Basic Probability Distributions, Part 3

Basic Probability Distributions Part 3 How can it be that mathematics, being after all a product of human thought independent of experience, is so admirably adapted to the objects of reality Albert Einstein Central Limit Theorem When the population from which we are selecting a random sample does not have a normal distribution, the central limit theorem is helpful in identifying the shape of the sampling distribution of x . CENTRAL LIMIT THEOREM In selecting random samples of size n from a population, the sampling distribution of the sample mean can be approximated by a normal distribution as the sample size becomes large. Sampling Distribution of x  Example: St. Andrew’s College Sampling Distribution x of for SAT Scores E( x ) = 1090 σ= x σ = n 80 = 14.6 30 x Sampling Distribution of x  Example: St. Andrew’s College What is the probability that a simple random sample of 30 applicants will provide an estimate of the population mean SAT score that is within +/−10 of the actual population mean µ ? In other words, what is the probability that x will be between 1080 and 1100? Sampling Distribution of x  Example: St. Andrew’s College Step 1: Calculate the z-value at the upper endpoint of the interval. z = (1100  1090)/14.6= .68 Step 2: Find the area under the curve to the left of the upper endpoint. =NORM.S.DIST(0.68,1) = 0.7517 Sampling Distribution of x  Example: St. Andrew’s College Sampling Distribution x of for SAT Scores σ x = 14.6 Area = .7517 1090 1100 x Sampling Distribution of x  Example: St. Andrew’s College Step 3: Calculate the z-value at the lower endpoint of the interval. z = (1080  1090)/14.6= - .68 Step 4: Find the area under the curve to the left of the lower endpoint. P(z < -.68) = 1- P(z < -.68) = 1- 0.7517 = 0.2483 Sampling Distribution of  for x SAT Scores Example: St. Andrew’s College Sampling Distribution x of for SAT Scores σ x = 14.6 Area = .2483 1080 1090 x Sampling Distribution of  for x SAT Scores Example: St. Andrew’s College Step 5: Calculate the area under the curve between the lower and upper endpoints of the interval. P(-.68 < z < .68) = P(z < .68)  P(z < -.68) = .7517  .2483 = .5034 The probability that the sample mean SAT score will be between 1080 and 1100 is: P(1080 < x < 1100) = .5034 Sampling Distribution of Sampling Distribution x of for SAT Scores for x SAT Scores σ x = 14.6 Area = .5034 1080 1090 1100 x We could have also said that for z = +0.68 we had Area = .7517 That is .7517-.5 = .2517 more than .5 Therefore, to get probability of -0.68 ≤ z ≤ +.68 = 2(.2517) = 5034 Relationship Between the Sample Size and the Sampling Distribution of x  Example: St. Andrew’s College • Suppose we select a simple random sample of 100 applicants instead of the 30 originally considered. • E( x) = µ regardless of the sample size. In our example, E( x ) remains at 1090. • Whenever the sample size is increased, the standard error of the mean σ x is decreased. With the increase in the sample size to n = 100, the standard error of the mean is decreased from 14.6 to: 𝜎𝜎 𝜎𝜎𝑋𝑋� = 𝑛𝑛= 80 =8 100 4 Relationship Between the Sample Size and the Sampling Distribution of x  Example: St. Andrew’s College With n = 100, 𝜎𝜎𝑋𝑋� =8 With n = 30, σ x = 14.6 E( x ) = 1090 x 4 Relationship Between the Sample Size and the Sampling Distribution of x  Example: St. Andrew’s College • Recall that when n = 30, P(1080 < x < 1100) = .5034. • We follow the same steps to solve for P(1080 < x < 1100) when n = 100 as we showed earlier when n = 30. • Now, with n = 100, P(1080 < x < 1100) = .7887. • Because the sampling distribution with n = 100 has a smaller standard error, the values of x have less variability and tend to be closer to the population mean than the values of x with n = 30. Relationship Between the Sample Size and the Sampling Distribution of x  Example: St. Andrew’s College With n = 100, Sampling Distribution x of for SAT Scores 𝜎𝜎𝑋𝑋� =8 Area = .7887 1080 1090 1100 x 4 Sampling Distribution of  p Making Inferences about a Population Proportion Population with proportion p=? The value of p is used to make inferences about the value of p. A simple random sample of n elements is selected from the population. The sample data provide a value for the sample proportion .p Sampling Distribution of p The sampling distribution of p is the probability distribution of all possible values of the sample proportion p . • Expected Value of p E ( p) = p where: p = the population proportion Sampling Distribution of p • Standard Deviation of p Finite Population N −n σp = N −1 p(1 − p) n Infinite Population σp = p (1 − p ) n • σ p is referred to as the standard error of the proportion. • ( N − n) / ( N − 1) is the finite population correction factor. Form of the Sampling Distribution of p The sampling distribution of p can be approximated by a normal distribution whenever the sample size is large enough to satisfy the two conditions: np > 5 and n(1 – p) > 5 . . . because when these conditions are satisfied, the probability distribution of x in the sample proportion, p = x/n, can be approximated by normal distribution (and because n is a constant). Sampling Distribution of  p Example: St. Andrew’s College Recall that 72% of the prospective students applying to St. Andrew’s College desire on-campus housing. What is the probability that a simple random sample of 30 applicants will provide an estimate of the population proportion of applicant desiring on-campus housing that is within plus or minus .05 of the actual population proportion? Sampling Distribution of  p Example: St. Andrew’s College For our example, with n = 30 and p = .72, the normal distribution is an acceptable approximation because: np = 30(.72) = 21.6 > 5 and n(1 - p) = 30(.28) = 8.4 > 5 Sampling Distribution of  p Example: St. Andrew’s College Sampling Distribution p of .72(1 − .72) σ p = .082 = 30 E( p ) = .72 p Sampling Distribution of  p Example: St. Andrew’s College Step 1: Calculate the z-value at the upper endpoint of the interval. z = (.77  .72)/.082 = .61 Step 2: Find the area under the curve to the left of the upper endpoint. P(z < .61) = .7291 Sampling Distribution of  p Example: St. Andrew’s College Sampling Distribution p of σ p = .082 Area = .7291 p .72 .77 Sampling Distribution of  p Example: St. Andrew’s College Step 3: Calculate the z-value at the lower endpoint of the interval. z = (.67  .72)/.082 = - .61 Step 4: Find the area under the curve to the left of the lower endpoint. P(z < -.61) = .2709 Sampling Distribution of  p Example: St. Andrew’s College Sampling Distribution p of σ p = .082 Area = .2709 .67 .72 p Sampling Distribution of  p Example: St. Andrew’s College Step 5: Calculate the area under the curve between the lower and upper endpoints of the interval. P(-.61 < z < .61) = P(z < .61)  P(z < -.61) = .7291  .2709 = .4582 We could have also computed it as 2(.7291-.5) = .4582 The probability that the sample proportion of applicants wanting on-campus housing will be within +/-.05 of the actual population proportion : P(.67 < p < .77) = .4582 Sampling Distribution of  p Example: St. Andrew’s College Sampling Distribution p of σ p = .082 Area = .4582 .67 .72 .77 p