1. Stat 301 – Day 17 Tests of Significance

2. Last Time – Sampling cont. Different types of sampling and nonsampling errors  Can only judge sampling bias if know the right answer and can see whether values of the statistic from repeated samples center at the population value (systematic tendency) Random sampling eliminates sampling bias and leads to a predictable pattern (probability distribution) in the sample results (sampling distribution)

3. Last Time We can use probability models to calculate how likely we are to obtain certain sample results  Hypergeometric: X = number of successes from a finite population of successes and failures (e.g., number of Kerry voters out of all N=705 freshmen)  Binomial: X = number of successes from a process with constant probability of success (e.g., number of infants choosing the helper toy, assuming each is equally likely to choose helper)

4. Investigation 3.3.1 (p. 215) C = number of measurements that are noncompliant (success)  “Bernoulli process” – measurements all have the same probability of success, independent   = probability of noncompliant measurement  n = 10 measurements in the year 2000 Made an initial conjecture:  <.10 (benefit of the doubt) If that is true, are we surprised to get 4 or more noncompliant measurements

5. Investigation 3.3.1 P(C > 4) where C follows a Binomial distribution with n = 10 and  =.10 equals.013 Give us fairly strong evidence that such a result did not arise from “random sampling error” alone “Reject” that initial conjecture .10

6. Terminology Detour (p. 219) Null hypothesis – the uninteresting conjecture  H 0 :  < 10 Alternative hypothesis – what you are hoping to show  H a :  > 10 The p-value is calculated assuming the null hypothesis is true. If the p-value is small (e.g., less than.05), we reject the null hypothesis.

7. Handout from Day 16 Suppose the probability a randomly selected voter from the population of all voters will pick Obama equals.5. What is the probability we would get a sample proportion of at least.52? Null hypothesis: Ho:  =.5 Alternative hypothesis Ha:  >.5 X = number of Obama voters in sample of 2774 interview Want P(X >.52  2774  1442)

8. Handout from Day 16 In this case, since the population is more than 20 times the size of the sample, we can use the Binomial distribution to approximate the hypergeometric distribution X is approximately Binomial with n = 2774 and  =.5. P(X > 1442) .02

9. Conclusions If 50% of population planned to vote for Obama, we would get a sample proportion of.52 or larger in about 2% of random samples from this population  We have strong evidence that this sample result came from a population with  >.5 This is a small p-value (.02 <.05), so we will reject the null hypothesis.

10. “Test of Significance” (p. 314) 1. Define the population parameter of interest 2. State the null and alternative hypotheses about the parameter Choose a probability distribution to model the behavior of the sample statistic 3. Calculate the p-value = probability of observing a result at least as extreme (according to alternative hypothesis) as found in the research study 4. Make a decision (reject, fail to reject) about the null hypothesis 5. State your conclusions in context

11. Yet another approach Often these sampling distributions are bell-shaped and symmetric, don’t even look that discrete… Not always…

12. Normal Distribution Characteristics: mound-shaped, symmetric, bell-shaped Parameters  Mean, , peak  Standard deviation, , inflection points f(x)f(x) N( ,  ) model  x  Area under curve = 1

13. Example Suppose we want to determine the probability of a randomly selected car model having a fuel capacity of 13 gallons or less. Suppose the distribution of fuel capacities follow a normal probability model with mean equal to 16.38 gallons and standard deviation 2.708 gallons. Want to find P(X < 13)

14. Minitab 15 Graph > Probability Distribution Plot > Normal

15. Example Probability is about.10 Interpretation: If we repeatedly sampled cars from this population (  = 16.38,  = 2.708), then we would select a car with fuel capacity at most 13 gallons about 10% of the time in the long run.

16. Exploration 4.2 Minitab 15 Probability distribution graph OR Normal Probability Calculator applet Data vs. model

17. For Thursday HW 4 (by Friday)  Include all graphs! PP 3.3.1 (p. 223) Read p. 280-284