Area 0.0.2 Area 0.96 Area 0.02 d: 6.64 =0.02. -Z12 Za/2 The area to the left of 20.02 is 0.98" (-0.96+0.02) Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> You select function after this step and proceed yourself 15. [Check your understanding] Find the margin of error for 96% confidence interval with standard error 6 -0.4 for simple random sample size of 100. Formula: 1. The margin of error (m)= (Critical Value) (Standard Error) 2. Find critical value using step to question No 15. 3. Find population Standard error using relation: 5 the 16. [Check Your Understanding] An IQ test was given to a simple random sample of 75 students at a certain college. The sample mean score was 105.2. Scores on this test are known to have a standard deviation of = 10. Construct a 90% confidence interval for the mean IQ score of students at this college. 99.5% Formula: 1. Confident Interval of population mean : (X-m)>u 7.3 ting Dist Formula for Z factor: Z: X- = 7. [Example 1: Area Under a Normal Curve (TI-84)} A study reported that the length of pregnancy from conception to birth is approximately normally distributed with mean = 272 days and standard deviation = 9 days. What proportion of last less than 265 days (i.e. area shaded under curve)?num Z=265 = 272ion mean days after conception aple (a, Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> You select function after this step and proceed yourself 8. [Example 2: Finding Normal Values (TI-84)] IQ scores are normally distributed a mean of 100 and a standard deviation of 15. Find the IQ scores that separate the middle 95% of the scores from the top and bottom 2.5%. 0.95 +0.025 +0.025 100 x=42 Use Formu S Area 0.025 =52 = 100 X1 X2 Area 0.025 Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> You select function after this step and proceed yourself Find critical value using step to question No 15. 8.3 Confidence Intervals for a Population Mean, Unknown 17. [Example: Critical value] A simple random sample of size 11 is drawn from a normal population. Find the critical value (CV) ta/2 for a 99% confident level. Use below table to find CV. Hint: Degrees of Freedom 0.10 0.05 Area in the Right Tail 0.01 0.025 0.005 0.0025 8 1.397 1.860 2.306 2.896 3.355 3.833 9 1.383 1.833 2.262 2.821 3.250 3.690 10 1.372 1.812 2.228 2.764 3.169 3.581 80% 90% 95% 98% 99% 99.5% Confidence Level Degree of freedom sample size -1 TI-84 Plus: STAT->2ND->VERS->DISTR->invT(.....,.....) 18. [Example 2: Confidence Interval (TI-84 Plus)] A sample of 123 people aged 18-22 reported the number of hours they spent on the internet in an average week. The sample mean was 8.20 hours, with a sample standard deviation of 9.84 hours. Assume this is a simple random sample from the population. Construct a 98% confidence interval for u, the population mean number of hours per week spent on the internet in that age group. Finished t Pre Degrees of Area in the Right Tail Freedom 0.025 0.01 0.005 0.0025 100 1.984 2.364 2.626 2.871 the 95% re 98% 99% 99.5% Confidence Level com to you Formula: Confident Interval of population mean u: (X-m)>u Area 9.36 Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> You select function after this step and proceed yourself Or Use A.2 table of the Book. 5. [Example 3: Finding z-Scores (TI-84 Plus)] Find the z-scores that bound the middle 96% of the area under the standard normal curve. Area 0.020 Area 0.96 after Area 0.02 0 Z1 Z2 Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> You select function after this step and proceed yourself Or Use A.2 table of the Book. 7.2 Applications of the Normal Distribution 6. [Convert values from a normal distribution to z-scores] Compact fluorescent bulbs are more energy efficient than incandescent bulbs, but they take longer to reach full brightness. The time that it takes for a compact fluorescent bulb to reach full brightness is normally distributed with mean 30.0 seconds and standard deviation 5.0 seconds. A randomly selected bulb takes 32.0 seconds to reach full brightness. Find and interpret the z- score for x = 32.0. P = 0.27 ample proportion 0.35='P Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> You select function after this step and proceed yourself step 8.1 Confidence Intervals for a Population Mean, Known 13. [Example: Confident Interval] A large sample has mean x = 7.1 and standard error = 2.3. Construct a 98% confident interval for the population mean u. Use below table to find the Critical value z of the 98% confident interval.nda Z 1.645 1.96 2.326 2.576 Formula: 90% 95% 98% 99% Confidence level The margin of error (m)= (Critical Value) (Standard Error) Confident Interval of population mean u: (X-m)>u b) Find the probability that the waiting time is greater than 4 minutes. c) Find the probability that the waiting time is between 2 and 5 minutes. 2. [Area under standard Normal Curve TI 84 Plus] Find the area to the left of z = -1.45. Z 0 Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> normalcdf Or Use A.2 table of the Book. 3. [Check you understanding) Find area between z = -1.96 and z = 1.96 -1.96 1.96 7.2 A Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> normalcdf Or Use A.2 table of the Book. Z tab 4. [Example 1: Finding z-Scores (Tables)] Find the z-score that has an area of 0.36 to its left. or S xtan Jn o the margin of error (m)= O S tan Jn with, ta/2 is critical value (CV) for a 99% confident level (obtain from the table). And the sample standard error= s/n TI-84 Plus: STAT-> Tests-> Tinterval-> STATS ( .............) 19. [Example: Sample Size] Scientists want to estimate the mean weight of mice after they have been fed a special diet. A sample of 70 mice have been weighed, and the standard deviation was s = 2 grams. Estimate the number of mice that must be weighed for a 98% confidence interval to have a margin of error of 0.4 grams. The Za/2 for 98% confident interval is 2.326. Formula: Za12 S n = m Finished test? Now is the time for feedback: Please write a reflection about this test: such as did you like the test format? Review questions helped? Were these questions covered in class? Did you got improved/worsened compared to previous tests? What can you do to do better in the test next time? Did you do great or not that great in this test? What should we change in the test? And in classroom lectures? Etc etc write whatever comes to mind. Everything comments that comes from your side helps me to make this course easy and enjoyable to you. Xtan n or S O the margin of error (m)= tan In O with, ta/2 is critical value (CV) for a 99% confident level (obtain from the table). O And the sample standard error= TI-84 Plus: STAT-> Tests-> Tinterval-> STATS (.. 19. [Example: Sample Size] Scientists want to estimate the mean weight of mice after they have been fed a special diet. A sample of 70 mice have been weighed, and the standard deviation was s = 2 grams. Estimate the number of mice that must be weighed for a 98% confidence interval to have a margin of error of 0.4 grams. The Za/2 for 98% confident interval is 2.326. Formula: n = m 7.3 Sampling Distributions and The Central Limit Theorem on after this (Find shaded 9. [Example: Sampling Distribution] 7. Among students at a certain college, the mean number of hours of television watched per week is = 11.0, and the standard deviation is = 4.0. A simple random sample of 18 students is chosen for a study of viewing habits. Let be the mean number hours of TV watched by the sampled students. Find the mean and the standard deviation of sample X. Formula: Use Formula to find sample information: stand 1. Sample mean (X)= Population mean () 2. Standard devitaion of Sample (ox_bar ) Standard devitaion of Population () Sample size (n) 10. [Use the Central Limit Theorem to compute probabilities involving sample means] popula 0 A population has mean = 50.0 and standard deviation = 13.0. A sample of size 110 is drawn. Find the probability that is between 42 and 52. The Step 1 Step 2 x = 42 = 50 2=52 Use Formula to find sample information: a. Sample mean (X)= Population mean (u) b. Standard devitaion of Sample (x_bar) = Standard devitaion of Population () Sample size (n) sample Hint: USE path in TI 84 Plus: STAT->2ND->VERS->DISTR-> You select function after this step and proceed yourself (Find shaded area of sample normal curve. 7.4 The Central Limit Theorem for Proportions 11. [Example: Sampling Distribution] = The soft-drink cups at a certain fast-food restaurant have tickets attached to them. Customers peel off the tickets to see whether they win a prize. The proportion of tickets 85 people purchase soft drinks between noon that are winners is p = 0.30. A total of n and 1:00 PM on a certain day. Let be the proportion that win a prize. Find the mean and standard deviation of sample proportion ( .). Formula: 1. The mean of the sampling distribution of population proportion: is denoted by Hp and equals the = p 2. The standard deviation of the sampling distribution of is denoted by p(1 - p) p = n p, is: 12. [Example 1: Using the Central Limit Theorem} According to a Harris poll, chocolate is the favorite ice cream flavor for 27% of Americans. If a sample of 200 Americans is taken, what is the probability that the sample proportion of those who prefer chocolate is greater than 0.35?