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 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. 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 Xtan n or S O O O the margin of error (m)= tan In with, ta/2 is critical value (CV) for a 99% confident level (obtain from the table). 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 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 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 or S xtal Jn" S O the margin of error (m)= t12 Jn" 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 Za/2 m S 7.1 The Standard Normal Curve 1. The waiting time at a bus stop for the next bus to arrive is uniformly distributed between 0 and 8 minutes. a) Find the probability that the waiting time is less than 2 minutes. 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.