INFERENTIAL STATISTICS REVISION

INFERENTIAL STATISTICS REVISION

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2016 LCHL Paper 2 Question 9 (a) (i) Data on earnings were published for a particular country. The data showed that the annual income of people in full-time employment was normally distributed with a mean of 39 400 and a standard deviation of 12 920. The government intends to impose a new tax on incomes over 60 000. Find the percentage of full-time workers who will be liable for this tax, correct to one decimal place. Calculate the z score. x μ σ x is a data point μ is the population mean σ is the population standard deviation Quick Sketch 39,400 60000 39400 12920 1.5944 μ = 39,400 σ = 12,920 x = 60,000 60,000 The tables only give values to the left of z but: P Z 1 P Z z P z > 1.59 = 1 P z < 1.59 = 1 0.9441 = 0.0559 5.6% 0.0559

2016 LCHL Paper 2 Question 9 (a) (ii) The government will also provide a subsidy to the lowest 10 % of income earners. Find the level of income at which the government will stop paying the subsidy, correct to the nearest euro. 39,400 Instead of calculating the lowest 10% we calculate those above 90% as the curve is symmetrical and the tables have no percentages less than 50%. 1 0.1 = 0.90 1.28 x μ σ = 1.28 Find the z score that corresponds to 0.90 x μ σ 0.10 Quick Sketch x 39400 = 1.28 12920 x 39400 = 1.28 12920 x = 22862.4 Less than 90% is the z score 1.28 BUT greater than 90% is 1.28 (same as less than 10%). x = 22,862

2016 LCHL Paper 2 Question 9 (a) (iii) Some time later a research institute surveyed a sample of 1000 full-time workers, randomly selected, and found that the mean annual income of the sample was 38 280. Test the hypothesis, at the 5% level of significance, that the mean annual income of full-time workers has changed since the national data were published. State the null hypothesis and the alternative hypothesis. Give your conclusion in the context of the question. 15 marks 11 for the z score. Was acceptable to use Confidence Interval Method State the null and alternate hypothesis H 0 : μ = 39400, the mean annual income of full time workers is still 39,400 H 0 : μ 39400, the mean annual income of full time workers is NOT still 39,400 Calculate the test statistic x μ σ n x = 38,280 μ = 39,400 σ = 12,920 n = 1,000 38280 39400 12920 1000 2.74 2.74 < 1.96 Reject the null hypothesis. Sketch of Critical Regions Give conclusion in the context of the question. There is enough evidence to reject the claim that the mean annual income has remained the same. Marking scheme: The mean income has changed.

2016 LCHL Paper 2 Question 9 (b) The research institute surveyed 400 full-time farmers, randomly selected from all the full-time farmers in the country, and found that the mean income for the sample was 26 974 and the standard deviation was 5120. Assuming that annual farm income is normally distributed in this country, create a 95% confidence interval for the mean income of full-time farmers. 10 marks 95% Confidence Interval for a Means x 1.96 σ n < μ < x + 1.96 σ n 26974 1.96 5120 5120 < μ < 26974 + 1.96 400 400 26974 501.76 < μ < 26974 + 501.76 26472.24 < μ < 27475.76 Margin of Error (Mean) ±1.96 σ n Note: σ is in the tables but you must n remember the ±1.96 which is specific to a 95% level of confidence.

2016 LCHL Paper 2 Question 9 (c) It is known that data on farm size are not normally distributed. The research institute could take many large random samples of farm size and create a sampling distribution of the means of all these samples. Give one reason why they might do this. Central Limit Theorem Mean μ x = μ Standard Deviation σ x = σ n The distribution of sample means will be normally distributed. The mean of the samples will be equal to the mean of the population (for n > 30) and the standard deviation will be σ n. Must know the above!

2016 LCHL Paper 2 Question 9 (d) The research institute also carried out a survey into the use of agricultural land. n farmers were surveyed. If the margin of error of the survey was 4 5 %, find the value of n. Without more information this is the only formula we could use. Syllabus probably needs a little clarity here. Margin or Error 1 n 1 n = 0.045 2 1 = n 0.045 n = 493.827 The institute interviewed 494 farmers.

2015 LCHL Paper 2 Question 2 (a) A survey of 100 shoppers, randomly selected from a large number of Saturday supermarket shoppers, showed that the mean shopping spend was 90.45. The standard deviation of this sample was 20.73. Find a 95% confidence interval for the mean amount spent in a supermarket on that Saturday. 10 marks Confidence Interval for a Mean x 1.96 σ n < μ < x + 1.96 σ n 90.45 1.96 20.73 20.73 < μ < 90.45 + 1.96 100 100 90.45 4.06308 < μ < 90.45 + 4.06308 86.38692 < μ < 94.51308 Margin of Error (Mean) ±1.96 σ n Note: σ is in the tables but you must n remember the ±1.96 which is specific to a 95% level of confidence. We can be 95% confident that the mean amount spent was in the range 86.39 < μ < 94.51

2015 LCHL Paper 2 Question 2 (b) A supermarket has claimed that the mean amount spent by shoppers on a Saturday is 94. Based on the survey, test the supermarket s claim using a 5% level of significance. Clearly state your null hypothesis, your alternative hypothesis, and your conclusion. Full marks for Confidence Interval Method EVEN if you already used it for part (a). That would be 20 marks for doing the same thing twice!! State the null and alternate hypothesis H 0 : μ = 94, the mean amount spent by shoppers on a Saturday is 94 H 1 : μ 94, the mean amount spent by shoppers on a Saturday is NOT 94 Calculate the test statistic x μ σ n x = 90.45 μ = 90 σ = 20.73 n = 100 90.45 94 20.73 100 1.71 1.71 > 1.96 Fail to reject H 0 Sketch of Critical Regions Give conclusion in the context of the question. We do not reject the null hypothesis as 1.71 is in the fail to reject region. This means that there is not sufficient evidence to reject the claim that the mean amount spent by shoppers is as the supermarket claims. Marking scheme: Outlined no conclusion.

2015 LCHL Paper 2 Question 2 (c) Find the p value of the test you performed in part (b) above and explain what this value represents in the context of the question. p value = 2 P z > z 1 where z 1 is the test statistic The probability of getting a value z < 1.71 is got from the tables. P z < 1.71 = 1 P z < 1.71 = 1 0.9564 = 0.0436 The probability of getting a value > 1.71 is also 0.0436 p value is the sum of these probabilities. 2 0.0436 = 0.0872 Compare this to the level of significance 5% or 0.05 0.0872 > 0.05 This is greater evidence to fail to reject the null hypothesis. What does the p value represent: If the supermarkets claim is true then our sample result would be this far (a measure of 94 90.45 = 3.55) from the true claim 8.72% of the time. We are unwilling to reject it as this % is too high (i.e. greater than 5%). p-value is easy to calculate but interpreting it has caused difficulty.

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