WARM – UP A local newspaper conducts a poll to predict the outcome of a Senate race. After conducting a random sample of 1200 voters, they find 52% support the republican candidate. a.) Check the assumptions. b.) Construct and Interpret a 95% Confidence Interval. SRS – Stated 1200(.52)=624≥ 10 1200(1 - .52)=576 ≥ 10 Population of voters in the state ≥ 10(1200) One Proportion Z– Conf. Int. We can be 95% confident that the true proportion of state voters who support the republican is between 0.492 and 0.548.

RED CARD = $20

CHAPTER 20 TESTING HYPOTHESES Tests of Significance – This is a method that assesses the evidence for a claim about a population by reporting the probability of the claim actually occurring by random chance. The Evidence in a Significance Test is called the P-Value. The P-Value is the probability, assuming the claim is TRUE, that natural sample variation would produce an observed test statistic this extreme as compared to the claimed population value. It is the probability of obtaining the statistic or one more extreme by pure chance or coincidence, given the claim is true.

The Claim will be written as a Hypothesis. The statement being tested in a test of significance is called the Null Hypothesis. You are assessing the strength of the evidence against the null hypothesis. The statement which expresses what we suspect is true is called the Alternative Hypothesis.

Hypotheses Think about the logic of jury trials: To prove someone is guilty, we start by assuming they are innocent (Hypothesis is true). We retain that hypothesis until the facts make it unlikely beyond a reasonable doubt. Then, and only then, we reject the hypothesis of innocence and declare the person guilty.

A Small P-Value (< 0.05) is evidence AGAINST the claim. When the data are consistent with the model from the null hypothesis, we are unable to reject the null hypothesis. (Remember… If the evidence is not strong enough to reject the presumption of innocent, the jury returns with a verdict of “not guilty.” The jury NEVER says that the defendant is innocent.) If P-value is high (>.05) we “fail to reject the null hypothesis.” since the observe statistic is likely given the null is true. If the P-value is low (≤.05) we “reject the null hypothesis,” since what we observed would be very unlikely were the null model true.

Assumption: The person is NOT Guilty. EX: CLAIM : US Criminal Defense System: According to the United States Criminal Defense system a person accused of a crime is presumed Innocent until PROVEN guilty. Assumption: The person is NOT Guilty. Alternative: The person is Guilty i) A murder (stabbing) has been committed in an apartment uptown. The defendant's fingerprint are all over the apartment, including on the murder weapon. Is he guilty? If the defendant lives there OR his girlfriend lives there, then the probability of his fingerprints being everywhere is HIGH! Meaning no evidence to declare him guilty. This could mean that there is a .5750 probability that the accused person’s fingerprints would be all over the crime scene even if he is innocent. (ie… he lives there or visits often). This means that fingerprints along is not enough to refute that fact that he is NOT guilty because there is good chance that his fingerprints would be there even if he was innocent. Again, with the identified fingerprint evidence, this means that there is a 0.006 probability that the accused person’s fingerprints would all over the crime scene, assuming his innocence. (ie… Should have never been near the crime scene). This means that fingerprints along is enough to refute that fact that he is NOT guilty because there is no reason why his fingerprints would be there if he was innocent. If the defendant has no affiliation to the apartment, then the probability of his fingerprints being everywhere is LOW! This is sufficient evidence to declare the defendant guilty.

Assumption: The person is NOT Guilty. EX: CLAIM : US Criminal Defense System: According to the United States Criminal Defense system a person accused of a crime is presumed Innocent until PROVEN guilty. Assumption: The person is NOT Guilty. Alternative: The person is Guilty i) A P-Value of 0.5750 says that there is a very good chance that the evidence obtained would have been obtained even if the assumption of innocence was True. This does NOT refute the Assumption! ii) A P-Value of 0.006 says that it is very unlikely, almost impossible, to have obtained the evidence if the assumption of innocence was True. BUT we have this evidence so the assumption must be WRONG. There is a 0.6% chance we obtain this evidence by pure chance. We Reject the Assumption. This could mean that there is a .5750 probability that the accused person’s fingerprints would be all over the crime scene even if he is innocent. (ie… he lives there or visits often). This means that fingerprints along is not enough to refute that fact that he is NOT guilty because there is good chance that his fingerprints would be there even if he was innocent. Again, with the identified fingerprint evidence, this means that there is a 0.006 probability that the accused person’s fingerprints would all over the crime scene, assuming his innocence. (ie… Should have never been near the crime scene). This means that fingerprints along is enough to refute that fact that he is NOT guilty because there is no reason why his fingerprints would be there if he was innocent.

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It is the probability of obtaining the statistic or one more extreme, given the claim is true.

Alternative: The SAT books raises scores 200pts. EX: CLAIM: SAT Review Material The Princeton SAT Review book says it will raise your score 200 points. A random sample of students use the book and their score difference has increased. Assumption: The Book has no effect. Student SAT scores stay the same (No Change). Alternative: The SAT books raises scores 200pts. i) A P-Value of 0.7850 says that there is a very good chance that score increase even if the book is not responsible for this increase . This does NOT refute the claim! ii) A P-Value of 0.0002 says that it is very rare (almost impossible) to have an increase in scores if the claim was actually true. BUT there was a point increase so the claim must be WRONG. This says that there is a 0.7850 probability that if multiple samples were obtained the average point increase you be all over the place. (ie… Sample #1 = 350 points, S#2 = 250 points, S#3 = 100 points). This high probability of inconsistency is NOT evidence against the claim. This says that there is a 0.0002 probability that if multiple samples were obtained the average point increase you be 120 every time. (ie… Sample #1 = 150 points, S#2 = 150 points, S#3 = 150 points). This very low probability of inconsistency means that using this Review book will consistently raise your score a value significantly lower than 200pt. This is evidence against the claim.

WARM - UP A company wants to estimate the true proportion of consumers that may buy the company’s newest product. What sample size must be collected in order to estimate this proportion within 3% points with 90% Confidence? Since no p was given, use the conservative p* = .5

p* = The Conservative p = 0.5 unless otherwise noted. WARM – UP Choosing a Sample Size (Chapter 19 cont.) Margin of Error: Solve for n: (Divide both sides by z* and then square both sides.) (Invert.) (Multiply by p(1 – p). ) p* = The Conservative p = 0.5 unless otherwise noted.

Choosing a Sample Size (Chapter 19 cont.) p* = The Conservative p = 0.5 unless otherwise noted.

EXAMPLE 1: A local newspaper conducts a poll to predict the outcome of a hotly contested state ballot initiative that would legalize gambling. Previous polling has revealed a of 55%. At least what sample size must be collected in order to estimate the proportion within 4% points with 90% Confidence? ALWAYS ROUND UP!!! What happens to the sample if you cut the margin of error in half?

Since no p was given, use the conservative p* = .5 EXAMPLE 2: A company wants to estimate the true proportion of consumers that may buy the company’s newest product. At least what sample size must be collected in order to estimate this proportion within 5% points with 95% Confidence? Since no p was given, use the conservative p* = .5