Chapter 8 Sampling Distributions Notes: Page 155, 156

Parameter A number that describes the population Symbols we will use for parameters include  - mean  – standard deviation  – proportion (p)  – y-intercept of LSRL  – slope of LSRL

Statistic A number that that can be computed from sample data without making use of any unknown parameter Symbols we will use for statistics include x – mean s  – standard deviation p  – proportion a  – y-intercept of LSRL b  – slope of LSRL

sampling distribution The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from the same population.

Consider the population of 5 fish in my pond – the length of fish (in inches): 2, 7, 10, 11, 14  x = 8.8  x = What is the mean and standard deviation of this population?

Let ’ s take samples of size 2 (n = 2) from this population: How many samples of size 2 are possible?  x = 8.8  x = C 2 = 10 Find all 10 of these samples and record the sample means. What is the mean and standard deviation of the sample means?

Repeat this procedure with sample size n = 3 How many samples of size 3 are possible? 5 C 3 = 10  x = 8.8  x = What is the mean and standard deviation of the sample means? Find all of these samples and record the sample means.

What do you notice? EQUALSThe mean of the sampling distribution EQUALS the mean of the population. As the sample size increases, the standard deviation of the sampling distribution decreases.  x  =  as n xx

unbiased A statistic used to estimate a parameter is unbiased if the mean of its sampling distribution is equal to the true value of the parameter being estimated. Remember the Jelly Blubbers? The judgmental samples were centered too high & were bias, while the randomly selected samples were centered over the true mean

Rule 1: Rule 2: This rule is approximately correct as long as no more than 5% (10%) of the population is included in the sample General Properties  x  =   n  x =

General Properties Rule 3: When the population distribution is normal, the sampling distribution of x is also normal for any sample size n.

General Properties Rule 4: Rule 4: Central Limit Theorem When n is sufficiently large, the sampling distribution of x is well approximated by a normal curve, even when the population distribution is not itself normal. CLT can safely be applied if n exceeds 30. How large is “ sufficiently large ” anyway?

EX) The army reports that the distribution of head circumference among soldiers is approximately normal with mean 22.8 inches and standard deviation of 1.1 inches. a) What is the probability that a randomly selected soldier ’ s head will have a circumference that is greater than 23.5 inches? P(X > 23.5) =.2623 = normalcdf(23.5, 10^99, 22,8, 1.1)

b) What is the probability that a random sample of five soldiers will have an average head circumference that is greater than 23.5 inches? What normal curve are you now working with? Do you expect the probability to be more or less than the answer to part (a)? Explain P(X > 23.5) =.0774  n  x = = 1.1 / sqrt 5 =.4919 Use: ncdf(23.5, 10^99, 22.8,.4919)

Suppose a team of biologists has been studying the Pinedale children ’ s fishing pond. Let x represent the length of a single trout taken at random from the pond. This group of biologists has determined that the length has a normal distribution with mean of 10.2 inches and standard deviation of 1.4 inches. What is the probability that a single trout taken at random from the pond is between 8 and 12 inches long? P(8 < X < 12) =.8427 = normalcdf(8, 12, 10.2, 1.4)

What is the probability that the mean length of five trout taken at random is between 8 and 12 inches long? P(8< x <12) =.9978 = normalcdf(8, 12, 10.2,.6261) Do you expect the probability to be more or less than the answer to part (a)? Explain x = inches = invnorm(.95, 10.2,.6261) What sample mean would be at the 95 th percentile? (Assume n = 5) σ x = σ/sqrt n = 1.4 / sqrt 5

A soft-drink bottler claims that, on average, cans contain 12 oz of soda. Let x denote the actual volume of soda in a randomly selected can. Suppose that x is normally distributed with  =.16 oz. Sixteen cans are randomly selected and a mean of 12.1 oz is calculated. What is the probability that the mean of 16 cans will exceed 12.1 oz? N(12,.16) P(x >12.1) =.0062 = normalcdf(12.1, 10^99, 12,.16/sqrt16)

A hot dog manufacturer asserts that one of its brands of hot dogs has an average fat content of 18 grams per hot dog with standard deviation of 1 gram. Consumers of this brand would probably not be disturbed if the mean was less than 18 grams, but would be unhappy if it exceeded 18 grams. An independent testing organization is asked to analyze a random sample of 36 hot dogs. Suppose the resulting sample mean is 18.4 grams. What is the probability that the sample mean is greater than 18.4 grams? P(x >18.4) =.0082 Ncdf(18.4, 10^99, 18, 1/sqrt36 N(18, 1) Because n exceeds 30, by Central Limit Theorem, can approximate to normal. 36 > 30

Does this result indicate that the manufacturer ’ s claim is incorrect? Yes, not likely to happen by chance alone. This probably indicates that the manufacturer’s claim is false. The probability of a sample having this mean is very unlikely if 18 is the true mean.

Homework: Yesterday: Page 156 and 157 Today: Page 158