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BASIC STATISTICS SAMPLING DISTRIBUTIONS Introduction Point Estimation Sampling Distribution of x Sampling Distribution for the Difference between Two Means Sampling Distribution of p̂ Sampling Distribution for the Difference between Two Proportions Introduction A sampling distribution is a distribution of all of the possible values of a sample statistic for a given size sample selected from a population. A sample is a portion of population. For example, suppose you sample 50 students from your college regarding their mean GPA. If you obtained many different samples of 50, you will compute a different mean for each sample. We are interested in the distribution of all potential mean GPA we might calculate for any given sample of 50 students. 4 The reason we select a sample is to collect data to answer a research question about a population. The sample results provide only estimates of the values of the population characteristics. The reason is simply that the sample contains only a portion of the population. With proper sampling methods, the sample results can provide “good” estimates of the population characteristics. Point Estimation Point estimation is a form of statistical inference. In point estimation we use the data from the sample to compute a value of a sample statistic that serves as an estimate of a population parameter. We refer to x as the point estimator of the population mean . s is the point estimator of the population standard deviation . p̂ is the point estimator of the population proportion p. 7 Relationships between the population distribution and the sampling distribution of the sample mean: mean of the sample means = population mean Population distribution Sampling distribution of the sample mean The dispersion narrower than the population distribution. tends to become a bellshaped and to approximate normal. Sampling Distribution of the Sample Mean, X The probability distribution of sample mean, X is called its sampling distribution. It list the various values that X can assume and the probability of each value of X . Mean and Standard Deviation of the Sample Mean, X Mean of sample mean = x x μ = mean of population Standard deviation (standard error) of sample mean = x x n σ = standard deviation for population n = sample size The Central Limit Theorem If all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution. The standard normal distribution; Z ( X ) n This approximation improves with larger samples. Sample Mean Sampling Distribution: If the Population is not Normal We can apply the Central Limit Theorem: ◦ Even if the population is not normal, sample means from the population will be approximately normal as long as the sample size is large enough. n↑ As the sample size gets large enough… the sampling distribution becomes almost normal regardless of shape of population Properties and Shape of the Sampling Distribution of the Sample Mean, X Sampling distribution of sample mean Properties and shape of sampling distribution 2 X ~ N , n - Sample size, n ≥ 30 (normally distributed) - If σ2 unknown, then it is estimated by s2 X ~ N , n - Sample size, n < 30 (sample is from normal population) - σ2 is known/ σ is known 2 T X 2 s n ~ tn 1 - Sample size, n < 30 - σ2 is unknown / σ is unknown - t-distribution with n-1 degree of freedom Example 1 Suppose a population has mean μ = 8 and standard deviation σ = 3. Suppose a random sample of size n = 36 is selected. What is the probability that the sample mean is between 7.8 and 8.2? Solution: Even if the population is not normally distributed, the central limit theorem can be used (n > 30) so the sampling distribution of x is approximately normal σ 3 with mean μ x = 8 and standard deviation σ x 0.5 n 36 P(7.8 X 8.2) P 7.8 - 8 X - μ 8.2 - 8 3 σ 3 36 n 36 P(-0.4 Z 0.4) 0.3108 Example 2 The amount of time required to change the oil and filter of any vehicles is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. 1. What is the standard error of the sample mean to be? 2. What is the probability of the sample mean between 45 and 52 minutes? 3. What is the probability of the sample mean between 39 and 48 minutes? 4. Find the two values between the middle 95% of all sample means. Solution: • X: the amount of time required to change the oil and filter of any vehicles X ~ N 45,102 n 16 • X 2 : the mean amount of time required to change the oil and 10 filter of any vehicles X ~ N 45, 16 a) Standard error = standard deviation, 52 45 45 45 b) P 45 X 52 P Z 2.5 2.5 P 0 Z 2.8 0.4974 10 2.5 16 48 45 39 45 c) P 39 X 48 P Z 2.5 2.5 P 2.4 Z 1.2 0.4918 0.3849 0.8767 P a X b 0.95 d) b 45 a 45 P Z 0.95 2.5 2.5 P za Z zb 0.95 from table: za 1.96 zb 1.96 a 45 1.96 a 40.1 2.5 b 45 1.96 b 49.9 2.5 Sampling Distribution for the Difference between Two Means - the sampling distribution of the difference between two sample means, the distribution of X 1 X 2 can be written as; Mean of X1 X 2 12 22 X1 X 2 ~ N 1 2 , n1 n2 - Hence, Z ( X 1 X 2 ) ( 1 2 ) 12 n1 22 n2 Variance of X1 X 2 Example 3 A taxi company purchased two brands of tires, brand A and brand B. It is known that the mean distance travelled before the tires wear out is 36300 km for brand A with standard deviation of 200 km while the mean distance travelled before the tires wear out is 36100 km for brand B with standard deviation of 300 km. A random sample of 36 tires of brand A and 49 tires of brand B are taken. What is the probability that the a) difference between the mean distance travelled before the tires of brand A and brand B wear out is at most 300 km? b) mean distance travelled by tires with brand A is larger than the mean distance travelled by tires with brand B before the tires wear out? Solution: X 1 : the mean distance travelled before the tires of brand A wear out X 2 : the mean distance travelled before the tires of brand B wear out 2002 3002 X 1 X 2 ~ N 36300 36100, 36 49 X 1 X 2 ~ N 200, 2947.846 a) P | X 1 X 2 | 300 P 300 X 1 X 2 300 300 200 300 200 P Z 2947.846 2947.846 P 9.21 Z 1.84 0.9671 b) P X 1 X 2 P X 1 X 2 0 0 200 PZ 2947.846 P Z 3.68 0.9999 Sampling Distribution of the Sample Proportion p̂ The population and sample proportion are denoted by p and , respectively, are calculated as, X p N and x pˆ n where N = total number of elements in the population; X = number of elements in the population that possess a specific characteristic; n = total number of elements in the sample; and x = number of elements in the sample that possess a specific characteristic. For the large values of n (n ≥ 30), the sampling distribution is very closely normally distributed. pq pˆ ~ N ( p, ) n Mean and Standard Deviation of Sample Proportion Mean of sample proportion = p̂ P̂ p Standard deviation of sample proportion = p̂ P̂ pq n Example 4 If the true proportion of voters who support Proposition A is p 0.40 what is the probability that a sample of size 200 yields a sample proportion between 0.40 and 0.45? Solution: σ pˆ p(1 p) 0.4(1 0.4) 0.03464 n 200 0.45 0.40 0.40 0.40 P(0.40 pˆ 0.45) P Z 0.03464 0.03464 P(0 Z 1.44) 0.4251 Example 5 The National Survey of Engagement shows about 87% of freshmen and seniors rate their college experience as “good” or “excellent”. p̂ Assume this result is true for the current population of freshmen and seniors. Let be the proportion of freshmen and seniors in a random sample of 900 who hold this view. Find the mean and standard deviation. Solution: Let p the proportion of all freshmen and seniors who rate their college experience as “good” or “excellent”. Then, p = 0.87 and q = 1 – p = 1 – 0.87 = 0.13 pˆ p 0.87 p̂ The mean of the sample distribution of is: p̂ pq 0.87(0.13) pˆ 0.011 n 900 The standard deviation of is: Sampling Distribution for the Difference between Two Proportions - the sampling distribution of the difference between two sample proportions, the distribution of Pˆ1 Pˆ2 can be written as; Mean of Pˆ1 Pˆ2 p1 1 p1 p2 1 p2 ˆ ˆ P1 P2 ~ N p1 p2 , n n 1 2 - Hence, ( Pˆ1 Pˆ2 ) ( p1 p2 ) Z p1q1 p2 q2 n1 n2 Variance of Pˆ1 Pˆ2 Example 6 A certain change in a process for manufacture of component parts was considered. It was found that 75 out of 1500 items from the existing procedure were found to be defective and 80 of 2000 items from the new procedure were found to be defective. If one random sample of size 49 items were taken from the existing procedure and a random sample of 64 items were taken from the new procedure, what is the probability that a) the proportion of the defective items from the new procedure exceeds the proportion of the defective items from the existing procedure? b) proportions differ by at most 0.015? c) the proportion of the defective items from the new procedure exceeds proportion of the defective items from the existing procedure by at least 0.02? Solution: PˆN :The proportion of defective items from the new procedure PˆE :The proportion of defective items from the existing procedure 80 0.04 2000 0.04(0.96) PˆN ~ N 0.04, 64 pN 75 0.05 1500 0.05(0.95) PˆE ~ N 0.05, 49 pE 0.05(0.95) 0.04(0.96) ˆ ˆ PN PE ~ N 0.04 0.05, 49 64 Pˆ Pˆ ~ N 0.01, 0.0016 N E a) P PˆN PˆE P PˆN PˆE 0 0 0.01 PZ 0.0016 P Z 0.25 0.4013 b) P | PˆN PˆE | 0.015 P 0.015 PˆN PˆE 0.015 0.015 0.01 0.015 0.01 P Z 0.0016 0.0016 P 0.125 Z 0.625 0.2838 c) P PˆN PˆE 0.02 P PˆN PˆE 0.02 0.02 0.01 PZ 0.0016 P Z 0.75 0.2266 Exercises: 1. Assume that the weights of all packages of a certain brand of cookies are normally distributed with a mean of 32 ounces and a standard deviation of 0.3 ounce. Find the probability that the sample mean weight of a random sample of 20 packages of this brand of cookies will be between 31.8 and 31.9 ounces. Answer: 0.0667 BQT 173 Institut Matematik Kejuruteraan, UniMAP 29 2. According to the BBMG Conscious Consumer Report, 51% of the adults surveyed said that they are willing to pay more for products with social and environmental benefits despite the current tough economic times (USA TODAY, June 8, 2009). Suppose this result is true for the current population of adult Americans. There is a sample 1050 adult Americans who will hold the said opinion. Find the probability that the value of sample proportion is between 0.53 and 0.55. Answer: 0.0921 3. It is known that 30% and 35% of the residents in Taman Sutera and Bandar Mas subscribe to New Straits Times newspaper respectively. If a random sample of 50 newspaper readers from Taman Sutera and 50 readers from Taman Mas were taken randomly, what is the probability that the sample proportion of New Straits Times subscribes in Taman Sutera is larger than Bandar Mas? Answer: 0.2981 End of Chapter 1