STA1510 - Basic Statistics Summary Notes.

STA1510 - Basic Statistics Summary Notes. Sampling distribution of the sample mean The sampling distribution of the sample mean is the distribution of all possible sample means if you selected all possible samples of a given size. This distribution follows a normal distribution with mean E(x) = µ and the standard deviation of the sample mean (the standard error) is n x σ σ = where σ is the population standard deviation and n is the 3 sample size. Changing a sample mean (x) to Z-standard normal values is called standardizing the sample mean. To standardize the sample mean (x) , we use the Z-value equal to the difference between the sample mean (x) and the population mean (µ) divided by the standard error ( ) σ x . That is, n x Z σ − µ = . Please note that in study unit 6, we standardized using σ − µ = X Z . The difference is that here we are dealing with the sample mean (x) and in study unit 6; we were dealing with a random variable (X ). Example 15 Given a normally distributed population of values with x = 76, n = 81and a population variance of 100, what is the standard error of the sample estimate? (1) 0.1447 (2) 0.2257 (3) 0.0342 (4) 1.1111 (5) 0.1915 Solution The standard error of the sample mean is given by 1.1111 9 10 81 100 = = = = n x σ σ Option (4) 7.2 Sampling distribution of the proportion In this section we discuss how the sample proportion ( p) is used to estimate the population proportion (π ) . A population proportion (π ) is the proportion of items in the entire population with the characteristics of interest while as a sample proportion ( p) is the proportion of items in the sample with the characteristics of interest. The sample proportion is expressed as Sample size Number of items of erest in a sample n x p int = = The sampling distribution of the proportion follows the binomial distribution. However, you can use the normal distribution to approximate the binomial distribution when nπ and n(1− π ) are each at least 5. In most cases in which inferences are made about proportions, the sample size (n)is large enough to meet the condition for using the normal approximation. When this is done, we substitute p for x , π for µ and ( ) n π 1− π for n σ . Hence, the sampling distribution of the proportion follows a normal distribution with mean E( p) = π and the standard deviation of the sample proportion (the standard error) is 4 ( ) n p π π σ − = 1 where π is the population proportion and n is the sample size. To standardize the sample proportion ( p) , we use the Z-value equal to the difference between the sample proportion ( p) and the population proportion (π ) divided by the standard error ( ) σ p . That is, ( ) n p Z π π π − − = 1 . Example 16 In a random sample of 350 employees at a company, 45% were in favour of a mandatory aids test. The standard error of p is (1) 0.0266 (2) 0.00142 (3) 0.24750 (4) 0.0132 (5) 0.4975 Solution The standard error of the sample proportion is ( ) ( ) 0.0266 0. 350 0.45 0.55 350 0.45 1 0.45 1 = = × = − = − = n p π π σ SELF ASSESSMENT EXERCISES – TEST YOUR KNOWLEDGE In this section we have selected questions based on whole study unit. Please attempt them before referring to the solutions. Question 1 The average cost of renting a compact size car in a major metropolitan area is R51.74 with a standard deviation of R7.48. Assume a normal distribution. If a sample of 20 cities is taken, the probability that the sample mean will fall between 47.74 and 55.74 is (1) 0.9832 (2) 0.4038 (3) 0.0956 5 (4) 0.4916 (5) 0.9823 Question 2 Fifty percent of all college students attend schools within 50 kilometres of their homes. In a sample of 500 college students, the probability that the sample proportion will be between 0.45 and 0.55 is (1) 0.4875 (2) 0.9818 (3) 0.4909 (4) 0.9750 (5) 0.5000 Question 3 Suppose that life of a particular brand of a calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. The probability that 16 randomly selected batteries from the population will have a sample mean life between 70 and 80 hours is equal to (1) 0.3830 (2) 0.9544 (3) 0.6170 (4) 0.0556 (5) 0.9736 Question 4 A sample of size 200 will be taken at random from an infinite population. Given that the population proportion is 0.60, the probability that the sample proportion will be greater than 0.58 is (1) 0.281 (2) 0.580 (3) 0.719 (4) 0.762 (5) 0.008 SOLUTIONS TO SELF ASSESSMENT EXERCISES Question 1 Given that µ = 51.74, σ = 7.48, n = 20 . Required is the P(47.74 x 55.74) . We begin by standardizing the sample mean to Z as follows: For x = 47.74 6 2.39 1.6728 4 20 7.48 47.74 51.74 = − − = − = − = n x Z σ µ For x = 55.74 2.39 1.6728 4 20 7.48 55.74 51.74 = = − = − = n x Z σ µ Now the P(47.74 x 55.74) changes to P(−2.39 Z 2.39). We can now use that normal table give in study unit 6. 0.9832 0.9916 0.0084 ( 2.39 2.39) ( 2.39) ( 2.39) = = − P − Z = P Z − P Z − Option (1) Question 2 Given that π = 50% = 0.50, n = 500 . Required is the P(0.45 p 0.55). We begin by standardizing the sample proportion to Z as follows: For p = 0.45 ( ) ( ) 2.24 0.0005 0.05 500 0.5 1 0.5 0.45 0.50 1 = − − = − − = − − = n p Z π π π For p = 0.55 ( ) ( ) 2.24 0.0005 0.05 500 0.5 1 0.5 0.55 0.50 1 = = − − = − − = n p Z π π π Now the P(0.45 p 0.55) changes to P(−2.24 Z 2.24). We can now use that normal table give in study unit 6. 0.9750 0.9875 0.0125 ( 2.24 2.24) ( 2.24) ( 2.24) = = − P − Z = P Z − P Z − Option (4) Question 3 Given that µ = 75, σ = 10, n = 16 . Required is the P(70 x 80). We begin by standardizing the sample mean to Z as follows: For x = 70 2.00 2.5 5 16 10 70 75 = − − = − = − = n x Z σ µ For x = 80 7 2.00 2.4 5 16 10 80 75 = = − = − = n x Z σ µ Now the P(70 x 80) changes to P(−2.00 Z 2.00). We can now use that normal table give in study unit 6. 0.9544 0.9772 0.0228 ( 2.00 2.00) ( 2.00) ( 2.00) = = − P − Z = P Z − P Z − Option (2) Question 4 Given that π = 0.60, n = 200. Required is the P( p 0.58). We begin by standardizing the sample proportion to Z as follows: For p = 0.58 ( ) ( ) 0.58 0.0012 0.02 200 0.6 1 0.6 0.58 0.60 1 = − − = − − = − − = n p Z π π π Now the P( p 0.58) changes to P(Z −0.58) . We can now use that normal table give in study unit 6. 0.719 1 0.2810 ( 0.58) 1 ( 0.58) = = − P Z − = − P Z − Option (3) 8 8 CONFIDENCE INTERVAL ESTIMATION STUDY UNIT 8 Key questions for this unit How do you estimate the confidence interval for the population mean when population standard deviation is known? How do you estimate the confidence interval for the population mean when population standard deviation is unknown? How do you estimate the confidence interval for the population proportion? In this study unit we estimate population parameters using either point estimates or interval estimates. A point estimates is the value of a single sample statistic. A confidence interval estimate is a range of numbers, called an interval, constructed around the point estimate. In this study unit, we study confidence for the population mean when the population standard deviation (σ ) is known, confidence for the population mean when the population standard deviation (σ ) is unknown and the confidence interval for the population proportion. The mind map for this is as follows: 9 8.1 Confidence interval estimation for the mean (σ known) If we wish to estimate the confidence interval for the mean when the population standard deviation is known, we use n x Z σ α 2 ± where, x is the sample mean, σ is the population standard deviation, 2 Z α is the multiplier from the Z-standard normal tables and n σ is the standard error of the point estimate. Example 17 Your statistics tutor wants you to determine a confidence interval estimate for the population mean score in the next exam. In the past the scores have been distributed normally with a mean of 74.2 and a standard deviation of 30.9. A 95% confidence interval if your class has 30 students, is (1) (68.72; 79.68) (2) (13.64; 34.76) (3) (64.92; 83.48) (4) (63.14; 85.26) (5) (62.66; 85.73) Solution Given that x = 74.2 , σ = 30.9 . The 95% confidence interval for the population mean with population standard deviation known is n x Z σ α 2 ± where 0.025 1.96 2 5% 2 Z α = Z = Z = from the Z-standard normal table given in section 6. The 1.96 is obtained as follows: We therefore have; 74.2 1.96(5.6415) 74.2 11.0574 30 30.9 74.2 1.96 2 ± ⇒ ± × ⇒ ± ⇒ ± n x Z σ α . The lower limit of the confidence interval is given by 74.2 −11.0574 = 63.1426 and the upper of the confidence interval is given by 74.2 +11.0574 = 85.2574 Option (4) 8.2 Confidence interval estimation for the mean (σ unknown) If we wish to estimate the confidence interval for the population mean when the population standard deviation is unknown, we use ( ) n s x t , n 1 2 − ± α where, x is the sample mean, sis the sample standard deviation, ,( 1) 2 n− tα is the multiplier from the student t-distribution table and 10 n s is the standard error of the point estimate. In this case we use the t-distribution table below: