Inference about Mean (σ Unknown)

Inference about Mean(σ Unknown) When σ is known, the sampling distribution for a sample mean is normal if conditions are satisfied. For many years, it was thought that when σ was unknown, this was still the case. However, because of the increased variability introduced by not knowing σ, the sampling distribution for a sample mean with unknown σ is not normal. This was discovered by W. S. Gosset, an Irish Brewery quality control inspector in 1908. He also discovered that when σ is unknown, we can still do inference using a sampling distribution model he discovered, the t-distribution. Section 9.1, Page 184

t-Distribution • The t-Distribution is unimodal and symmetric • It has more area in that tails and is flatter in the middle than the normal distribution • It is a family of distributions, with a different curve for each μ, σ, and df (degrees of freedom) df=n-1, where n is the sample size. Critical Values for 95% Confidence Interval Known σ: 1.960 from Normal Distribution Unknown σ with 3 df: 3.18 from t-Distribution Unknown σ with 10 df: 2.23 from t-Distribution Unknown σ with 1000 df: 1.962 from t-Distribution Section 9.1, Page 185

t-Distribution 1. Sample is large enough if n ≥ 30 2. If sample is not large enough, no analysis can be done. Section 9.1, Page 185

Confidence Interval(Unknown σ) Confidence Interval = Sample Mean ± Margin of Error ± Critical Value × Standard Error of where s is the standard deviation of the sample and n is the sample size, and df is the degrees of freedom, n-1. Conditions: The population must be normal or the sample is large (n ≥ 30) Section 9.1, Page 187

Confidence Interval (Unknown σ)Illustrative Problem- TI 83 Add-in Programs C.I. =Sample Mean ± Critical Value * Standard Error PRGM – CRITVAL – ENTER 2: T-DIST CONF LEVEL = .95 df = 19 Answer: 2.0930 PRGM – STDERROR-ENTER 4: 1-MEAN n=20 Sx=1.76 Answer: .3935 C.I. = 6.87 ± 2.0930*.3935 = 6.87 ± .8236 =(6.87 - .8236, 6.87 + .8236) = (6.05, 7.69) Section 9.1, Page 187

Confidence Interval (Unknown σ)Illustrative Problem- TI 83 Black Box Program STAT – TESTS – 8:Tinterval Inpt: Stats 6.87 Sx: 1.76 n: 20 C-Level: .95 Calculate Answer: (6.05, 7.69) Section 9.1, Page 187

Problems • Find the 98% confidence interval. • Find the critical value • Find the margin of error. • Find the standard error. • What assumption must we make about the the population to have a t-sampling distribution. • What are the proper words to describe the confidence interval? • If you wanted to have a margin of error of one minute and the 98% confidence interval for this data, how large must the sample be? Problems, Page 205

Problems • Find the 93% confidence interval. • Find the critical value • Find the margin of error. • Find the standard error. • What assumption must we make about the population to have a t-sampling distribution. • What are the proper words to describe the confidence interval? • If you wanted to have a margin of error of 25 lbs. and a confidence level of 93% for this data, how many students would you have to monitor? Section 9.1, Page 205

Hypotheses Test (Unknown σ)Illustrative Problem- TI-83 Add-In The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution. Ho: μ = 4.9 (≤) (no higher than) Ha: μ > 4.9 (higher than) t-Distributiondf = 21 P-value = .2158 PRGM – TDIST – ENTER LOWER BOUND = 5.1 UPPER BOUND = 2ND EE99 MEAN = 4.9 df = 21 Answer: .2158 (Evidence is not sufficient to support the EPA’s claim.) Section 9.1, Page 188

Hypotheses Test (Unknown σ)Illustrative Problem- TI-83 Black Box The EPA wanted to show that the mean carbon monoxide level is higher than 4.9 parts per million. Does a random sample of 22 readings (sample mean = 5.1, s=1.17) present sufficient evidence to support the EPA’s claim? Use α = .05. Assume the readings have approximately a normal distribution. Ho: μ = 4.9 (no higher than) Ha: μ > 4.9 (higher than) There is not sufficient evidence to support the EPA claim. Section 9.1, Page 188

Problems • Find the p-value. • State your conclusion. • What is the name of the probability model used for the sampling distribution • What is the mean of the sampling distribution? • What is the value of the standard error? • If your conclusion is in error, what type of error is it? Problems, Page 209

Problems • Find the p-value. • State your conclusion. • What is the name of the probability model used for the sampling distribution • What is the mean of the sampling distribution? • What is the value of the standard error? • If your conclusion is in error, what type of error is it? Problems, Page 205

Problems • State the appropriate hypotheses. • Find the p-value. • State your conclusion. • What is the name of the probability model used for the sampling distribution • What is the mean of the sampling distribution? • What is the value of the standard error? • If your conclusion is in error, what type of error is it? Problems, Page 206

Inference about Proportions A binomial experiment is a experiment that has only two outcomes. Binomial experiments relate to categorical variables. Consider the variable, Supports Obama. The variable has two categories, yes and no. We will consider the yes category as a “success” and the no category a “failure”. Suppose we have a population of 1000 voters, and 550 support Obama. We define proportion of success for the population, p = 550/1000 = .55. The proportion of failures is q = 1-p = .45. Suppose we take a sample to estimate p, the true proportion of voters that support Obama. We take a random sample of 100 voters and 53 support Obama. Our sample proportion is p’ = 53/100=.53. Our sample q’ = 1-p’ = .47. Section 9.2, Page 192

Sampling Distribution for a Proportion • In practice, the following conditions will insure the sampling distribution for a proportion is normal. • The sample size is greater than 20 • The product np and nq are both greater than 5. Where we do not know p, we substitute p’, np’ = # success in sample must be > 5 and nq’ =n(1-p’) = # failures in sample must be > 5. • The sample consists of less than 10% of the population. Section 9.2, Page 192

Confidence Interval for a Proportion, p Given a sample proportion p’, that has a normal sampling distribution, the confidence interval for the true population proportion, p, is: sample proportion ± margin of error = sample proportion ± critical value × standard error of p’ where q’ = proportion of failures = 1-p’ and n is the sample size. Section 9.2, Page 193

Confidence Interval for ProportionIllustrative Example TI-83 Add-in Programs In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles. Check conditions for normal sampling condition: n = 200 > 20 # successes = 17 > 5 # failures = 200 – 17 = 183 > 5 Conditions are satisfied. PRGM: CRITVAL 1 – CONF LEVEL = .90 Answer: 1.6449 PRGM: STDERROR – 1:1 PROP – p’ = 17/200; n = 200 Answer: .0197 Confidence Interval = 17/200 ± 1.6449*.0197 = 0.0850 ± 0.0324 = (.0850 - .0324, .0850 + .0324)= (.0526, .1174) Section 9.2, Page 193

Confidence Interval for ProportionIllustrative Example TI-83 Black Box Program In a discussion about the cars that fellow students drive, several statements were made about types, ages, makes colors, and so on. Dana decided he wanted to estimate the proportion of convertibles students drive, so he randomly identified 200 cars in the student parking lot and found 17 to be convertibles. Find the 90% confidence Interval for p, the true proportion of convertibles. STAT – TESTS – A:1 – PROPZInt x: 17 (The number of success in p’: must be integer) n: 200 C-Level: .90 Calculate Answer: (.0526, .1174) We are 90% confident that the true proportion of convertibles is in the interval. Section 9.2, Page 193

Problems Problems, Page 206

Confidence Interval for ProportionRequired Sample Size ME = Critical Value × Standard Error If we have a good estimate of p, we use it. If we have no good estimate of p, we estimate p = .5. This will produce the largest sample for the given conditions. If p’ turns out to be different that p, our ME will be less than we initially required. Section 9.2, Page 194

Required Sample SizeIllustrative Problem TI-83 Consider a manufacturer that purchases bolts from a supplier who claims the bolts are approximately 5% defective. How large a sample do we need to estimate the true proportion to be within ± .02 with 90% confidence? PRGM – SAMPLSIZ – 1:PROPORTION CONF LEVEL = .90 ME = .02 p EST = .05 (Problem estimate) Answer: 322 Section 9.2, Page 195

Problems Problems, Page 208

Hypotheses Test - One ProportionIllustrative Problem – TI-83 Add-In Ho: p = .61 Ha: p > .61 p-value Sampling Distribution for p’ p’ = 235/350 = .6714 p=.61 PRGM – NORMDIST – 1 LOWER BOUND = .6714 – UPPERBOUND = 2ND EE99 MEAN = .61 ; Answer: p-value = .0093 (Reject Ho, We proved that more than 61% sleep more that 7 or more hours) Section 9.2, Page 196

Hypotheses Test - One ProportionIllustrative Problem – TI-83 Black Box Ho: p = .61 Ha: p > .61 STAT – TESTS – 5:1-PropZTest po: .61 x: 235 (# of successes in p’; must be integer) n: 350 prop > po Calculate Answer: p-value = .0092 (Ho is rejected) Section 9.2, Page 196

Problems • Check the conditions for a normal sampling distribution. • State the hypotheses. • Find the p-value. • State your conclusion • If you make an error in your conclusion, what type is it? • Find the mean of the sampling distribution. • Find the standard error of the sampling distribution. Problems, Page 207

Inference about Mean (σ Unknown)