Suppose M₁ and M₂ are two models where from prior knowledge we believe P(M₁) = 0.25 and P(M₂) = 0.75. We then observe a data set D. Given each model the likelihood of D is P(D|M₁) 0.5 and P(DM₂) = 0.01. Given data D which model has a higher accuracy, or probability to be correct?
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- Read through this scenario and look at the data that was collected. State the null and all possible research hypotheses. Review the results below (I used SPSS) and answer the questions that follow. Scenario: A researcher wants to see if gender and / or income affects the total amount of help given to a stranger who is sitting on the side of a busy road with a sign asking for help. The independent variables are gender, income, and the interaction of gender and income. The dependent variable is total help. He wants to know if one or both factors – or the interaction of the two - affects the total amount of help offered. Because he is analyzing two independent variables (gender and income), he used a factorial ANOVA. His results show the main effect of each of the independent variables on the dependent variable (total help) and the interaction effect. The researcher is using a 95% confidence interval which means that he wants to be at least 95% sure that his independent variables…Why is it that the correct answer is D? How can we know that the data appear to be skewed in the given problem? What are the hint words in the situation that we can say it is skewed?The table below shows the number of state-registered automatic weapons and the murder rate for several Northwestern states. xx 11.5 8.3 6.9 3.7 2.4 2.3 2.7 0.4 yy 13.7 10.9 9.7 7 6 5.9 6.5 4.2 xx = thousands of automatic weaponsyy = murders per 100,000 residentsThis data can be modeled by the equation y=0.84x+3.97.y=0.84x+3.97. Use this equation to answer the following;Special Note: I suggest you verify this equation by performing linear regression on your calculator.A) How many murders per 100,000 residents can be expected in a state with 10.5 thousand automatic weapons?Answer = Round to 3 decimal places.B) How many murders per 100,000 residents can be expected in a state with 1 thousand automatic weapons?Answer = Round to 3 decimal places.
- A researcher hypothesizes that opera music will have an effect on test scores (but she is unsure whether they will increase or decrease). On a standardized reasoning test, known to have mu = 70 and sigma = 10, she has a sample of 30 subjects complete the test while listening to opera music. She then tallies up their scores, and finds a mean score for her sample of 66. Go through the steps of hypothesis testing and make a conclusion for the researcher about her experiment. Use a significance level of .05 (two-tailed). Sketch the chance distribution and shade your critical region.The table below shows the number of state-registered automatic weapons and the murder rate for several Northwestern states. xx 11.5 8.4 6.9 3.9 2.7 2.8 2.2 0.6 yy 13.7 11 10 7 6.3 6.4 6.2 4.8 xx = thousands of automatic weaponsyy = murders per 100,000 residentsThis data can be modeled by the equation y=0.82x+4.16.y=0.82x+4.16. Use this equation to answer the following; Special Note: I suggest you verify this equation by performing linear regression on your calculator.A) How many murders per 100,000 residents can be expected in a state with 3.9 thousand automatic weapons?Answer = Round to 3 decimal places.B) How many murders per 100,000 residents can be expected in a state with 1.7 thousand automatic weapons?Answer = Round to 3 decimal places.#31). Both pictures are the same problem.
- The table below shows the number of state-registered automatic weapons and the murder rate for several Northwestern states. xx 11.5 8.5 7 3.7 2.7 2.5 2.6 0.8 yy 13.8 11.1 10.1 7.2 6.4 6.1 6.1 4.8 xx = thousands of automatic weaponsyy = murders per 100,000 residentsThis data can be modeled by the equation ˆy=0.85x+4.05.y^=0.85x+4.05. Use this equation to answer the following.A) How many murders per 100,000 residents can be expected in a state with 6.7 thousand automatic weapons?Answer = Round to 3 decimal places.B) How many murders per 100,000 residents can be expected in a state with 3.3 thousand automatic weapons?Answer = Round to 3 decimal places.Chapter 11-2. The following table presents the numbers of cucumber seeds, by brand, that germinated or failed to germinate 14 days after planting. Brand A Brand B Brand C Brand D 27 Germinated 28 33 36 Failed to Germinate 5 17 4 5 Can you conclude that the the germination rate is related to the brand of seed? Use the a = 0.05 level of significance. Reject Ho. It appears that the germination rates differ among the seed brands. Do not reject Ho. There is insufficient evidence to conclude that the germination rate differs among the seed brands.Suppose the observational units in a study are the patients arriving at an emergency room on a given day, 11 Nov 2019. For the following indicate whether it can legitimately be considered a variable or not: Whether or not men have to wait longer than women yes no
- Suppose the observational units in a study are the patients arriving at an emergency room on a given day, 11 Nov 2019. For the following indicate whether it can legitimately be considered a variable or not: Whether or not men have to wait longer than women yes noPlease answer number 17.A weight-loss program wants to test how well their program is working. The company selects a simple random sample of 51 individual that have been using their program for 15 months. For each individual person, the company records the individual's weight when they started the program 15 months ago as an x-value. The subject's current weight is recorded as a y-value. Therefore, a data point such as (205, 190) would be for a specific person and it would indicate that the individual started the program weighing 205 pounds and currently weighs 190 pounds. In other words, they lost 15 pounds. When the company performed a regression analysis, they found a correlation coefficient of r = 0.707. This clearly shows there is strong correlation, which got the company excited. However, when they showed their data to a statistics professor, the professor pointed out that correlation was not the right tool to show that their program was effective. Correlation will NOT show whether or not there is…