Introduction to Sampling Distributions Chapter 18 Introduction to Sampling Distributions Sampling Distributions for Proportions Central Limit Theorem

Parameter and statistic Statistic is a numerical descriptive measure of a sample Parameter numerical descriptive measure of a population. 𝑝 or P-hat is the mean of a sample.

Standard deviation of p-hat & 𝑿 𝜎 𝑝 = 𝑝∗𝑞 𝑛 𝜎 𝑥 = 𝜎 𝑛 (you might think of x-bar being the same as p-hat)

Assumptions Sample is independent Sample is random Sample size is sufficiently large (>30 usually) Sample is < 10% of the population Success/Failure– sample size will guarantee 10 success and 10 failures

Sampling Distributions for Proportions Allow us to work with the proportion of successes rather than the actual number of successes in binomial experiments.

Sampling Distribution of the Proportion n= number of binomial trials r = number of successes p = probability of success on each trial q = 1 - p = probability of failure on each trial

Sampling Distribution of the Proportion If np > 5 and nq > 5 then p-hat = r/n can be approximated by a normal random variable (x) with:

The Standard Error for

Suppose 12% of the population is in favor of a new park. Two hundred citizen are surveyed. What is standard deviation?

Is it appropriate to the normal distribution? 12% of the population is in favor of a new park. p = 0.12, q= 0.88 Two hundred citizen are surveyed. n = 200 Both np and nq are greater than five.

Find the mean and the standard deviation

Central limit theorem (in everyday terms) The CLT states that if we take a sufficiently large number of samples of a population (even if the population is skewed), the mean of the samples will be normally distributed and approach the mean of the population.

The average time to go through a tunnel is 11 minutes with a SD of 2 minutes. A sample of to cars took 11.5 minutes. What is the probability of this occurring? 𝜎 𝑥 = 2 50 = .283 𝑧= 11.5 −11 .283 =1.767 This only occurs 3.9% of the time

Approximately 13% of the population Is left handed. A 200 seat auditorium Is equipped with 15 “lefty” seats. In a class of 90, what is the chance there Will not be enough seats for “lefties”? 𝑆𝐷 𝑝 = .13∗.87 90 = .035 15 90 = .167 𝑧= .167−.13 .035 =1.05 2nd distr. ,<2>,1.05,99) = 14.7%