AP Statistics Unit 5Addie Lunn, Taylor Lyon, Caroline Resetar

Chapter 18Sampling Distribution Models

Sampling Distributions• A sampling distribution model for how a sample proportion varies

from sample to sample allows us to quantify that variation and how likely it is that we’d observe a sample proportion in any particular material.

• Using a normal modelo µ for meano Ơ for standard deviation

The Central Limit Theorem (CLT)

• The mean of a random sample is a random variable whose sampling distribution can be approximated by a Normal model. The larger the sample, the better the approximation will be.

Conditions for Normality • In order to perform this test the following must be true:

o Normal Model• Unimodal• Symmetric• Bell shaped

o Conditions Random <10% np≥10 nq≥10 Independent Large enough

Formulas• When only given sample size:

o µ()=Po Ơ=

• When given sample size and standard deviation:o µ()=Po ơ()=

Practice Problem • Groovy M&M’s are supposed to make up 30% of the candies sold. In

a large bag of 250 M&M’s what is the probability that we get at least 25% groovy candies?o Normal cdf(.25,,.3, .029) =.96 = = .029

Chapter 19Confidence Intervals and Proportions

Standard Error• Whenever we estimate the standard deviation of a sampling

distribution, we call it a standard error.o Formulas

• Sample Proportion

oSE()= • Sample Mean

oSE()=

Confidence Intervals

• By the 68-95-99.7% rule we know that…o About 68% of all samples will have ’s within 1 SE of po About 95% of all samples will have ’s within 2 SE of po About 99.7% of all samples will have ’s within 3 SE of p

• Each Confidence Interval uses a sample statistic to estimate a population parameter.

• Example:o There is a 95% chance that p is no more than 2 SE away from

Margin of Error• We claim, with 95% confidence, that the interval )

o The extent of the interval on either side of p is called the margin of error.

• We find the critical value by using inverse norm in the calculator

• ME = z*• Your critical value is represented by the z-score in the equation

Conditions Random Independent < 10% np≥ 10 nq ≥10

Practice Problem• A person interviews 50 of the 750 seniors in her high school and

finds that 36 plan to go to the prom. Construct and interpret a 90% confidence interval.o µ = 36/50 =.72 N(.72,.063)o = =0.063 Prop z 1 interval (36,50, .90) = 0.62- 0.82o I’m 95% confident that the proportion mean of .72 is between .62 and .82.

Chapter 20Testing Hypothesis About Proportions

Hypothesis• Our staring hypothesis is the null hypothesis• The null hypothesis that we denote is Ho • The Alternative Hypothesis is Ha• We retain that hypothesis until the facts make it unlikely beyond a

reasonable doubt.• We reject the null if there is no evidence to support the null.• The null hypothesis specifies a population model parameter of

interest and proposes a value for that parameter.• Nothing can be proven.

P-Values• We retain the null if the p-value is greater than 0.05P-value ≥ 0.05• We reject the null if the p-value is less then 0.05P-Value ≤ 0.05• The P-value is a conditional probability, meaning it is the probability

that the observed results could have happened/occurred.

Alternative Alternatives• Ha: Parameter < hypothesized value (one-sided)• Ha: Parameter > hypothesized value (one-sided) - a one-sided alternative focuses on deviations from the null hypothesis value in only one direction• Ha: Parameter ≠ hypothesized value (two-sided alternative) - for two-sided alternatives, the p-value is the probability of deviation in either direction from the null hypothesis value

Conditions Random Independent <10% np≥10 nq≥ 10

PRACTICE PROBLEM• According to the Law School Admission Council, in the fall of 2006,

63% OF LAW School applicants were accepted into law school. The Training program LSAT claims that 168 of the 240 students trained in 2006 were admitted to law school.o Has LSAT demonstrated real improvement over the national Average?

• Ho= .63 N(.63,.031)• Ha> .63 Normal CDF(.7, 1, .63,.031)• SE = √pq/n =√.63× .37/240 = .031 p-value = 0.012

• We reject the null, since the p-value is significantly less. Therefore there is no evidence to support that the LSAT demonstrated any improvement.

Chapter 21More About Tests and Intervals

Zero in on the null• To perform a hypothesis test, the null must be a statement about

the value of a parameter for a model.• We then use that value to figure out the probability that the

observed sample statistic might occur.

P-values• A large p-value doesn’t prove that the null hypothesis is true, but it

offers no evidence that it is not true. So when we see a large p-value, all we can say is that we “don’t reject the null hypothesis.” (we retain the null hypothesis)

• When we see a small p-value, we could continue to believe the null hypothesis and conclude that we just witnessed a rare event. But instead, we trust the data and use it as evidence to reject the null hypothesis.

Alpha Levels and Confidence Intervals

• The Alpha level is denoted as α• Common alpha levels are 0.10, 0.05, 0.01.• Because confidence intervals are two-sided, they correspond to

two-sided tests-in general, a confidence interval with a confidence level of C%

corresponds to a two-sided hypothesis test with an α level of 100 – C%• One-sided test: alpha level of 100-C%/2

Type 1 and Type 2 errors

Chapter 22Comparing two Proportions

Comparing Two Proportions• Comparisons between two percentages are much more common

than questions about isolated percentages.• We want to know how the two groups differ, such as a treatment is

better than a placebo

Two proportion z-Interval• When the conditions are met, we are ready to find the confidence

interval for the difference of two proportions• Confidence Interval

o (P1-P2) ± z* X SE (P1-P2)

• The Critical Value z* depends on the particular confidence level, C, that you specify.

formulas• Standard Deviation of the difference between to sample proportions

SD=• Standard Error

o SE=

Conditions Random <10% Independent np≥10 nq≥10