WARM – UP Quiz Review An insurance company checks police records on 582 accidents selected at random. Teenagers were involved in 91 of them. a.) Find the 95% Confidence Interval for the TRUE % of accidents that involved teens. (Be sure to interpret your interval and check the assumptions!) What does p represent?

a. ). An insurance company checks police records on 582 a.) An insurance company checks police records on 582 accidents selected at random. Teenagers were involved in 91 of them. Find the 95% Confidence Interval for the % of accidents that involved teens. (Follow all steps) One Proportion Z– Conf. Int. We can be 95% confident that the true proportion of accidents involving teenage drivers is between 0.127 and 0.186. SRS – Stated Appr. Normal: 582(.1564) ≥ 10 582(1 - .1564) ≥ 10 91.025 490.975 3. Population of National Auto Accidents ≥ 10(582)

a.) 95% n = 582: (0.127 and 0.186) b.) An insurance company checks police records on 582 accidents selected at random. Teenagers were involved in 91 of them. Find only the numerical values for the lower and upper bound for a 99% Confidence Interval for the % of accidents that involved teens. 99% = (0.117 and 0.195) c.) An insurance company checks police records on 2328 accidents selected at random. (The sample size was quadrupled.) Teenagers were involved in 364 of them. Find only the numerical values for the lower and upper bound for a 95% Confidence Interval for the % of accidents that involved teens. 95% n=2328: (0.141 and 0.171)

a.) 95% n = 582: (0.127 and 0.186) b.) 99% n = 582: (0.117 and 0.195) c.) 95% n=2328: (0.141 and 0.171) What happens to the width of the Confidence interval if you… Increase your Confidence Level? 2. Increase your sample size? Interval gets wider Interval gets Narrower.

The Difference between Confidence Level & Intervals Vitamin D from natural sunlight is important for strong healthy bones. But with children increasingly staying inside to play Video games the risk of Vitamin D deficiency is on the rise. A recent study of 2700 random children found 20% deficient. 1.) Interpret what a 98% Level means. 2.) Find and Interpret the 98% Confidence Interval. 1.) A 98% Confidence Level means that in Repeated Sampling, 98% of all random samples will produce intervals that contain the true proportion, p.

The Difference between Confidence Level & Intervals Vitamin D from natural sunlight is important for strong healthy bones. But with children increasingly staying inside to play Video games the risk of Vitamin D deficiency is on the rise. A recent study of 2700 random children found 20% deficient. 2.) Find and Interpret the 98% Confidence Interval. 2.) p = The true proportion of children deficient in Vitamin D. SRS – Stated Population of Children ≥ 10(2700) 2700(.20) ≥ 10 2700(1 - .20) ≥ 10 One Proportion Z– Conf. Int. We can be 98% confident that the true proportion of Vitamin D deficient children is between 18.2% and 21.8%.

WARM – UP Quiz Review The heights of people in a certain population are normally distributed with a mean of 72 in and a standard deviation of 2.5 in. An SRS of 36 people are selected from the population. a.) Describe the Distribution. b.) Verify the conditions are met. c.) What is the probability that the sample will yield an average height less than 68 inches?

2.) Find and Interpret the 98% Confidence Interval. EXAMPLE: Vitamin D from natural sunlight is important for strong healthy bones. But with children increasingly staying inside to play Video games the risk of Vitamin D deficiency is on the rise. A recent study of 2700 random children found 20% deficient. 1.) Interpret what a 98% Level means. 2.) Find and Interpret the 98% Confidence Interval. Check the Assumptions/Conditions….. SRS – Stated Population of children ≥ 10(2700) 2700(.20) ≥ 10 2700(1 - .20) ≥ 10

Draw the Sampling Distribution. Review EXAMPLE #1: An SRS of 500 students leads to a result of 12 National Merit Students. Draw the Sampling Distribution. Check the Assumptions. Find the Probability that 12 or more of 500 students would be National Merits given that the true proportion = 0.01. SRS – Stated Approx. Norm: 500(.024)=12 ≥ 10 500(1 - .024)=488 ≥ 10 Population of Students ≥ 10(500)

Review EXAMPLE #2: An SRS of 50 students leads to an average SAT score of 2100 with σ = 100. Draw the Sampling Distribution. Check the Assumptions. 3. Find the Probability that 50 students will have an average score greater than 2100 if the true mean SAT score = 2000. SRS – Stated Appr. Normal due to Large n and C.L.T.

Quiz Review EXAMPLE #3: A national study was interested in the proportion of high school students that take AP Statistics. An SRS of 196 students are selected where 20 indicated they are in Stats. a.) Find the margin of error for a 95% Confidence Interval

Quiz Review EXAMPLE #4: -Central Limit Theorem A Large Random Sample will produce an approximately normal distribution, regardless of Population. -z score for proportion -z score for means