1. Hypothesis Tests Chapter 7 Prof. Felix Apfaltrer fapfaltrer@bmcc.cuny.edu Office:N518 Phone: 212-220 74 21 Office hours: Tue, Thu 10:30am-12:00 pm

2. Statistical Methods Descriptive Statistics Inferential Statistics Estimation Hypothesis Testing

3. Population Mean  X = 20 Random sample Hypothesis Testing Reject hypothesis! Not close. I believe the population mean age is 50 (hypothesis).

4. What’s a Hypothesis? 1.A Belief about a Population Parameter –Parameter Is Population Mean, Proportion, Variance –Must Be Stated Before Analysis I believe the mean GPA of this class is 3.5! © 1984-1994 T/Maker Co.

5. Definition  In statistics, a hypothesis is a claim or statement about a parameter (or property) of a population.  A hypothesis test (or test of significance) is a standard procedure for testing a claim about a parameter (or property) of a population. If, under a given assumption, the probability of a particular observed event is exceptionally small (as , we conclude that the assumption is probably not correct. Rare Event Rule for Inferential Statistics

6. Null Hypothesis 1.What Is Tested 2.Has Serious Outcome If Incorrect Decision Made 3.Always Has a Sign:  =   or  4.Designated H 0 (pronounced H-oh, or H-zero) 5.Specified as H 0 :   Some Numeric Value – Specified with = Sign Even if , or  – Example, H 0 :   3 Alternative Hypothesis 1. Opposite of Null Hypothesis 1. Opposite of Null Hypothesis 2.Always Has Inequality Sign:  2.Always Has Inequality Sign: , , or  3.Designated H a 3.Designated H a 4.Specified H a : < Some Value 4.Specified H a :  < Some Value Example, H a : < 3 Example, H a :  < 3

7. Example: ProCare Industries, Ltd., once provided a product called “Gender Choice,” which, according to advertising claims, allowed couples to “increase your chances of having a boy up to 85%, a girl up to 80%.” Gender Choice was available in blue packages for couples wanting a baby boy and (you guessed it) pink packages for couples wanting a baby girl. Suppose we conduct an experiment with 100 couples who want to have baby girls, and they all follow the Gender Choice “easy-to- use in-home system” described in the pink package. For the purpose of testing the claim of an increased likelihood for girls, we will assume that Gender Choice has no effect. Using common sense and no formal statistical methods, what should we conclude about the assumption of no effect from Gender Choice if 100 couples using Gender Choice have 100 babies consisting of a) 52 girls?; b) 97 girls? a) We normally expect around 50 girls in 100 births. The results of 52 girls is close to 50, so we should not conclude that the Gender Choice product is effective. If the 100 couples used no special method of gender selection, the result of 52 girls could easily occur by chance. The assumption of no effect from Gender Choice appears to be correct. There isn’t sufficient evidence to say that Gender Choice is effective.

8. b) The result of 97 girls in 100 births is extremely unlikely to occur by chance. We could explain the occurrence of 97 girls in one of two ways: Either an extremely rare event has occurred by chance, or Gender Choice is effective. The extremely low probability of getting 97 girls is strong evidence against the assumption that Gender Choice has no effect. It does appear to be effective. In either case, the null hypothesis would be: H 0 : p=0.5 (the proportion of girls is the usual) While the alternative hypothesis would be H a : p>0.5 (the proportion of girls is indeed improved by the product)

9. Test statistics Is a value computed from the sample datea, used to make a decision about rejecting the null hypothesis: 1.for proportion: Example: A survey of 880 randomly selected drivers showed that 56% admitted to running red lights. Find the value of the test statistic for the claim that the majority of adults run red lights. Example: A survey of 880 randomly selected drivers showed that 56% admitted to running red lights. Find the value of the test statistic for the claim that the majority of adults run red lights. H0: p=0.5 Ha: p>0.5 z= x - μ σ/ n t = x - μ s/ n z= p - p (pq)/ n = 0.56 – 0.5 (0.5*0.5)/ 880 = 3.56 z= p - p (pq)/ n 2.for mean

10. Type I and II errors Type I and type II errors True State or Nature Null hypothesis trueNull hypothesis false Reject null Type I error Decision hypothesis Correct decision  Accept null Type II error hypothesis Correct decision 

11. Claim about mean,  not known Body temperatures: A premed student in a stats class does a project. She collects sample data to test the claim that the mean body temp is less than 98.6F. Collects data from 12 people: 98.0 97.5 98.6 98.8 98.0 98.5 98.6 99.4 98.4 98.7 98.6 97.6 No outliers Make histogram Statistics: –n=12, –xbar=98.39 –s=0.535 Xbar<98.6 H 0 :  H a :  Significance level  =0.05 t = x - μ s/ n = 98.39 – 98.6 0.535/ 12 = - 1.36 The critical value is t= - 1.76, n-1=11 degrees of freedom Left-tailed with  =0.05 Test statistic does not fall in Critical region Accept H0