Confidence Intervals

How confident are you? Guess my weight… within 50 pounds Shoot a basketball… at a swimming pool at a large trash can at a carnival hoop

% What happens to your confidence as the interval gets smaller? The smaller the interval, the lower your confidence. % % % %

Point Estimate Use a statistic to estimate a parameter Simple, but not always very precise  Lots of variation in the sampling distribution

estimate + margin of error Confidence Interval Use an interval to estimate a parameter Formula: (on green sheet) estimate + margin of error

Margin of Error How accurate we believe the estimate is Smaller margin of error = more precise estimate of the true parameter Formula: based on confidence level

Confidence Level Success rate of the method used to construct the interval ____% of the time, the intervals we construct this way will contain the true parameter

Critical Value: z* ("z-star") Represents confidence level z* = upper z-score for confidence % in standard normal curve Confidence level tail area z* .05 1.645 .025 1.96 .005 2.576 .05 z*=1.645 .025 .005 z*=1.96 z*=2.576 90% 95% 99%

Confidence Interval for a Proportion: standard deviation of the statistic critical value point estimate margin of error

What does it mean to be 95% confident? 95% chance that p is contained in the confidence interval The probability that the interval contains p is 95% Constructing intervals with this method will produce intervals that contain p 95% of the time

Steps for a confidence interval Check Conditions Randomization: SRS from population, or Randomly assigned treatments Sampling dist. is (approx.) normal: np > 10 and n(1 – p) > 10 (Since we don’t know p, use p ) Independence: Population > 10n

Steps for a confidence interval 2) Calculate the interval 3) Write statements about the interval in context: We are _____% confident that the true proportion of context is between ______ and ______. If we made lots of intervals this way, _____% of them would contain the true proportion.

1. a) A Gallup Poll found that 38% of a random sample of 1012 adults said that they believe in ghosts. Find a 95% confidence interval for the true proportion of all adults who believe in ghosts. Conditions: SRS of adults np = 384.56 > 10, n(1 – p) = 627.44 > 10  Samp. dist. is approx. normal There are at least 10(1012) = 10,120 adults. We are 95% confident that the true proportion of adults who believe in ghosts is between .35 and .41. If we made lots of intervals this way, 95% of them would contain the true proportion.

b) Construct a 90% confidence interval for the same data. c) Construct a 99% confidence interval for the same data. d) How does the confidence level impact the width of the interval? (.3553, .4055) (.3411, .4198) Higher confidence  wider interval (To be more confident, we have to cast a wider net)

How can we make the margin of error smaller? z* smaller (lower confidence level) p(1 – p) smaller n larger (to cut the margin of error in half, n must be ___ times as big) Can we really change this? 4

2. a) A survey of a random sample of 1000 college freshmen found that 47% carry a credit card balance from month to month. Construct a 90% confidence interval for the proportion of all college freshmen who carry a credit card balance from month to month. Conditions: SRS of college freshmen np = 470 > 10, n(1 – p) = 530 > 10  Samp. dist. is approx. normal There are at least 10(1000) = 10,000 college freshmen. We are 90% confident that the true proportion of college freshmen carry a balance from month to month is between .445 and .497. If we made lots of intervals this way, 90% of them would contain the true proportion.

b) Does this interval provide evidence that the proportion of college freshmen who carry a balance is below 50%? Yes, since 50% is not included in the confidence interval.

3. In a nationwide survey, among a random sample of 526 businesses, 137 reported firing workers for misuse of the internet. Construct a 99% confidence interval for the true proportion of U.S. businesses who have fired workers for misuse of the internet. Conditions: SRS of businesses np = 137 > 10, n(1 – p) = 389 > 10  samp. dist. is approx. normal There are at least 5,620 businesses  independence We are 95% confident that the true proportion of businesses who have fired workers for misuse of the internet is between .21 and .31. If we made lots of intervals this way, 95% of them would contain the true proportion.

1. a) A Gallup Poll found that 38% of a random sample of 1012 adults said that they believe in ghosts. Find a 95% confidence interval for the true proportion of all adults who believe in ghosts.

b) Construct a 90% confidence interval for the same data. c) Construct a 99% confidence interval for the same data. d) How does the confidence level impact the width of the interval? (.3553, .4055) (.3411, .4198) Higher confidence  wider interval (To be more confident, we have to cast a wider net)

How can we make the margin of error smaller? z* smaller (lower confidence level) p(1 – p) smaller n larger (to cut the margin of error in half, n must be ___ times as big) Can we really change this? 4

Necessary Sample Size If we want a certain margin of error, we can figure out how big a sample we need: But since we haven't taken a sample yet, we don't have a p or a p-hat to use!

What p-hat (p) do we use in a sample size calculation? p(1 – p) .1(.9) = .09 .2(.8) = .16 .3(.7) = .21 .4(.6) = .24 .5(.5) = .25 By using .5 for p-hat, we are using the worst-case scenario and the largest possible SD in our calculations.

Always round sample size up! 4. Gallup plans to survey Americans to determine the proportion who approve of attempts to clone humans. What sample size is necessary to be within 4% of the true proportion with 95% confidence? Use p = .5 Always round sample size up!

Confidence Interval for a Mean: standard deviation of the statistic critical value point estimate margin of error

Conditions for Means SRS or Randomly assigned treatments Sampling dist. is (approx.) normal: Population is normal, or n > 30 (CLT), or An appropriate graph shows it’s roughly normal

 It’s very unlikely that we would know σ but not μ 1. A test for the level of potassium in blood is not perfectly precise. Repeated measurements for the same person on different days vary normally. A random sample of three patients has a mean of 3.2 and a standard deviation of 0.2. What is a 90% confidence interval for the true mean potassium level? What is σ??  It’s very unlikely that we would know σ but not μ

Student’s t-distribution Very similar to normal distribution  Symmetrical, bell-shaped  Area equals 1 Graph examples of t-curves vs. normal curve: Y1: normalpdf(x), Y2: tpdf(x, 2) Y3: tpdf(x, 5) (use "-0") Window: x = [-4,4], xscl =1; y = [0,.5], yscl =1 Change Y3: tpdf(x, 30)

How does t compare to normal? Shorter & more spread out More area in the tails Based on df (degrees of freedom) = n – 1 As n increases, t-dist. becomes more like standard normal

How to find t-values Green sheet t-table Confidence level at bottom, df on the side df = n – 1 Find these t*: 90% confidence when n = 5 95% confidence when n = 15 t* = 2.132 t* = 2.145

Formula: Margin of error Standard deviation of statistic Critical value point estimate Margin of error

1. A test for the level of potassium in blood is not perfectly precise 1. A test for the level of potassium in blood is not perfectly precise. Repeated measurements for the same person on different days vary normally. A random sample of three patients has a mean of 3.2 and a standard deviation of 0.2. What is a 90% confidence interval for the true mean potassium level? Conditions: SRS of patients Potassium level is normal  normal samp. dist. We are 90% confident that the true mean potassium level is between 2.86 and 3.54. If we made lots of intervals this way, 90% of them would contain the true mean.

Normal prob. plot is linear  normal samp. dist. 2. A random sample of 7 high school students yields the following SAT scores: 950 1130 1260 1090 1310 1420 1190 Construct and interpret a 95% confidence interval for the true mean SAT score. Conditions: SRS of students Normal prob. plot is linear  normal samp. dist. We are 95% confident that the true mean SAT score for high school students is between 1049.9 and 1335.8. If we made lots of intervals this way, 95% of them would contain the true mean.

n > 30  normal samp. dist. (CLT) 3. A random sample of 50 high school students has a mean SAT score of 1250, with a standard deviation of 105. Find a 95% confidence interval for the mean SAT score of all high school students. Conditions: SRS of students n > 30  normal samp. dist. (CLT) We are 95% confident that the true mean SAT score for high school students is between 1220.2 and 1279.8. If we made lots of intervals this way, 95% of them would contain the true mean.

4. The heights of GBHS male students are normally distributed with s = 2.5 inches. How large a sample is necessary to estimate the true mean height within 0.75 inches with 95% confidence? Use z*, since we know σ n = 43 male students

CIs deal with area in the tails – the area most affected by skewness Robustness CIs deal with area in the tails – the area most affected by skewness A robust inference procedure doesn’t change much if the conditions are violated t-procedures can be used with some skewness, as long as there are no outliers Larger n  can handle more skewness t-distributions already have more area in the tails, so if a distribution has some skewness, the tail area is not greatly affected.

6. A medical researcher measured the pulse rate of a random sample of 20 adults and found a mean pulse of 72.69 beats per minute with a standard deviation of 3.86 beats per minute. Assume pulse rate is normally distributed. Compute a 95% confidence interval for the true mean pulse rate of adults. (70.883, 74.497)

Another medical researcher claims that the true mean pulse rate for adults is 72 beats per minute. Does our evidence refute this claim? Explain. Since the interval contains 72 bpm, our evidence does not refute this claim.

7. Consumer Reports tested 14 randomly selected brands of vanilla yogurt and found the following numbers of calories per serving: 160 200 220 230 120 180 140 130 170 190 80 120 100 170 Compute a 98% confidence interval for the average calorie content per serving of vanilla yogurt. (126.16, 189.56)

A diet guide claims you get 120 calories from a serving of vanilla yogurt. What does our evidence indicate? Since the interval does not contain 120, we have strong evidence to suggest that the true mean is not 120.