1. BA5001 Business Decision Making Autumn 2014 Session 4 – Lecture Estimation Module Leader: Dr Francisca Tej

2. Agenda The concept of ‘Estimation’ Estimation of a population mean Estimation of proportion

3. Let us recall that the purpose of inferential statistics is to draw conclusions about a whole population on the basis of information that has been collected in a sample. Two issues are closely related. The first issue is to do with precision of the results. Generalising from sample to population inevitably involves an element of guesswork. There is always a margin of error into our statement.

4. The second issue is to do with randomness of the sample. We could be very unlucky and select a random sample that includes a large number of exceptional cases. This sample cannot be representative even if collected at random. This type of samples is very rare in the population but a reality nevertheless. Therefore, with every inference we associate a probability of error, which can be set at 10%, 5% or even 1%. Note: In this module, we set this probability of error at 5% unless we specify it.

5. 1. Estimation of a population mean Example: The manager of food processing company informed the department of health and hygiene that the sugar content is normally distributed among all their batches of baked but did not reveal the statistics for sugar content. A food inspector, suspicious of the manager’s reluctance to publish the statistics for sugar content, decided to investigate it. She selected a random sample of 36 separate kilograms of baked beans from different batches and measured the amount of sugar used using a chemical procedure. The table shows the sugar content (per kg of baked beans):

6. 1. Estimation of a population mean This sample yields the following statistics: Mean = 23.22 grams Standard deviation = 1.775 n=36 Standard error = Std. Deviation/Square root of sample size = 1.775/6 = 0.296

7. According to the sampling distribution, the distribution of sample means is normal with a mean equal to the population average and a standard deviation equal to the population standard deviation divided by the square root of the sample size(Standard Error). We also know that if we selected all random samples of the same size, we would expect 95% of them to have a mean value is not that far off the population mean and in fact it is, at most, 1.96 standard errors on either side of the true population mean.

8. Estimation from one sample is made by reversing this argument. Our sample mean of 23.22 grams, one of many sample means, is at most, 1.96 standard errors away from the population mean in 95% of cases. 1.96 standard error = 1.96 x 0.296 = 0.58016 = 0.58 This quantity is called the margin of error. So, the population mean, i.e. the average sugar content of all baked beans, has a 95% chance to between: 23.22 – 0.58 = 22.64grams and 23.22 + 0.58 = 23.80grams The confidence interval for population mean is [22.64 to 23.80]

9. Graphical representation Therefore, if we know the sample mean is 23.22, we could guess that the population mean would not be more than further away in 95% of cases. The health researcher could report this finding as: 95% of Sample s If our sample mean is located here with a mean = 23.22 µ

10. The average sugar content in baked beans in this factory is estimated to be 23.22 grams, with a margin of error of 0.58 grams at a risk of error of 5%. Or, The average sugar content in baked beans in this factory is estimated, with a 95% confidence level, to be between 22.64 grams and 23.80 grams.

11. Equally, we can say that our sample mean of 23.22 grams, is at most, 2.58 standard errors away from the population mean in 99% of cases. The margin of error is, 2.58 x 0.2958 = 0.763164 = 0.76 and the reporting statement would be; The average sugar content in baked beans in this factory is estimated to be 23.22 grams, with a margin of error of 0.76 grams at a risk of error of 1%.

12. Note: One notes that the margin of error is greater when we impose on ourselves a smaller risk of error. (This is a logical result. To be more precise about the value of the population mean, we need to have a greater margin of error. Ultimately, to locate the population mean with no risk of error, we need a margin of error that is very big, which is not terribly helpful.)

13. t-scores In our example here, we have assumed that the population data is normally distributed. This is not always true. If normality is not proven, then as we said in the previous topic, the t-scores are used instead. These are obtained from the t-table at the end of this lecture notes. Degree of Freedom: df= (n-1) – Degree of freedom is the number of values in the final calculation of a statistic that are free to vary – The number of independent ways by which a dynamic system can move without violating any constraint imposed on it, is called degree of freedom.

14. t-scores In the t-table, the t-scores depend on the degree of freedom (df= n-1 ) and the value of half the risk of error. For example, if the sample size is 36 and the risk of error is 5%, the t-score is 2.0301 In our example above, the margin of error is 2.0301 x 0.2958 = 0.6grams.

15. The resulting confidence interval is: There is not a great deal of difference. However, if the sample size is very small, then the results are markedly different. AN D As a general rule, if the population data is not normally distributed and the sample size is 30 or larger, use the z-scores. If the sample size is less than 30, use the t-scores. Note: 23.22 is the mean value

16. Using SPSS to find the confident interval for population mean First, enter the data into SPSS and then go to Analyze, Descriptive statistics, then Explore. Fill in the boxes as below. Click Continue and then OK.

17. Note: If the sample size is below 30 then use t test. If more than 30 then use z test. Note that SPSS generate the confidence interval by using the t-scores (and not the z-scores) Confiden ce interval

18. Next class Estimation of population proportion Computing session

19. Recap of last class Standard Error = Standard Deviation/ Sample Size = / n Margin of error = 1.96 x Standard Error [For 95% Confidence Level) Confidence Interval for Population Mean = Value 1 = Sample Mean – Margin of Error Value 2 = Sample Mean + Margin of Error Note: Margin of error is a small amount that is allowed for in case of miscalculation or change of circumstances

20. Estimation of population proportion Estimating the population inevitably involves a variable that is quantitative (or scale) in nature. Sometimes, we have a variable that is nominal or ordinal and you want to estimate a population proportion.

21. Example Question: A survey among 260 online customers showed that 45% remained unsatisfied about the delivery service of their purchased goods. Estimate the proportion of all unsatisfied online customers.

22. Solution The sampling distribution of all possible random samples of size 260 is normally distributed. Therefore 95% of all samples have their proportions within 1.96 standard errors of the population (or true) proportion. The standard error = Where p is the proportion we want to estimate (i.e. the proportion of unsatisfied customers) q is the remaining proportion (i.e. the proportion of satisfied customers)

23. Note Here that there are only two proportions, the proportion of those who are unsatisfied and the proportion of those satisfied. We can split any number of proportions into two proportions as long as we know the one of interest.

24. p = 0.45 q = 1 - 0.45 = 0.55 Standard Error = = [(0.45 x 0.55)/260] = 0.031 Margin of Error = 1.96 x 0.031 = 0.06 (which is 6%)

25. The confidence statement would read: A survey conducted on a random sample of 260 online customers showed that 45% of all online customers are unsatisfied by the delivery service, with a margin of error of 6%, at a 95% level of confidence. Therefore; The interval Estimate Value 1 = 0.45 – 0.06 = 0.39 Value 2 = 0.45 + 0.06 = 0.51 Statement: A survey conducted on a random sample of 260 online customers showed that the percentage of all online customers that are unsatisfied by the delivery service is somewhere between 39% and 51%, at a 95% level of confidence.

26. Question: Verify that the point estimate of the same proportion, at a 99% level of confidence, carries a margin of error of almost 8%

27. Computer Session Using SPSS to generate the confidence interval Open “Delivery Service” file from “Customer Satisfaction Folder” First generate Frequency Table followed by Explore.