Lesson 05 chapter 8 hypothesis testing
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This document contains a PowerPoint presentation on inductive statistics covering topics like probability distributions, sampling distributions, estimation, hypothesis testing for means and proportions, and two-sample hypothesis tests. It provides an overview of the chapters that will be covered, examples of hypothesis tests for means and proportions when the population standard deviation is known and unknown, and examples of independent and dependent two-sample hypothesis tests for differences in means and proportions with both large and small sample sizes. Step-by-step explanations are given for conducting hypothesis tests.
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Lesson 05 chapter 8 hypothesis testing
- 1. Inductive Statistics Dr. Ning DING [email_address] I.007 IBS, Hanze You’d better use the full-screen mode to view this PPT file.
- 2. Table of Contents Review: Chapter 5 Probability Distribution Chapter 6 Sampling Distribution Chapter 7 Estimation Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 3. Chapter 5: Probability Distribution Normal Distribution continuous z=1.00 P=0.3413 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 4. Chapter 5: Probability Distribution Normal Distribution continuous z=±1.00 P=0.6826 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 5. Chapter 6 Sampling Distribution Infinite population Finite population Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 7. Chapter 8 Testing Hypotheses-Summary Ch 8 Example P.417 H 0 H 1 There is no difference between the sample mean and the hypothesized population mean. There is a difference between the sample mean and the hypothesized population mean. H 0 : µ = 10 H 1 : µ > 15 H 1 : µ < 2 H 1 : µ ≠ 15 For example: Mean Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Two-tailed test One-tailed test
- 8. Chapter 8 Testing Hypotheses: Practice 8-28 Step 1: List the known variables Step 2: Formulate Hypotheses Step 3: Calculate the standard error Step 5: Calculate the z value 0.05 P=0.45 z=-1.645 Step 4: Visualize the confidence level With acceptance region accept H 0 so, new bulb producing is good! Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test σ =18.4 n=20 954
- 9. Ch 8 No. Example P.433 Example: Step 1: List the known variables Step 2: Formulate Hypotheses Step 3: Calculate the standard error The HR director thinks that the average aptitude test is 90. The manager sampled 20 tests and found the mean score is 80 with standard deviation 11. If he wants to test the hypothesis at the 0.10 level of significance, what is the procedure? Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 10. Ch 8 No. Example P.433 Example: Step 4: Visualize the confidence level Step 5: Calcuate the t value The HR director thinks that the average aptitude test is 90. The manager sampled 20 tests and found the mean score is 80 with standard deviation 11. If he wants to test the hypothesis at the 0.10 level of significance, what is the procedure? Appendix Table 2 t=-1.729 +1.729 Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 11. Ch 8 No. Example P.433 Step 4: Visualize the confidence level Appendix Table 2 Confidence Interval df 12 0.05 0.10 1.782 Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 12. Chapter 8 Testing Hypotheses-Summary Ch 8 Example P.417 H 0 H 1 There is no difference between the sample mean and the hypothesized population mean. There is a difference between the sample mean and the hypothesized population mean. H 0 : µ = 10 H 1 : µ > 15 H 1 : µ < 2 H 1 : µ ≠ 15 For example: Mean Proportion Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Two-tailed test One-tailed test
- 13. Chapter 8 Testing Hypotheses: Proportion Ch 8 Example P.427 HR director tell the CEO that the promotability of the employees is 80%. The president sampled 150 employees and found that 70% are promotable. The CEO wants to test at the 0.05 significance level the hypothesis that 0.8 of the employees are promotable. Example: Step 1: List the known variables Step 2: Formulate Hypotheses Step 3: Calculate the standard error Proportion Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 14. Chapter 8 Testing Hypotheses: Proportion Ch 8 Example P.427 HR director tell the CEO that the promotability of the employees is 80%. The president sampled 150 employees and found that 70% are promotable. The CEO wants to test at the 0.05 significance level the hypothesis that 0.8 of the employees are promotable. Example: Step 4: Visualize the confidence level Step 5: Calculate the z score Proportion Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 15. Chapter 8 Testing Hypotheses: Practice Ch 8 SC 8-9 P.431 . Step 4: Visualize the confidence level Step 5: Calculate the z score Step 1: List the known variables Step 2: Formulate Hypotheses Step 3: Calculate the standard error SC 8-9 Proportion Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 16. Chapter 8 Testing Hypotheses: Measuring Power of a Hypothesis Test True Not True Accept Reject H 0 Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Type I Error Type II Error
- 17. Chapter 8 Testing Hypotheses--Summary Test Hypotheses for the Mean Test Hypotheses for the Proportion σ is known σ is unknown n <30 & σ is unknown large sample small sample Review : Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 19. Chapter 9 Testing Hypotheses: Two-Sample Tests Let’s compare ! Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test 80
- 20. Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics Independent Samples Dependent Samples Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 21. Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics-Independent σ is known: σ is unknown: H 0 H 1 n <30 & σ is unknown Two-tailed test One-tailed test
- 22. Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples 9.1.1 Difference between means: Large Samples Example: Ch 9 Example P.456 Whether the hourly wages of semiskilled workers are the same between females and males. The survey showed: Step 1: Formulate hypotheses Two-tailed Test Step 2: Find the Estimated Standard Error of Difference Estimated Standard Error of Difference Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Gender Mean hourly wages from sample Standard Deviation of Sample Sample size Female $8.95 $.40 200 Male $9.10 $.60 175
- 23. Chapter 9 Testing Hypotheses: Two-Sample Tests : Two-Independent Samples z=-1.96 +1.96 Step 3: Visualize and find the z values Step 2: Find the Standard Error Ch 9 Example P.456 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 24. Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples 9.1.1 Difference between means: Large Samples Example: Ch 9 Example P.456 Whether the hourly wages of female semiskilled workers are lower than that of males. The survey showed: Step 1: Formulate hypotheses One-tailed Test P=0.45 z=-1.645 Step 2: Visualize and find the z values Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Gender Mean hourly wages from sample Standard Deviation of Sample Sample size Female $8.95 $.40 200 Male $9.10 $.60 175
- 25. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No.9-2 P.460 Step 1: Formulate hypotheses 9-2 P=0.48 z= Step 2: Find the Standard Error Step 3: Visualize and Calculate the z scores - 2.05 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 26. Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples 9.1.2 Difference between means: Small Samples Example: Ch 9 Example P.462 Which program is more effective in raising sensitivity? The survey showed: Step 1: Formulate hypotheses One-tailed Test Step 2: Find the Pooled Estimate of σ 2 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Program Mean Sensitivity from sample Estimated Standard Deviation of Sample Sample size Formal 92 15 12 Informal 84 19 15
- 27. Chapter 9 Testing Hypotheses: Two-Sample Tests: Two-Independent Samples 9.1.2 Difference between means: Small Samples Ch 9 Example P.462 t=1.708 Step 3: Calculate the standard error Step 4: Visualize and find the t scores df =(12-1)+(15-1)=25 Areas in both tails combined=0.10 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Program Mean Sensitivity from sample Estimated Standard Deviation of Sample Sample size Formal 92 15 12 Informal 84 19 15
- 28. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-9 P.466 Step 1: Formulate hypotheses 9-9 Step 2: Find the Pooled Estimate of σ 2 Step 3: Calculate the standard error Step 4: Visualize and Find the t scores One-tailed Test df = 16 area=0.10 t=1.746 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Gender Mean Standard Deviation Sample size Female 12.8 1.0667 10 Male 11.625 1.4107 8
- 29. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 30. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Will the participant lose more than 17 pounds after the weight-reducing program? The survey data is: Step 1: Formulate Hypotheses One-tailed Test Example: Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 31. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 2: Calculate the estimated standard deviation of the population difference Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 32. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 3: Find the Standard Error of the population difference Step 4: Calculate the t value Step 5: Visualize and get the t values df = 10-1=9 area = 0.10 t=1.833 One-tailed Test reject H 0 significant difference Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 33. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-15 P.474 Step 4: Visalize and Calculate the t values t=-1.895 9-15 Step 3: Find the Standard Error of the population difference Step 1: Formulate Hypotheses Step 2: Calculate the estimated standard deviation of the population difference df=7 area=0.10 reject H 0 sig difference Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test One-tailed Test
- 34. Summary Review: Chapter 5 Probability Distribution Chapter 6 Sampling Distribution Chapter 7 Estimation Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test
- 35. Connection with BRM (Business Research Methods)
- 36. Connection with BRM (Business Research Methods) P.354
- 37. The Normal Distribution SPSS Tips The data can be downloaded from: Blackboard – Inductive Statsitics STA2—SPSS-- Week 5 Correlation and Regression.sav
- 38. The Normal Distribution SPSS Tips Our research data is as below: in our research, we are interested in the relationship between the mean response time and the total number correct for 30 puzzles. We obtained scores on 25 adults who are between the ages of 70 and 80 and are not cognitively impaired. Please run the SPSS analysis to explore the relationship between the two variables, Latency and Accuracy. Variable Description Latency Mean response time for 30 puzzles Accuracy Total number correct for 30 puzzles
- 39. The Normal Distribution SPSS Tips Step 1: Click Analyze Correlate Bivariate
- 40. The Normal Distribution SPSS Tips Step 2: Double click on the variables to move to the Variables box
- 41. The Normal Distribution SPSS Tips Step 3: Check it is a two- or one-tailed test and click Options
- 42. The Normal Distribution SPSS Tips Step 4: Click Means and Standard Deviations
- 43. The Normal Distribution SPSS Tips
- 44. The Normal Distribution SPSS Tips Step 5: Click Graph Legacy Dialogs Scatter/Dot...
- 45. The Normal Distribution SPSS Tips Step 6: Choose Scatter/Dot Simple scatterplot
- 46. The Normal Distribution SPSS Tips Step 7: Choose variables for X,Y axis respectively.
- 47. The Normal Distribution SPSS Tips Now you know something about the correlation, but how can you get the regression line as below?
- 48. The Normal Distribution SPSS Tips The correlation between latency and accuracy is -.545, indicating the greater the latency the less the accuracy. The p value of .005 indicates we reject at the .05 level the null hypothesis that latency and accuracy are linearly unrelated in the population. An examination of the bivariate scatterplot supports the conclusion that there is a fairly strong negative linear relationship between the two variables. Interpretation:
Editor's Notes
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
- Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.