Data set Description

Question: Write an essay on Data set Description. Answer: The data set given on the responses have been 100,000 sample but every student is allotted 100 sample from which three possible endings are analysed namely ending 1, 2 and 3. However, the questions that are used for the analysis are Do they like the TV show? and How much are they willing to pay for the DVD? b) The other questions that could be asked on this guideline are the contrasting view of two endings whether Ending 1 and 2 or Ending 2 or 3 or Ending 1 and 3 and based on the analysis on which ending they will pay more. Liking for the TV show a) Proportions phat1 = = 15/33 = 0.454545 phat2 = = 15/27 = 0.518519 phat3 = = 28/40 = 0.7 b) Proportions differences phat1 phat2 = = 0.454545 - 0.518519 = -0.06397 phat2 phat3 = = 0.518519 0.7 = -0.18148 c) Chart for proportions of three endings of a show Ending 1 Ending 2 Ending 3 Yes 54.55% 51.85% 70.00% No 45.45% 48.15% 30.00% a) Mean of the DVD pay on each ending xÃÅ'†¦ 1 on ending 1= 4.3939 xÃÅ'†¦ 2 on ending 2= 5.07407 xÃÅ'†¦ 3 on ending 3= 7.305 b) Two sample difference xÃÅ'†¦ 1- xÃÅ'†¦ 2 = 4.3939 - 5.07407 = -0.68013 xÃÅ'†¦ 2- xÃÅ'†¦ 3 = 5.07407 7.305 = -2.23093 a) Histograms b) Ending 1 Ending 1 Summary Average 4.393939 Standard Deviation 4.689479 Minimum 0 Quartile 1 0.4 Median 1 Quartile 3 9 Maximum 12 c) Ending 2 d) Ending 3 Endings on TV Show The ending of the TV show states that the maximum responses has been achieved on ending 3 followed by ending 2 and ending 1. However, according to the results, the respondents who reviewed Ending 1 did not like the TV show which constituted to be 54.44% on the sample of 33. However, on ending 2, the results were mixed as 48.15% did not like the TV show. Lastly, on ending maximum respondents like the TV show which constituted to be 70% on the sample of 40 If evaluated on responses, the sequence of proportion on each ending has been ending 1 DVD Pay Ending 1 Results The mean has been lowest on this ending for DVD pay that is 4.3939. The minimum a person would pay is 0 and is constituted to be 54.55% on ending 1 Maximum Responses The maximum responses on DVD pay has been on $12 with 4 respondents followed by 2 respondents each on $5, $6, $10 and $11. Histogram The histogram is normally distributed yet unimodal in nature. Ending 2 Results The mean on this ending for DVD pay has been 5.407. The minimum a person would pay is 0 and the respondents who constitute to pay are 13 constituting to be 48.15%. Maximum Responses The maximum responses on DVD pay has been on $10 with 5 respondents followed by 4 respondents on $7. Histogram The histogram is normally distributed yet unimodal in nature. Ending 3 Results The mean on this ending for DVD pay has been 7.305. The minimum a person would pay is 0 and the respondents who constitute to pay are 12 constituting to be 30%. Maximum Responses The maximum responses on DVD pay has been on $9 and $11 with 5 respondents followed by 4 respondents each on $10 and $12. Histogram The histogram is normally distributed yet bimodal in nature. Survey Problem When a business analysis is taken place then the maximum importance is given to the respondents who are responding to the given survey. A problem may arise if the people answering do not represent the population. As a result, the results will be biased and will not yield accurate results even the sample data is large. A specific issue can be highlighted that a survey can yield low response rate if the person who is surveyed is the person who has a lot of spare time (Ott and Longnecker 2015). Hypothesis testing 1a) Variable ending type is independent of the variable like TV show yes/no i.e., Ho: = (Not independent) H1: - 0 (Independent) (Florescue 2014). Given: n1 = 33 and n2 = 27, = 0.4545 and = 0.518519, p* = (x1 + x2)/( n1 + n2) = (15 +14)/ (33 + 27) = 29/60 = 0.4833 Sample Error =) =) =) == 0.12920 | Z | = (phat1)/ Sample Error = (0.454545 0.518519) / 0.12920 = 0.49515 p value = Prob (z 0.49515) + Prob (z 0.49515) = 0.18793 + 0.18793 = 0.37586 As seen that the Z statistics is lower than the table estimation of Z that delineates that the null hypothesis is acknowledged. Notwithstanding, it can be summed up that TV appearing with ending 1 and 2 is not autonomous of the responses on likeness "Yes or No." Howsoever, it can be said that there is reliance of the TV show and closure on ending 1 and 2 particularly. b) Test the claim that there is a difference in the amount people would pay for ending 1 and ending 2 Ho: 1 = 2 (No difference) H1: 1 2 (There is difference) (Pituch, Whittaker and Stevens 2015). Given: n1 = 33 and n2 = 27, xÃÅ'†¦1 = 4.39394 and xÃÅ'†¦2 = 5.07407, 1 = 5 and 2 = 4.9 Standard deviation = = = Standard deviation = 1.28328 Sample Error = s * () = 1.28328 * = 1.28328* 0.259499 = 0.333010 | Z | = (xÃÅ'†¦1 -xÃÅ'†¦2)/ Sample Error = (4.39394 5.07407) / 0.333010= 0.68013/ 0.33301= 2.0423 According to the results, Z statistics is higher than the table value of Z ay 95% interval then it illustrates that alternate hypothesis is accepted . However, it can be evaluated that there is significant difference in the amount of people pay for DBD pay in ending 1 and 2. 2a) Variable ending type is independent of the variable like TV show yes/no i.e., Ho: = (Not independent) H1: - 0 (Independent) Given: n2 = 27 and n2 = 40, = 0.518915 and = 0.7, p* = (x2 + x3)/( n3 + n2) = (14+28)/ (27 + 40) = 42/67= 0.62686 Sample Error =) =) =) == 0.120460 | Z | = (phat2)/ Sample Error = (0.518915 0. 7) / 0.120460= 1.50327 p value = Prob (z 1.50327) + Prob (z 1.50327) = 0.0630 + 0.0630 = 0.1260 As seen that the Z statistics is lower than the table estimation of Z that delineates that the null hypothesis is acknowledged. Notwithstanding, it can be summed up that TV appearing with ending 2 and 3 is not autonomous of the responses on likeness "Yes or No." Howsoever, it can be said that there is reliance of the TV show and closure on ending 1 and 2 particularly. b) Test the claim that there is a difference in the amount people would pay for ending 2 and ending 3 Ho: 2 = 3 (No difference) H1: 2 3 (There is difference) Given: n2 = 27 and n3 = 40, xÃÅ'†¦2 = 5.074074 and xÃÅ'†¦3 = 7.305, 2 = 4.9 and 3 = 4.6 Standard deviation = = = Standard deviation = 1.1764 Sample Error = s * () = 1.17641 * = 1.1764* 0.249072 = 0.293008 | Z | = (xÃÅ'†¦1 -xÃÅ'†¦2)/ Sample Error = (7.305 5.074074) / 0.293008= 2.23093/ 0.293008 = 7.613 According to Z statistics, it can be reviewed that the Z statistics is higher than the critical vale of Z at 95% level that is Concept of Sample distribution The sample distribution can be worked out in light of the sample estimates on the case of deductive thinking. Be that as it may, the hypothesis can be worked out taking into account the population were the sample is taken. The sample can be founded on the current hypothesis from which the testing of hypothesis has and evaluated on the sample (Franklin, Allison and Gorman 2014). p1-p2 is considered to be an estimate of phat1-phat2 (the difference between the proportions of population for ending 1 and 2) p2-p3 is considered to be an estimate of phat2-phat3 (the difference between the proportions of populations for ending 2 and 3) 1- 2 is considered to be an estimate of xbar1-xbar2 (the difference between the population means of DVD pay of ending 1 and ending 2) 2- 3 is considered to be an estimate of xbar2-xbar3 (the difference between the Population means of DVD pay of ending 2 and ending 3). Beside the sample is normally distributed, the normal distribution is given by the Normal Q-Q plots, the z distribution is confirmed by finding the z scores (Fukunaga 2013). a) phat1 phat2 b) phat2 phat3 c) xÃÅ'†¦1 - xÃÅ'†¦2 d) xÃÅ'†¦2 - xÃÅ'†¦3 Conclusion To finish up, it can be concluded that the results have been shown significant difference in the likeness of TV show in ending 3. Hence, the hypothesis states that DVD pay has significant difference on all the endings. However, the results analyse that the sample taken has been more responsive to ending of TV show and the amount of DVD pay. References Florescu, I., 2014.Probability and Stochastic Processes. John Wiley Sons. Franklin, R.D., Allison, D.B. and Gorman, B.S., 2014.Design and analysis of single-case research. Psychology Press. Fukunaga, K., 2013.Introduction to statistical pattern recognition. Academic press. Ott, R.L. and Longnecker, M., 2015.An introduction to statistical methods and data analysis. Nelson Education. Pituch, K.A., Whittaker, T.A. and Stevens, J.P., 2015.Intermediate statistics: A modern approach. Routledge.