Stratagem Method of Experimental Transactions a Dispatch on Study Methodologies - Assignment Example

Name: Course: Tutor: Date: Question 1 a) The data is normally distributed as is shown from the graph shown above. The mean, median and mode are almost equal, i.e. mean is 99.965, median is 100 and mode is 98. More than 68% of the data is distributed around the mean mean 99.96597291 median 100 mode 98 max 180 min 31 count 6054 b) From the graph of proportion, we can get the number of people with IQs between 130 and 144. No. of people with IQs between 130-144= 131, total population= 6054 Prop= (131/6054)* 100= 2.1635% c) Proportion of the population which falls above 144 No. of people with IQs above 144= 10 Prop= (10/6054)* 100= 0.1658% d) Finding the individuals who are above the 99.9th percentile will give the minimum IQ of a genius. To get the minimum IQ from the 99.9th percentile, the following steps are put into consideration. i) Arrange the data in ascending order i.e. from the lowest IQ to the highest ii) Find the 99.9th position by Calculating the 99.9th percentile: 99.9/100= 0.999, 0.999*6054= 6047.946 iii) The answer above refers to the position of the person with the minimum IQ to be considered a genius. Since there cannot be 6047.946 position, we round off to a whole figure to get the exact position which is 6048. iv) We can therefore say that the minimum IQ is found from the person at position 6048. From the arranged data (ascending) in excel, find the person in position 6048, which gives the minimum IQ. v) The 6048 position is occupied by an IQ of 146. vi) Therefore, we can safely conclude that, if geniuses were being determined by the 99.9th percentile, the minimum would be an IQ of146. e) 0.95095095 f) 0.02002002 g) Probability that was calculated in part c) used the entire population as the sample space, whereas the probability in f) used a sample space that is a sample of the population, i.e. 150. Question two a) The following is the procedure involved in the construction of a 95% confidence interval Obtain the mean and the standard deviation of the data in the sample. The mean and the standard deviation in this case are to be calculated. However, the sample mean is given as 56,475 hours. The variance in this question is given as 25 million (hours)2, the standard deviation is, therefore, the square root of 25 million hours=5000 hours. The second step is to obtain the degrees of freedom by subtracting 1 from the sample size. In this case, the sample size in this case is 30 televisions. The degree of freedom is equal to 29. Then subtract the confidence level from 1 and divide by 2, that is; 1-.95=0.05/2=0.025. Use the degrees of freedom and the level of confidence, α to obtain the z value. Then divide the standard deviation we obtained earlier with the square root of the sample size and then multiply by the z value. This gives us the margin of error. This should give us the confidence interval. Computations Standard deviation sqrt 25000000=5000, z-vale from tables=1.96. the margin of error=zα/2 *s.d/sqrt n= 1.96*5000/sqrt 30. After carrying out the computations, the result will be=912.871. The CI will be obtained by adding and subtracting the margin of error to the mean, that is [56475+/-912.871] the CI is therefore, [55562.129, 57387.871] b) There is every reason to challenge the claim of the manufacturer because the mean he provides, 58000, is outside the confidence interval. c) Television screen failed after 52,500 hours., given that the true mean is 58000, the probability of the same thing happening again is 52500/58000=0.905 Part B a) The characteristics of a normal distribution among many others include; the mean, mode and median are equal, the area under the curve is equal to 1, at least 98% of the data will fall within the standard deviations, the normal distribution is symmetric around the mean and 68% of the area of a normal distribution is within one standard deviation. In this case the mean, median and mode are not equal but 72% 0f the data lies within one standard deviation (6.4/8.8=.727), therefore, the distribution is normal based on this fact. b) Yes, the confidence interval can be determined. This is because all the parameters required in its construction are given, or at least can be calculated. The parameters are mean, the standard deviation, the degrees of freedom and the level of significance. c) α=1-.97=.03/2=.015, degrees of freedom =797-1=796, z value from tables = 1.37, the margin of error= zα/2 *s.d/sqrt n= 1.37*6.4/sqrt 797 =0.31. The CI= [8.8+/-0.31] = [8.49, 9.11] Question three total $3,228,724 std dev 19142.7203 mean $124,182 max $174,036 min $90,144 a) H0= the company’s weekly sales have declined following the launch of the competitor’s new products H1= the decline in the company’s sales has nothing to do with the launching of the new products by the company’s competitors b) The most appropriate test-statistic in this case is the t-statistic, its formula is t=μ/s/sqrt n. from the data in excel, the generated standard deviation is 19142.7203, the mean as generated by excel is 124,182, the sample size is 26 weeks. The t-statistic therefore, =124182/ (19142.7203/sqrt 26)=33.08. The tabular value of the t statistic is 2.131 at 0.05 level of significance. c) To find the critical value at 2.5% confidence level, use 25 degrees of freedom to obtain 13.12 d) Since the calculated value is> than the critical value, reject the null hypothesis or accept the alternative hypothesis e) The conclusion here is to reject the null hypothesis and conclude that the decline in the company’s sales has nothing to do with the launching of the new products by the company’s competitors f) At 1% level of significance, the critical value is 34.213, in this case, fail to reject the null hypothesis or simply accept the null hypothesis. Question four a) The scatter plot showing of MCG attendance (dependent variable) against combined club membership (independent variable) b) The correlation coefficient of this data as generated by excel computer package is 0.3117. This shows a positive correlation between MCG attendance and combined club membership. The correlation coefficient in this particular case was generated/calculated by excel computer package. c) The estimate regression equation takes the form . The numerical equivalent of this regression equation as generated by excel computer package y=1.2073x+9621.6. d) The slope of this regression equation is 1.2073. This means when x increases with one, y increases with 1.2073. This is to say when the number of combined club membership (independent variable) increases by 1, the number of MCG attendance (dependent variable) increases 1.2073 times. e) Statistical significance is a statement or a phrase that that advocates for the unlikelihood of a positive result. The criterion used in this case to test for statistical significance in this case is 5%. Combined club membership a statistically significant predictor of match attendance at the MCG given that it raises the likelihood of rejecting the null hypothesis at 5% level of significance g) To predict the attendance at a match between Adelaide (25,000 members) and Western Bulldogs (16,500 members), we use the regression equation;  which translated into y=1.2073x+9621.6. To compute this, take x=16500, using the regression equation, attendance = 1.2073*16500+9621.6=29,542.05 which translates to 29543 people. f) The goodness of fit attempts to show the discrepancy between the supposed values and the values that have been observed. It therefore, aims at testing how well a statistical model estimates a set of data. The data in this case, there is an observed positive correlation between the variables but the discrepancy between the expected data and the observed data is large. This is to say the model is not the best estimate of the data. I) To improve this model’s ability to predict the attendance at an AFL match held at the MCG, a statistician has to do some of the following adjustments on the statistical model; based on residual plots, the statistician should update the function. Another way/means of validating a statistical model is by ensuring all the errors are accounted for using a non-normal distribution. And finally, the researcher/statistician must ensure that he/she accounts for all the non-constant variations across the sample data. Part B a) The graph as generated by excel computer package is as shown: b) The estimated regression equation linear regression equation is y=0.0122x=20.737 as generated by the excel computer package. The equation gives the slope and the intercept. The correlation coefficient of the data as generated by excel computer package is an estimated at 0.6196. a) The slope in this case shows that a change in time from one year to the other leads to a change of 0.0122times in the mean annual temperatures. The y intercept shows the initial temperature at time=0; that is the start time of the experiment. d) Statistical significance is indicated by a p value of less than 0.05. there is no statistical significant trend in the data on Sydney temperatures since the p value in this case is greater than the p value of 0.05. Statistical significance is a statement or a phrase that that advocates for the unlikelihood of a positive result. e) The model on Sydney temperatures has depicted goodness of fit. There is minimal discrepancy between the observed values and the expected values. This shows that the statistical model used in this case is valid f) To forecast the mean annual Sydney temperature for 2015, we use our generated model: y=0.0122x+20.737, from 2012 to 2015 is a period of 3 years, therefore, in 2015, the mean annual temperatures will be 0.0122*3+20.737=20.7736 degrees Celsius. c) To forecast the mean annual Sydney temperature for 2030, use the model generated by the excel computer package: y=0.0122x+20.737, in 2030, there a difference of 18 years, therefore, we substitute in our predictor model. The temperatures in 2030 will be 0.0122*18+20.737=20.9566 degrees Celsius h) The forecast that is more reliable is the forecast of the year 2030 because the length of time is long enough and therefore, takes into account the experimental variations. d) Yes, this analysis shows that human activities lead to global warming. Works Cited Moballeghi, Mostafa and Moghadam, Golnessa. How do we measure use of scientific Journals? A note on research methodologies. 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1-.97=.03/2=.015, degrees of freedom =797-1=796, z value from tables = 1.37, the margin of error= zα/2 *s.d/sqrt n= 1.37*6.4/sqrt 797 =0.31. The CI= [8.8+/-0.31] = [8.49, 9.11] Question three total $3,228,724 std dev 19142.7203 mean $124,182 max $174,036 min $90,144 a) H0= the company’s weekly sales have declined following the launch of the competitor’s new products H1= the decline in the company’s sales has nothing to do with the launching of the new products by the company’s competitors b) The most appropriate test-statistic in this case is the t-statistic, its formula is t=μ/s/sqrt n.

from the data in excel, the generated standard deviation is 19142.7203, the mean as generated by excel is 124,182, the sample size is 26 weeks. The t-statistic therefore, =124182/ (19142.7203/sqrt 26)=33.08. The tabular value of the t statistic is 2.131 at 0.05 level of significance. c) To find the critical value at 2.5% confidence level, use 25 degrees of freedom to obtain 13.12 d) Since the calculated value is> than the critical value, reject the null hypothesis or accept the alternative hypothesis e) The conclusion here is to reject the null hypothesis and conclude that the decline in the company’s sales has nothing to do with the launching of the new products by the company’s competitors f) At 1% level of significance, the critical value is 34.

213, in this case, fail to reject the null hypothesis or simply accept the null hypothesis. Question four a) The scatter plot showing of MCG attendance (dependent variable) against combined club membership (independent variable) b) The correlation coefficient of this data as generated by excel computer package is 0.3117. This shows a positive correlation between MCG attendance and combined club membership. The correlation coefficient in this particular case was generated/calculated by excel computer package.

c) The estimate regression equation takes the form . The numerical equivalent of this regression equation as generated by excel computer package y=1.2073x+9621.6. d) The slope of this regression equation is 1.2073. This means when x increases with one, y increases with 1.2073. This is to say when the number of combined club membership (independent variable) increases by 1, the number of MCG attendance (dependent variable) increases 1.2073 times. e) Statistical significance is a statement or a phrase that that advocates for the unlikelihood of a positive result.

The criterion used in this case to test for statistical significance in this case is 5%. Combined club membership a statistically significant predictor of match attendance at the MCG given that it raises the likelihood of rejecting the null hypothesis at 5% level of significance g) To predict the attendance at a match between Adelaide (25,000 members) and Western Bulldogs (16,500 members), we use the regression equation;  which translated into y=1.2073x+9621.6. To compute this, take x=16500, using the regression equation, attendance = 1.

2073*16500+9621.6=29,542.05 which translates to 29543 people. f) The goodness of fit attempts to show the discrepancy between the supposed values and the values that have been observed. It therefore, aims at testing how well a statistical model estimates a set of data. The data in this case, there is an observed positive correlation between the variables but the discrepancy between the expected data and the observed data is large. This is to say the model is not the best estimate of the data. I) To improve this model’s ability to predict the attendance at an AFL match held at the MCG, a statistician has to do some of the following adjustments on the statistical model; based on residual plots, the statistician should update the function.

Another way/means of validating a statistical model is by ensuring all the errors are accounted for using a non-normal distribution. And finally, the researcher/statistician must ensure that he/she accounts for all the non-constant variations across the sample data.

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