 decimal places). Yes No Too little About right Too much Compute the value of the test statistic. (Round your answer to 2 decimal places.) X2: 3. Compute the p-value. (Round your answer to 4 decimal places.) p-value = In , they randomly sampled 260 female voters, and 310 male voters. They collected data on the respondent's opinion on building a new sports stadium. We want to know whether there is good evidence that one's gender influences whether a person is for or against the new stadium. Use a: = 0.01. _ For Bond Issue Against Bond Issue m- _-_IIEI a) What is the correct null hypothesis? 3P1:P2 :P3 :P4 : Gender and Stadium Attitudes are dependent. : Gender and Stadium Attitudes are Independent. 1191751327519375104 c) Chi Square Test Statistic = 1 decimal place Degree's of Freedom = d) The p-value = 3 decimal places