Problem #1 Neurological signs and symptoms of ciguatera poisoning as the start of treatment and 2.5 hours after treatment with mannitol.

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1 Ho (null hypothesis) Ha (alternative hypothesis) Problem #1 Neurological signs and symptoms of ciguatera poisoning as the start of treatment and 2.5 hours after treatment with mannitol. Hypothesis: Ho: µ number of neurological signs and symptoms measured at baseline = µ number of neurological signs at 2.5 hours after treatment with mannitol. Ha: µ number of neurological signs and symptoms measured at baseline µ number of neurological signs at 2.5 hours after treatment with mannitol. Statistical Procedure, Tests and Assumptions: Test: Paired Sample T-test Rationale: This test is being used as there are two observations being completed on the same test subject. Assumptions: 1. Normally distributed population. 2. Dependent observations 3. Random Samples - simple random sampling of the test subjects 4. Equality of Variances between the groups. Results: In reviewing the output data from SPSS the significance level is.021, which is less than.05 set, which suggests that the results are not happening by chance less than 5% of the time. We will reject the null hypothesis (which is µ number of neurological signs and symptoms measured at baseline = µ number of neurological signs at 2.5 hours after treatment with mannitol) and conclude that there is a difference in signs and symptoms after mannitol is administered; specifically with this data 2.5 hours.

2 You can see in the diagram above the mean of the signs and symptoms after 2.5 hours is less than the mean of the number of signs and symptoms at baseline. Inference: If the data was generalized to the rest of the population, it could be stated that the administration of the drug mannitol reduces the number of neurological signs and symptoms related to ciguatera poisoning. As stated in the results sections, Ho will be rejected as the data supports that there is a difference in signs and symptoms after mannitol is administered Problem #2 Wait times bases on time of day. Hypothesis: Ho: µ wait time in the morning = µ wait time in the afternoon = µ wait time in the evening. Ha: µ wait time of at least one group is to be different in the groups: morning; afternoon; or evening. Statistical Procedure, Tests and Assumptions: Test: Analysis of Variance Rationale: This test is being chosen as there is a categorical independent variable (time of day), as well as a continuous dependent variable (wait time). We are comparing multiple groups; more than two categorical variables are used in this test. We wish to measure the amount of variability within the groups and between the groups. Assumptions: 1. Normality of the data 2. Independence of observations of test subjects 3. Simple random samples are used 4. Equality of Variances groups have a constant variance within them. Results: In reviewing the output of the data after running the Analysis of Variance test, one will note that the sig. level is.000 for the group (time of day), which is less than.05. This tells us that this data is happening by chance less than 5% of the time, which dictates that we reject the null hypothesis (which states that each group is equal to the other) in favor of alternative hypothesis (which states that at least one group is different). Because the Test of Between-Subjects Effects table does not depict the sig. level for each group, we will need to run Post Hoc tests, specifically the Tukey test, to ascertain which groups indicate a significant variance. After running the Tukey test, one will see that the morning group, compared to the other two groups has a sig. level lower than 5%; specifically.000 for this group. The sig. levels for the evening and afternoon groups are greater than.05. This indicates that the other groups are statistically equivalent to each other, excluding the morning group.

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4 Inference: The Ho µ wait time in the morning = µ wait time in the afternoon = µ wait time in the evening will be rejected. The morning group has a higher wait time than the evening and afternoon groups. Problem #3 Mice exposure time to Nitrogen dioxide (N02) Hypothesis: Ho: µ percent of serum fluorescence on day 10 = µ percent of serum fluorescence on day 12= µ percent of serum fluorescence on day 14. Ha: at least one serum fluorescence level taken on day 10, day 12, or day 14 is to another. Statistical Procedure, Tests and Assumptions: Test: Randomized Block Design Rationale: This test is chosen as repeated measures/tests are being on the same test subjects and we need to remove the variability among subjects; also there are more than two groups involved in the data. Assumptions: 1. Normality - there is normal distribution of the data 2. Independence - each observation is independent of the other observation 3. Random Samples - simple random sampling of the test subjects 4. Equality of Variances groups have a constant variance within them Results: In reviewing the output results of the Randomized Block Design below, we note the significance level of.013 is less than the confidence level of.05 set. This indicates that there is a significant difference in the data related to number of days after exposure to Nitrogen dioxide and levels of serum fluorescence, thus the null hypothesis (which is µ percent of serum fluorescence on day 10 = µ percent of serum fluorescence on day 12= µ percent of serum fluorescence on day 14) is rejected. We can drill down to the specific effects by running Post Hoc Tukey Test. This will show each of the significance levels. In reviewing the values taken on each of the days (10, 12, and 14), it can be seen that the Sig. levels are all greater than.05 (alpha), which indicates that there is a great chance that the data is happening not by chance much of the time.

5 Inference: The Ho: µ percent of serum fluorescence on day 10 = µ percent of serum fluorescence on day 12= µ percent of serum fluorescence on day 14 will be rejected. Yes, the effect does depend on exposure time. In looking at the data, the longer the exposure time, the lower the average serum fluorescence rate.

6 Problem #4 Average time to Accelerate Is the average time to accelerate from 0 to 60 across all cars significantly different than 15? Hypothesis: Ho: µ time for the cards to accelerate from 0 to 60 seconds = 15 seconds. Ha: µ time for the cards to accelerate from 0 to 60 seconds 15 seconds. Statistical Procedure, Tests and Assumptions: Test: One Sample T-test Rationale: This test is being run as there is only one sample taken, the variable is continuous, and we are looking to see if the sample observations were drawn from a population with a specific mean value (15). Assumptions: 1. Continuous response variables on a ratio or interval scale. 2. Normal distribution. 3. Simple random sampling. 4. Observations are independent of one another. Results: In reviewing the output, the sig level is noted to be.000, which is less than.05 confidence level set; which suggests that the results are not happening by chance less than 5% of the time. The null hypothesis (time for the cars to accelerate from 0 to 60 seconds = 15 seconds) will not be rejected. In looking at the One-Sample Statistic table, you can see that the average wait time is seconds.

7 Inference: The average time to accelerate from 0 to 60 seconds across all cars is NOT significantly different than 15. As previously indicated, the Ho will not be rejected. Problem #5 Prediction of time to accelerate from 0 60 (MPG, ED, HP, # Cyl) Hypothesis: Ho: There is no relationship between time to accelerate and miles per gallon (MPG), engine displacement (ED), horsepower (HP), and number of cylinders (#cyl) Ha: There is a relationship between time to accelerate miles per gallon (MPG), engine displacement (ED), horsepower (HP), and number of cylinders (#cyl) Statistical Procedure, Tests and Assumptions: Test: Linear Regression Rationale: We have a continuous dependent variable (acceleration time) with a set of independent variables (miles per gallon, engine displacement, horsepower, number of cylinders) and we are attempting to predict acceleration time from 0 to 60 - based on the continuous variables listed above. Assumptions: 1. Normality: sample form a normal population; assumes errors are normally distributed. 2. Independence: samples are independent of one another. 3. Simple random sampling. 4. Equality of variance: the variation around regression equation is constant along the continuum. 5. Linearity: predictors are linearly related. In looking at the Model Summary, we see that the r² is.541. This indicates how much variation within acceleration is based on the independent variables (miles per gallon, engine displacement, horsepower, and number of cylinders). This number is significantly different than 0 and we are explaining about 54% of the variation.

8 In viewing the Anova table, we note that here is a sig. level of.000. This is less than.05, which indicates that the data is predicting something that what we are attempting to prove is better than just random guessing. If the sig. level was greater than.05, there would be no evidence to suggest that we were predicting anything, and we would not continue our review of the regression coefficients as there would be nothing to interpret. In reviewing the Coefficients table, we can analyze the relationship between acceleration (dependent variable) and miles per gallon, engine displacement, horsepower, number of cylinders (independent predictors). When looking at this table, we need to decide if we need all the predictors. It is noted that # of cyl significance level is.608. If we remove this predictor, we will not lose any predictability in the model. Upon removal, we will buy back a degree of freedom and obtain a more parsimonious (powerful) model.

9 Upon reviewing the Scatterplot, (residual plot) we are looking to be sure there is no distinctive pattern on the plot. No pattern is noted. If there is random variation above and below zero this indicates that linearity is an appropriate model. With this model, there is evidence to suggest that the variation is constant. After I reran the model upon removing the non-significant variable - # of cylinders- when analyzing the coefficients table, it is noted that all sig. levels are less than.05 which indicates that we now have predictors that are of significance.

10 In reexamining the fit of the model, upon reviewing the Scatterplot, (residual plot) we are looking to be sure there is no distinctive pattern on the plot. No pattern is noted. If there is random variation above and below zero this indicates that linearity is an appropriate model. With this model, there is evidence to suggest that the variation is constant. After removing the non-significant variable, our regression equation now reads as: Acceleration Time = (mpg) *.010 (ed) * (hp) The Ho, that there is no relationship between time to accelerate and miles per gallon (MPG), engine displacement (ED), horsepower (HP), and number of cylinders (#cyl) is rejected, as there is an indication that there is a relationship between these predictors as evidenced by the model.

11 Problem #6 Self Esteem Prediction Hypothesis: Ho: There is no relationship between self-esteem and locus of control, alienation, and social ability collectively. Ha: There is a relationship between self-esteem and locus of control, alienation and social ability. Statistical Procedure, Tests and Assumptions: Test: Linear Regression Rationale: We have a continuous dependent variable (self-esteem) with a set of independent variables (locus of control, alienation, and social ability) and we are attempting to predict - associate self-esteem prediction - based on the continuous variables listed above. Assumptions: 6. Normality: sample form a normal population; assumes errors are normally distributed. 7. Independence: samples are independent of one another. 8. Simple random sampling. 9. Equality of variance: the variation around regression equation is constant along the continuum. 10. Linearity: predictors are linearly related. Results: In reviewing the Correlations matrix, one can see that there is a inverse relationship between self-esteem and locus of control (-.516). There is a positive relationship between self-esteem and alienation (.701). There is a positive relationship between self-esteem and social ability (.422).

12 In reviewing the Model Summary table and looking at the multiple predictors R =.729 and this refers to the multiple correlations coefficient which refers to the total associations collectively between all of the independent variables and the dependent variable. R² =.531 and this reflects the proportion of variance accounted for by the model how much variation exists within self-esteem, based on the independent variables. We are explaining about 51% of the variation of self-esteem by locus of control, alienation, and social ability. The Anova Table is used in testing the null hypothesis that no relationship exists between the dependent variable and the set of independent variables. The significance level of.000 indicates we are doing more than just randomly guessing that there is a relationship. If the sig. level was greater than.05, we would stop the testing and analysis at this point as there would be no need to interpret the regression coefficient as there would be nothing to interpret we are not predicting anything. However, because the sig level is less than.05, the testing and analysis will continue because there is some indication that there is a relationship to begin with. The Coefficients table is the table that provides us with the most valuable information as it formally documents the relationship between the dependent variable and the set of independent predictors. The sig. levels for the variables are reviewed: Locus of Control =.608 Alienation =.000 Social Ability =.183

13 What we are looking to determine here is if we need to keep all of the independent variables. We look at each variable s significance level to determine if the variable should be kept. We want to determine if it tests the null hypothesis that is whether that predictor is significantly different from zero is it adding anything to the model. Alienation and social ability have a sig. level of less than.05, which indicates that there is evidence to reject the null hypothesis that they are significantly different from zero they are providing significant predicting power to the model. Because locus of control has a sig. level of.608, if we remove this independent predictor from the model, we will not lose any predictability in the dependent variable. We are trying to end up with a more parsimonious model and by removing the non-affective predictors, this will release a degree of freedom back into the error term and make for a stronger prediction; our model will be more powerful. Scatter Plot Interpretation: In reviewing the scatter plot (residual plot) we are looking to be sure there is no distinctive pattern on the plot. No pattern is noted. If there is random variation above and below zero this indicates that linearity is an appropriate model. With this model, there is evidence to suggest that the variation is constant.

14 At this point, we are going to remove the predictor of locus of control and rerun the data as that predictor is not providing any predictability. In looking at the r² we see that the number has changed from.531 to now read.529. We currently have a more significant model as the table of coefficients only includes significant predictors. They are explaining about 53% of variation related to self-esteem scores based on the predictors of alienation and social ability. Based on the assessments scores of alienation and social ability, if we know that we have these particular predictors, we can predict what the self-esteem score will be. If we know the assessment scores of alienation and social ability, we can predict the assessment score for self-esteem. Before we interpret the data, we need to be sure that we adhere to particular model assumptions, thus we need to review the histogram and scatter plots. Both indicate there are no problems with the data sets. None suggested.

15 Inference: Can one predict self-esteem based on alienation and social ability? The regression equation will give us a predicted self-esteem score based on the scores of alienation and social ability.

16 Self-esteem = (alienation) (social ability) This would reflect a change in self-esteem scores based on the scores of alienation and social ability. For instance if social ability score went up 10, the self-esteem score would go up (on average) 1.7 and would = The Ho that there is no relationship between self-esteem and locus of control, alienation, and social ability collectively is rejected as the model indicates that there is a relationship. Problem #7 Ductal carcinoma and family history of breast cancer. Hypothesis: Ho: There is no relationship between family history of breast cancer and ductal cancer. Ha: there is a relationship between family history of breast cancer and ductal cancer. Statistical Procedure, Tests and Assumptions: Test: Chi-square (AKA: Cross Tab Analysis) Rationale: This test is being run as there is a categorical dependent variable and a categorical independent variable. Assumptions: 1. No normality or equality of variance assumption 2. Simple random sampling. 3. Observations are independent of one another. Results: In looking at the Case Processing Summary table we can see that data was entered correctly by viewing the totals N=210. By viewing the Cross Tabulation table it appears that SPSS is interpreting the data the right way (observe this by viewing the counts in each category). This table also indicates that positive family history breast cancer and positive for Ductal Carcinoma is 17%, while negative family history breast cancer and positive for Ductal Carcinoma is roughly 6%.

17 The Person Chi-Square significance level of.029 is less then.05 which indicates that this is happening by chance less then 5% of the time. This is happening because there is a relationship between a positive family history of breast cancer and Ductal Carcinoma. In calculating the odds ratio having Ductal carcinoma based on whether you have a positive family history of breast cancer, you have 3.4 greater odds to have Ductal carcinoma if you have a positive family history of breast cancer.

18 Inference: The Ho that there is no relationship between family history of breast cancer and ductal cancer will be rejected as the data does suggest that there is a relationship between the two.

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