1. Introduction to Statistical Inferences Inference means making a statement about a population based on an analysis of a random sample taken from the population. Types of Inferences: Estimation of a parameter, such a the mean. We make an estimate and calculate a margin of error for the estimate. For example, the mean age of shoreline students is 28.5 years with a Margin of Error of ± 3 years. Hypothesis Testing. We test the truth of a statement about a population. We test the statement that the water quality meets quality standards. Both types of inference rely on the use of Sampling Distributions. 1Section 8.1, Page 152

2. Confidence Interval for Mean, μ with known σ 2Section 8.1, Page 154 A random sample of 36 rivets is selected and each is tested for shearing strength. The sample mean = 924 lbs, σ = 18. Our point estimate for the mean shearing strength for the entire population would be μ =924 lbs. Because this is just one sample, it is unlikely that the sample mean of 924 lbs. exactly equals the true mean of the population. How close is the sample mean to the true mean of the population? We will use the sample mean to develop and confidence interval or range of numbers for the plausible value of the true mean.

3. Confidence Interval for Mean, μ with known σ 3Section 8.1, Page 153 Because the sampling distribution is normal, we know that the area between μ – 6 and μ+6 contains 95% of all the sample means. If I pick a sample mean at random and construct an interval ( -6, +6), there is a 95% chance that this will be within 6 units of the true mean, and the interval will therefore contain the true mean. For our example, = 924, our 95% confidence interval is (918, 930)

4. Confidence Interval for Mean, μ with known σ 4Section 8.2, Page 156 When a sampling distribution for a sample mean, is normal, then a confidence interval for μ, the true mean as follows: Confidence Interval = ± Margin of Error Margin of Error = Critical Value × Standard Error. The critical value sometimes referred to as confidence coefficient is the number of standard error units in the Margin of Error for a given confidence level. We will use the notation z(α) to refer to to the critical value for the confidence level α. The equation for the confidence interval for confidence α: ± Margin of Error = ± z(α) ×

5. Constructing An Interval TI-83 Add-in Programs A random sample of 100 commuting students was obtained. The resulting sample mean was 10.22 miles. (σ = 6 miles) Find the 95% confidence interval for the true mean. Check conditions: Since the sample size is ≥ 30, the sampling distribution will be normal, even if the population is not normal. C. I. = sample mean ± margin of error = sample mean ± critical value * standard error. Find the critical value for 95% confidence. PRGM – 1:CRITVAL – ENTER 1 ENTER -.95 - ANSWER: CR VALUE = 1.96 Find the Standard Error of the Sampling Distribution PRGM – STDERROR-ENTER 4:1 MEAN : 6 : 100 : ANSWER: SE =.60 C.I. = 10.22 ± 1.96 * 0.60 = 10.22 ± 1.176 C.I. = (10.22 – 1.176, 10.22 + 1.176 = (9.04, 11.40) 5Section 8.2, Page 158

6. Constructing An Interval Black Box Program A random sample of 100 commuting students was obtained. The resulting sample mean was 10.22 miles. (σ = 6 miles) Find the 95% confidence interval for the true mean. STAT - TESTS – 7:Zinterval – ENTER STATS : σ = 6 ; = 10.22 ; n=100 ; C-Level =.95 Answer: (9.04, 11.40) Using this output we can say: 1. We are 95% confident that the true mean commute distance is in the the interval. 2. If we were to take 100 different samples, and construct 100 different confidence intervals, approximately 95 of them would contain the true mean commute distance. Find the Margin of Error for the confidence interval. ME =.5(width of interval) =.5( 11.40 - 9.04) = 1.18. 6Section 8.2, Page 158

7. Problems 7Problems, Page 50

8. Problems a.What is the variable being studied? b.Find the 90% confidence interval estimate for the mean speed. c.Find the 95% confidence interval estimate for the mean speed. d.Which interval is larger. Why? 8Problems, Page 178

9. Sample Size TI-83 Add-in Program To solve this problem, we need a relationship between sample size and the variables given. The margin of error (ME) is such a relationship. We solve using the TI-83: PGRM – SAMPLSIZ – ENTER – 3: KNOWN σ x ; CONF LEVEL =.99; ME = 75; σ x = 900; Answer: n = 956 9Section 8.2, Page 160

10. Problems 10Problems, Page 179 the standard deviation is 5 seconds.

11. Hypothesis Testing H o : The average GPA of students who take statistics is 3.30 (or more) H a : The average GPA of students who take statistics is less than 3.30. Sample evidence in the form of the sample mean of a sample of students will try to prove H a is true. If H a is true, then H o is false. 11Section 8.3, Page 162 The evidence for H a is the sample mean

12. Writing Hypotheses State authorities suspect the the manager of investment fund is guilty of embezzling money for his own use. In our system of justice, a presumption of innocence is essential to a trial procedure. H o : Manager is innocent H a : Manager is not innocent The state will present evidence in trial to try to prove H a. 12Section 8.3, Page 162

13. Problems 13Problems, Page 179

14. Problems 14Problems, Page 179

15. Hypothesis Test of Mean μ (σ Known) Illustrative Problem Problem: An aircraft manufacturer must demonstrate that its rivets meet the required specifications. One of the specs is: “The mean shearing strength of all such rivets, μ, is at least 925 lbs. (σ=18). Each time the manufacturer buys rivets, it is concerned that the mean strength might be less than the 925-lb pound specification. A random sample of 50 rivets is selected. The sample mean is 921.18 and n = 50. STEP 1: The set up a.Describe the parameter of interest. The parameter of interest is μ, the population mean. b.Write the Hypotheses. Ho: μ = 925 (The mean is at least 925) Ha: μ < 925 (The mean is less than 925) 15Section 8.4, Page 167

16. Illustrative Problem (2) STEP 2: Check assumptions for Normal Sampling Distribution Since σ is known, we will need a normal sampling distribution. The sampling distribution will be normal if the population is normal, or if the sample size is ≥ 30. Since the sample size is 50, the Central Limit Theorem insures that that the sampling distribution is normal STEP 3: The evidence for H a The evidence for Ha is that the sample mean is 921.18 lbs. This is less than the H o value of 925 lbs. The are two possible explanations for the difference between the sample mean and the H o mean: 1. Samples are subject to sampling variation. H o is true and the sample mean difference is explained by natural sampling variation. 2. The difference is too great to be reasonably explained by sampling variation. The difference is explained by the fact that H o is not true. 16Section 8.4, PLage 169

17. Illustrative Problem (3) STEP 4. The probability distribution. We will use the probability distribution to calculate the probability that if H o, is true, the difference between the evidence, and is due to sampling variation. 17Section 8.4, Page 169 PRGM – NORMDIST -1 LOWER BOUND = -2 ND EE 99 UPPER BOUND = 921.18 MEAN = 925 ANSWER: AREA = p-value = 0.0667. p-value = area =0.0667. Sampling Distribution

18. Illustrative Problem (4) Using Black Box Program to Calculate p-value Problem: An aircraft manufacturer must demonstrate that its rivets meet the required specifications. One of the specs is: “The mean shearing strength of all such rivets, μ, is at least 925 lbs (σ=18). Each time the manufacturer buys rivets, it is concerned that the mean strength might be less than the 925-lb pound specification. A random sample of 50 rivets is selected. The sample mean is 921.18 and n = 50. STAT-TESTS-1:Ztest Input: Stats μ o : 925 (This is the Ho parameter value, μ=925) σ: 18 : 921.18 n: 50 μ: <μ 0 (The is the alternate Hypotheses) Calculate Answer: P =0.0667 = p-value 18Section 8.4, Page 169

19. Illustrative Problem (5) STEP 5. Decision We only have two choices for a decision: 1. We reject H o 2. We fail to reject H o Recall that there were only two possibilities that could explain the difference between the H o mean and the sample mean: a. H o is true and the difference is due to sampling variation. b. H o is not true. The p-value tells us how likely it is that a. is the correct explanation for the evidence. If a. is unlikely – it has a very small probability of occurring- then we conclude b. must be the correct explanation for the evidence, the sample mean. Decision Criteria (Significance Level for p-value) If the p-value falls below the significance level, α, then a. is considered too unlikely, and we reject Ho, and conclude Ha is true. If α is not specifically stated in the problem, it is assumed to be 0.05. Since our problem has a p-value of 0.0667 > 0.05, we fail to reject H o. 19Section 8.4, Page 177

20. What does the P-value really mean? The p-value is a probability! It is the probability that, if H 0 is true, the difference between the H 0 value and the sample statistic is due to sampling variation. If the p-value is very small, then the difference between the H 0 value and the sample statistic is unlikely due to sampling variation, so we must conclude that sampling variation is an unlikely explanation for the difference. We therefore conclude that H A must be true. Sometimes, in the press, we see that a study was inconclusive because the study results are likely caused by chance. Or, that the study results are conclusive because the results are unlikely due to chance. In this case, “chance” means normal sampling variation. We also say that the results of the study are not “statistically significant.” There is nothing really “statistically significant” when the null hypothesis is not rejected. On the other hand, when null hypothesis is rejected, it is a “big deal” and we say the results are “statistically significant.” 20Section 8.4

21. Problems a.Write the appropriate hypotheses. b.What condition must be met? Is it met? Explain. c.What is its mean and standard error of the sampling distribution? d.Find the p-value. e.What is your decision? Explain. 21Problems, Page 180

22. Problems a.Write the appropriate hypotheses. b.What condition must be met? Is it met? Explain. c.Sketch the sampling distribution and show its mean and standard deviation? d.Find the p-value. e.What is your decision? Explain. 22Problems, Page 181

23. Two Tailed Test In this problem, sample evidence larger than the mean or evidence smaller than the mean can cause us to reject the null hypothesis. The appropriate hypotheses are: Ho: μ = 82 (The new test mean test value is 82) Ha: μ ≠ 82 (The new test mean is either larger than 82 or smaller than 82) 23Section 8.4, Page 171

24. Two Tailed Test Continued μ=82 Left Tail: PRGM - NORMDIST 1 LOWER BOUND = -2 ND EE99 UPPER BOUND = 79 MEAN = 82 ANSWER: 0.0122 p-value = Left tail area + right tail area = 2*Left tail area =0.0244. Since the p-value is less than 0.05, we reject the null hypothesis and conclude the alternative hypothesis is true – the mean of the new test is different than the mean of the old test. P-value = sum of the two symmetrical areas 24Section 8.4, Page 171

25. Problems Test the claim that the BMI of the cardiovascular technologists is different than the BMI of the general population. Use α =.05. Assume the population of the BMI of the cardiovascular technologists is normal. a.State the necessary hypotheses. b.Is the sampling distribution normal. Why? c.Find the p-value. d.State your conclusion and your reason for it. 25Problems Page 181

26. Types of Errors The probability of a Type I error is the α level or significance level. Recall that we reject H o if the p- value is 5% or less. If the p-value =5%, there is a 5% chance that the scenario of H o true and the evidence is due to sampling variation is the correct scenario. In the long run, we will make an error rejecting H o 5% of the time. We can reduce the probability of a type I error by reducing the α level to 1%. If the p-value =1%, there is a 1% chance that the scenario of H o true and the evidence is due to sampling variation is the correct scenario. In the long run, we will make a type I error only 1% of the time. Reducing the α level will reduce the probability of a type I error, but it will increase the probability of a type II error, fail to reject a false H o. 26Section 8.3, Page 162 Type I Error: Reject a true H 0. Type II Error: Failure to reject a false H 0.

27. Problems 27Problems, Page 179

28. Problems 28Problems, Page 179