ESTIMATION-OF-PARAMETERS ON THE SUBJECT STATISTICS
- 2. Very often, we want to describe particular characteristics of an entire population. The information might be needed for decision-making. However, collecting data from the population is impractical even if it is possible to do so.
- 3. Researchers often use a random sample to undertake this. This results may not exactly be equal the population parameter but there are acceptable procedures to approximate population values.
- 4. An estimate is a value or a range of values that approximate a parameter. It is based on sample statistics computed from a sample data. Estimation is the process of determining parameter values.
- 5. A good estimator has the following properties: 1.When the mean of a sample statistic from a large number of different random samples equals the true population parameter, then the sample statistic is an unbiased estimate of the population parameter. 2.Across many repeated samples, the estimates are not very far from the true parameter value.
- 6. Population parameters are usually unknown fixed values. But, there is a way to determine them. There are two ways of determining them under estimation: 1.POINT ESTIMATE – it is a specific numerical value of a population parameter. The sample mean is the best point estimate of the population. 2.INTERVAL ESTIMATE – it is a range of values that may contain the parameter of a population.
- 7. POINT ESTIMATION It is assumed that the population is normally distributed when the sample mean is used as a point estimate of the population mean. Sample variance, s2, will be the point estimate for population variance, σ2.
- 8. POINT ESTIMATION Example #1: A teacher wanted to determine the average height of Grade 9 students in their school. What he did was to go to one of the eight sections in Grade 9 and then took their heights. He computed for the mean height of the students and got 165 cm. Is the computed mean a good point estimate or not? Assumed that the population distribution is normal.
- 9. POINT ESTIMATION Example #2: Consider the population consisting of values (6, 2, 8, 9, 3). Find the population mean and the point estimates considering sample sizes n = 2, 3, and 4 from the same population without replacement. Observation x 1 6 2 2 3 8 4 9 5 3
- 10. POINT ESTIMATION Example #3: Mr. Santiago’s company sells bottled coconut juice. He claims that a bottle contains 500 ml of such juice. A consumer group wanted to know if his claim is true. They took six random samples of 10 such bottles and obtained the capacity, in ml, of each bottle. The result is shown in the table. What is the point estimate of the population mean? Sample 1 500 498 497 503 499 497 497 497 497 495 Sample 2 500 500 495 494 498 500 500 500 500 497 Sample 3 497 497 502 496 497 497 497 497 497 495 Sample 4 501 495 500 497 497 500 500 495 497 497 Sample 5 502 497 497 499 496 497 497 499 500 500 Sample 6 496 497 496 495 497 497 500 500 496 497
- 11. INTERVAL ESTIMATION Instead of pointing to a single number, it is better to approximate the population parameter by determining a range of values within which the population mean is most likely to be located. This range of values is called confidence interval. The confidence levels of 90%, 95%, and 99% are usually chosen.
- 12. INTERVAL ESTIMATION Confidence level refers to the probability that the confidence interval contains the true population parameter. 𝑐𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙 = 1 − 𝛼 100% Where 𝛼 = probability that the confidence interval does not contain the true population parameter. This corresponds to the level of significance. Say, a 95% confidence level implies that the probability of the confidence interval containing the true population parameter is 95%.
- 13. INTERVAL ESTIMATION Critical value (or the tabular value) is the value that indicates the point beyond which lies the rejection region. This region does not contain the true population parameter.
- 14. INTERVAL ESTIMATION INTERVAL ESTIMATE OF POPULATION MEAN WITH KNOWN VARIANCE The formula for interval estimate of population mean when population variance is known and 𝑛 ≥ 30: ҧ 𝑥 − 𝐸 < 𝜇 < ҧ 𝑥 + 𝐸 𝐸 = 𝑧𝛼 2 𝜎 𝑛 Where: 𝜇 = population mean ҧ 𝑥 = mean of a random sample E = margin of error ҧ 𝑥 − 𝐸 = the lower confidence limit ҧ 𝑥 + 𝐸 = the upper confidence limit 𝜎 = population standard deviation 𝑧𝛼 2 = z value at 1 − 𝛼 2 confidence level = the confidence coefficient n = sample size
- 15. INTERVAL ESTIMATION Say, if 𝛼 = 0.05, the distribution is The mean of a random sample size of n is usually different from the population mean 𝜇. The difference which is added to and subtracted from the sample mean in the computation of confidence interval is considered the margin of error.
- 16. The values at each end of the interval are called confidence limits. Between these limits lies the true population parameter. If 𝑛 ≥ 30, z-statistic is to be used with the numerical values of parameter. While if 𝑛 < 30, t-statistic is to be used with the numerical values of statistic. INTERVAL ESTIMATION
- 17. Example #4: The mean score of a random sample of 49 Grade 10 students who took the first periodic test is calculated to be 78. The population variance is known to be 0.16. Find the 95% confidence interval for the mean of the entire Grade 10 students. INTERVAL ESTIMATION
- 18. INTERPRETATION: The researcher is 95% confident that the sample mean of 78 differs from the population mean by no more than 0.112. Also, the researcher is 95% confident that the population mean is between 77.89 and 77.11 when the mean of the sample is 78. INTERVAL ESTIMATION
- 19. Example #5: Assuming normality, use the given confidence level and the sample data below to find the following: a.Margin of error b.Confidence interval for estimating the population parameter Given data: Confidence level = 99% n = 50 ҧ 𝑥 = 18,000 𝜎 =2,500 INTERVAL ESTIMATION
- 20. INTERPRETATION: The researcher is 99% confident that the sample mean differs from the population mean by no more than 910.40. The value of the population mean is within the interval 17,089.60 and 18,910.40 INTERVAL ESTIMATION
- 21. ESTIMATING THE DIFFERENCE BETWEEN TWO POPULATION MEANS The point estimate for the difference of two populations means 𝜇1 − 𝜇2 is ҧ 𝑥1 − ҧ 𝑥2. To obtain this point estimate, select two independent random samples, one from each population with sizes 𝑛1 and 𝑛2, then, compute the difference between their means.
- 22. For interval estimate, when 𝜎1 2 and 𝜎2 2 are known, the confidence interval would be ҧ 𝑥1 − ҧ 𝑥2 − 𝐸 < 𝜇1 − 𝜇2 < ҧ 𝑥1 − ҧ 𝑥2 + 𝐸 𝐸 = 𝑧𝛼 2 𝜎1 2 𝑛1 + 𝜎2 2 𝑛2
- 23. Example #6: Two groups of student in Grade 10 were subjected to two different teaching techniques. After a month, they were given exactly the same test. A random sample of 60 students were selected in the first group and another random sample of 50 students were selected in the second group. The sampled students in the first group made an average of 84 with a standard deviation of 8, while the second group made an average of 78 with a standard deviation of 6. Find a 95% confidence interval for the difference in the population means.
- 24. Example #7: Independent random samples were selected from two populations. Find the 90% confidence interval for estimating the difference in the population means. Population 1 Population 2 Sample Mean 34 38 Population Variance 5 7 Sample Size 40 46
- 25. It is possible that the difference between the two population means could be negative. This indicates that the mean of population 1 could be less than that of population 2.
- 26. Interval Estimate of Population Mean with Unknown Variance If the population standard deviation is unknown and the sample size is small (n < 30), the confidence interval would be ҧ 𝑥 − 𝑡𝛼 2 𝑠 𝑛 < 𝜇 < ҧ 𝑥 + 𝑡𝛼 2 𝑠 𝑛 Where ҧ 𝑥 = mean of a random sample of size n n = sample size s = sample standard deviation 𝑡𝛼 2 = t-value at 1 − 𝛼 2 100% confidence level
- 27. Example #8: The mean and standard deviation of the content of a sample of 10 similar containers are 10.5 liters and 0.352 , respectively. Find a 95% confidence interval and its margin of error for the actual mean content.
- 28. Example #9: The following data in the table are randomly selected scores in Filipino of twelve Grade 10 students. Find a 99% confidence interval for the mean score or all grade 10 students, assuming that the students’ score is approximately normally distributed. 75 81 65 83 76 80 80 70 85 71 77 69
- 29. Confidence Interval for the Difference Between Two Population Means The confidence interval for 𝜇1 − 𝜇2, if the two populations are assumed to have equal but unknown population variances, and n1 < 30 and n2 < 30 is: ҧ 𝑥1 − ҧ 𝑥2 − 𝑡𝛼 2 (𝑆𝑝) 1 𝑛1 + 1 𝑛2 < 𝜇1 − 𝜇2 < ҧ 𝑥1 − ҧ 𝑥2 + 𝑡𝛼 2 (𝑆𝑝) 1 𝑛1 + 1 𝑛2 Where 𝑡𝛼 2 = t-value with df = n1 + n2 - 2 at 1 − 𝛼 2 100% confidence level Sp = pooled sample standard deviation = 𝑛1 −1 𝑠1 2+(𝑛2 −1)𝑠2 2 𝑛1 +𝑛2 −2
- 30. Example #10: A group of 15 students was taught using a new method of teaching math. A second group of 12 students was taught using the traditional method of teaching math. At the end of the grading period, the same examination was given to each group. With the given data in the table, find a 95% confidence interval for the difference between population means. Assume that the population is approximately normally distributed with equal variances. Population 1 2 Sample size 15 12 Sample mean 88 80 Sample standard deviation 8 5
- 31. Confidence Interval for Two Populations that have Unequal Unknown Variances The confidence interval for 𝜇1 − 𝜇2, if the two populations are assumed to have unequal unknown population variances, and n1 < 30 and n2 < 30 is: ҧ 𝑥1 − ҧ 𝑥2 − 𝑡𝛼 2 𝑠1 2 𝑛1 + 𝑠2 2 𝑛2 < 𝜇1 − 𝜇2 < ҧ 𝑥1 − ҧ 𝑥2 + 𝑡𝛼 2 𝑠1 2 𝑛1 + 𝑠2 2 𝑛2 Formula for the degrees of freedom: 𝑑𝑓 = 𝑠1 2 𝑛1 + 𝑠2 2 𝑛2 2 𝑠1 2 𝑛1 2 (𝑛1 − 1) + 𝑠2 2 𝑛2 2 (𝑛2 − 1)
- 32. Example #11: Independent random samples were selected from population 1 and 2 which passes the following data in the table. Find a 95% confidence interval for estimating the difference of the true population means assuming that the population variances are unknown and unequal. Population 1 2 Sample size 20 15 Sample mean 6.95 4.92 Sample standard deviation 2.5 1.2
- 33. Example #12: Find a 95% confidence interval for estimating the difference of the true population means with the following samples taken from two different population with unknown unequal variances. (s1 = 6.60 and s2 = 4.18) Sample 1 Sample 2 66 86 78 77 85 75 76 83 79 76 83 86 88 85 81 80 76 82 75 78 90