Estimating Means and Proportions Using Sample Means and Proportions To Make Inferences About Population Parameters

Estimating Means Almost Always We Do Not Know the Value of the Population Parameters of Interest. Hence, We Must Estimate Their Values Based on Sample Information. Estimators: –X-bar –P -hat –s 2

Sampling Distribution Of The Mean Shape: Approximately Normal with sufficiently large sample size (n). Mean: The mean of this distribution is the same as the mean of the population from which the sample is drawn. Variance: The variance of this distribution is equal to the population variance divided by the sample size (n). Justification: Central Limit Theorem. Implications: If we draw a random sample of sufficient size we can estimate the mean and variance and make probability statements about X-bar.

Vending Machine Example Revisited Recall the cup was 7oz., the population mean and standard deviation was 5oz. and.75oz. respectively. Suppose 30 students decide to get coffee during the break. What is the probability that the average cup will be less than 4 oz. ?

Vending Machine Continued Step 1: P (X-bar < 4oz.) Step 2: Z = (4 - 5)/.75/(sq.. root of 30) –Z = -7.3 Draw the picture P (Z < -7.3) <.001 Interpretation: It’s very unlikely that that the average of 30 students will be less than 4oz.

Introduction To Point And Interval Estimation Suppose we do not know the average amount of coffee dispensed but we do know the standard deviation (.75oz.). We draw a sample of 30 students and compute the sample mean. It turns out to be 5.25oz. Point Estimate: Our best estimate of the population mean is the sample mean 5.25oz.

Computing An Interval Estimate Suppose we want to express that we’re not really sure what the population mean is and would rather put an upper and lower bound on our estimate. Confidence Interval (see formula):

Inferences About Proportions Suppose that X = the number of occurrences of a particular event of interest (e.g., people voting for a candidate, coin turning up heads, people buying a product). P-hat: p = x/n Mean of X = np Variance of X = npq (q = 1-p) Mean of p-hat = P Variance of p-hat = pq/n

Inferences About Proportions Continued For large sample size (n) p-hat is approximately normally distributed (Central Limit Theorem). Z = (p-hat - P)/sq. root of PQ/n Interval Estimate: See formula given in class

What Proportion Of Prospects Will Buy Suppose we are selling insurance and we approach 40 prospects on a given day. Sales records indicate that on average about 20% of prospects buy ( that is P =.2). What is the chance that the proportion of these 40 prospects that buy will be less than 10%? Suppose we were actually able to sell 25%. What is the 95% Confidence Interval in this situation?