Table 1 uses the data in students.mwx to compare the number of M140 students in Wales with those in the other areas. The

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Table 1 Uses The Data In Students Mwx To Compare The Number Of M140 Students In Wales With Those In The Other Areas The 1 (151.11 KiB) Viewed 62 times

Table 1 uses the data in students.mwx to compare the number of M140 students in Wales with those in the other areas. The first row of data gives the number of M140 students, by year, for Wales. The row labelled 'Other areas' is the total number of M140 students, by year, for the East of England, the South West of England, Scotland and Northern Ireland combined. Table 1 Numbers of M140 students 2013 2014 2015 2016 2017 2018 2019 Total Wales Other areas 54 387 55 373 51 359 31 315 42 355 62 316 83 359 378 2464 Total 441 428 410 346 397 378 442 2842 (a) Give three reasons why Table 1 is correctly described as a contingency table. Justify each reason with reference to this particular contingency table. (b) Using Table 1, calculate the following probabilities. (i) The probability that a randomly chosen M140 student studied in Wales in 2017 (ii) The probability that a randomly chosen M140 student studied in Wales. (iii) The probability that a randomly chosen M140 student studied in Wales and/or studied M140 in 2017. (c) The M140 module team is interested in the following question: Does the distribution of students in Wales across the years differ from the combined distribution across the other four areas across the years? The module team decides to use the x2 test for contingency tables with the data in Table 1 and produces the Minitab output in Figure 1. Rows: Area Columns: year 2013 2014 2015 2016 2017 2018 2019 All Wales 378 54 58,66 55 56.93 51 54.53 31 46,02 42 52.80 62 50.28 83 58.79 Other areas 387 373 359 315 355 316 359 2464 382.34 371.07 355.47 299.98 344.20 327.72 383.21 All 441 428 410 346 397 378 442 2842 Cell Contents Count Expected count Chi-Square Test Chi-Square DF P-value Pearson 23.623 6 0.001 Likelihood Ratio 23.140 6 0.001 Figure 1 Minitab output for a x2 test for contingency tables page 4 of 7 (i) Write down suitable null and alternative hypotheses when using the X² test for contingency tables to try to answer the module team's question. (ii) Explain how the expected value for students in Wales in 2013 is calculated. (iii) What is the value of the test statistic? (iv) Explain why the output says the number of degrees of freedom is 6. (v) The p-value is 0.001. What does that tell you in terms of the null hypothesis? (vi) What is your conclusion from the test? Question 6 16 marks