Introduction to inference Confidence Intervals Statistical inference provides methods for drawing conclusions about a population from sample data. 10.1 Estimating with confidence SAT σ = 100 n = 500 µ = 461 For sample (σ x = σ/ ) σ x = 100/ = 4.5 µ - 9 µ µ + 9 95 % 95% of the samples of size 500 will capture µ between x 9 461 9 = 452 461 + 9 = 470 95% between 452 and 470 *******We are 95% confident that the true mean of the SAT for California falls between 452 and 470. ***** Margin of error how accurate we believe our guess is.
Confidence interval A level C confidence interval for a parameter has two parts 1. An interval calculated from the data, usually of the form Estimate margin of error x 2 standard deviations 2. A confidence level, C, which gives the probability that the interval will capture the true parameter value in repeated samples. C.9 or 90 % Page 541 picture (need to look at) Homework read pages 542-543 do problems 1-4
Confidence interval for a population mean with known σ Conditions for constructing a confidence interval for µ 1. Data comes from SRS of the population of interest 2. Sampling distribution of x is approximately normal..10 0.8.10-1.28 1.28 _.05 0.9.05_ -1.645 1.645 Need to know Confidence Tail area Z* 80%.10 1.28 90%.05 1.645 95%.025 1.960 99%.005 2.576
Can find the z* at the bottom of the table in the back of book. Z* is called critical value (on the handout z table Z* is noted as ) Critical values The number z* with probability p lying to its right under the standard normal curve is called the upper p critical value of the standard normal distribution Probability p Z* Confidence interval for a population mean Choose an SRS of size n from population having unknown µ and known σ. A level C confidence interval for µ is X z* ( ) Where z* is the value with an area C between z* and z* under the standard normal curve.
Confidence intervals 1. Identify population of interest and the parameter 2. Choose the appropriate inference procedure. Verify the conditions for using the procedure. 3. If conditions are met, do procedure CI = estimate margin of error 4. Interpret results ---Context!!!!! Problems chapter 10 page 550-551 8 10 Margin of error gets smaller when Z* gets smaller σ gets smaller n gets larger choosing sample size m = z*( ) 95% CI m 5 σ = 43 5 (1.96) (43/ ) 5 (1.96) (43) 5 84.28 16.856 N 284.125 n 285
Cautions page 553 Homework problems pages 552-557 12, 13 σ = 3.2, 14 σ=0.60, 20 c, 22 a,b Null Hypothesis states there is no change or effect on the population Alternate Hypothesis there is a change Null Hypothesis H o : µ = # (This is what you are really trying to disprove) Alternate Hypothesis H a : µ # (this is really what you want) α=level of significance 95% confidence interval α =.05 99% confidence interval α =.01 If mean is in the range than we say at the 5% significance level we fail to reject the claim that (whatever the null is) If mean is outside of the range then we say at the 5% significance level we reject the claim that (whatever the null is) Problems 79a, b,c (change wording to is not equal to the published threshold) 80, 87
Worksheet 1,2 and 6 Inference for the mean of a population with unknown σ Last section- We did not know the true mean but claimed to know the standard deviation for the population. This section- We do not know the population mean or standard deviation of the mean Conditions 1. SRS 2. Normal distribution We will now use S for standard deviation instead of σ Standard error of the statistic is When σ is known we use Z-table When we switch to normal distribution) we switch to t-distribution (this does not have a How to use t-table 80% CI with n = 25 99% CI with n = 12 95% CI with n = 62 90% CI with n = 148
t confidence intervals and tests Confidence intervals x t * Construct a 95% CI x= 1.329 n= 46 s= 0.484 x t * ( ) 1.329 (2.021)( ) 1.329.1442 1.1848 1.4732 We are 95% confident that the true mean level of nitrogen oxides emitted by this type of light duty engine is between 1.1848 and 1.4732 grams/mi Formula Estimate t * SE stimate x
11.4 and 11.9a-c Confidence interval (matched pair) x t*(s/ ) Mean difference Standard deviation of the differences (X 1 -X 2 ) t*(s/ ) This is the list you want x and s Test 1 Test 2 Test 1 - Test 2 Need to know these Using t-procedures SRS-more important than normal n < 15 use t-procedure if close to normal n n 15 use t-procedure except is strong outlier or strong skewness 40 can always use t-procedure Do problems chapter 11 13a,b,d, 15a,b,c
Review problems 11.17, 11.18, 11.19 Review 11.27, 11.28, 11.29, 11.31 (95% CI) confidence interval (x1 x2) t*( degrees of freedom = n 1 for the smallest n problems 11.40 a,c, 11.40 b, 11.42 a,b,c,d, 11.43 (90% CI) Review 11.47a, 11.53 a, CI 99%, c, 11.55 b, 11.56 c, 11.63 b, ****11.64 a e****, 11.72 b,c Chapter 12 Inference for proportions Inference for a population proportion P= Tonya wants to estimate what proportion of the students in her dormitory like the dorm food. She interviews an SRS of 50 of the 175 students living in the dormitory. She finds that 14 think the dorm food is good.
Population: Students in dorm Parameter: percent of students who like dorm food P= 14/50 =.28 Standard deviation of p= Conditions 1. SRS 2. Population 10 times (sample) 3. np 10 4. n(1-p) 10 CI: P z* 95% CI with p =.5069 n = 4040.5069 (1.96).5069 0.0154 0.4915-0.5223 95% confident that the possibility of getting a head is between.4915 and.5223
m= z* When p is not known we can use p* to find the margin of error m= z* p*=.5 ME no greater than 3% How larger of a sample do you need? M=.03= (.03) 2 = 2.0009=.0009n =.25 n= 277.77 n Homework pages 689-697 do problems 8,11, and 15
Two sample proportion P 1 P 2 (find from sample mean) (find from sample mean) 2 Independent samples Confidence Intervals for p 1 - p 2 (p 1 -p 2 ) z* + ) Pop 10 (sample) N 1 p 1 n 2 p 2 n 2 (1-p 2 ) 5 n 1 ( 1-p 1 ) 5 Homework 22-24