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For this second assignment you will have to use the OECD database to investigate possible changes in the policy preferences of Western welfare states. You have to develop (i) an indicator to measure how much a country spends on unemployment benefits per registered unemployed; (ii) an indicator that measures how much it spends on old age pensions per person of 65 or more. Subsequently you have to relate both indicators in a scatter plot and a line of best fit. As you will learn throughout this block, it is important that the countries in your plot remain identifiable.
(1) As a first step you have to retrieve data from the statistics portal of the OECD to develop your two indicators: expenditure data and data on the number of potential recipients (i.e. the number of unemployed and the number of people aged 65 or more). You have to choose two points in time (and in order to be able to observe a meaningful trend, those points should not be too close to one another. You must be able to obtain data for at least 12 OECD countries, amongst which the Netherlands (you can choose the other countries, but you will see that you will be constrained by the availability of data). Part of the exercise is that you learn to find your way in the maze of links on the website.
(2) Next you must figure out a way to represent graphically the relationship between your indicators so that you can capture both different expenditure patterns between countries, as well as changes within the same country over time. Hint: you can best make three scatter plots plotting (i) your two indicators against one another for year x; (ii) your two indicators for year y; and (iii) changes in your two indicators between year x and year y. For each plot you should add a trendline. You can best draw scatter plots using Excel or if you prefer SPSS. Make sure that the points in your plot are clearly labelled (in case you do not succeed to add labels using the software, do it manually).
Question: (a) Suppose that 100 tires made by Bridgestone last on average 21,819 miles with a standard deviation of 1,295 miles. Test the null hypothesis µ = 22, 000 miles again
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