Chapter 8: Estimating with Confidence

Chapter 8: Estimating with Confidence

What’s a sample statistic & what’s a population parameter?
Each group come up with an example (in context) of a population parameter and a sample statistic that are related. Consider µ, p, σ, 𝑥 , 𝑝 , s … Five minutes and then share out

Last chapter… In chapter 7, we assumed that we knew the true value of a parameter and then asked questions about the distribution of the statistic used to estimate that parameter We asked questions like, “Assuming that the average number of cookies eaten every week by each adult in the entire U.S. is µ = 10, what is the probability that if we took a SRS of 100 U.S. adults 𝑥 would be greater than 12?”

Now this chapter… Now, we no longer pretend to know the true value of a parameter; we know only the value of the sample statistic which helps us estimate the value of the population parameter “If the mean weight of a sample of 10 adult females is 𝑥 = 123.8, what are the plausible values of µ, the mean weight of all adult females ?

The Mystery Mean … Teams of 3 or 4
Try to estimate the mystery value of the population mean, µ (that I have chosen); I will represent the mystery number as M mean (randNorm (M, 20, 16)) Tells calculator to choose an SRS of 16 observations from a Normal population with mean M and standard deviation 20 and then compute the mean, 𝑥 of those 16 sample values. Is the sample mean shown likely to be equal to the mystery mean M? Why or why not?

The Mystery Mean … mean (randNorm (M, 20, 16))
Tells calculator to choose an SRS of 16 observations from a Normal population with mean M and standard deviation 20 and then compute the mean, 𝑥 of those 16 sample values. Now for the challenge… your group must determine an interval of reasonable values for the population mean µ. Use what we learned about sampling distributions in Chapter 7. Write your interval on the board.

The Mystery Mean … mean (randNorm (M, 20, 16))
Tells calculator to choose an SRS of 16 observations from a Normal population with mean M and standard deviation 20 and then compute the mean, 𝑥 of those 16 sample values. For our interval, would it have been helpful if we said our interval of plausible values could be anywhere from to 1000? What size/width interval are we interested in guessing?

The Mystery Mean … Now that we’ve completed the Mystery Mean activity, do you want to know the true value of M? In real life, we need to conduct a census to find out the true value of a parameter (what are some examples of parameters?). A census (many times) is not even a possibility; and if it is, it is very costly So many times we don’t EVER know the true value of a parameter; we must be satisfied with knowing/using a sample statistic (what are some examples of sample statistics?)

In this chapter… … and the next, we will learn the two most common types of formal statistical inference This chapter: confidence intervals (estimates the value of a parameter) Next chapter: significance testing/hypothesis testing (assesses the evidence for a claim about a parameter) Both are based on the sampling distributions of statistics, i.e., what would happen if we used the inference method many, many times?

Statistical Inference...
Statistical inference provides methods for drawing conclusions about a population based on sample data (like the height sampling distribution, the test scores sampling distribution, the Math 140 data sampling distributions, etc.) Methods used for statistical inference assume that the data was produced by properly randomized design (either random sampling (SRS) or random assignment)

Point estimator and point estimate…
For Mystery Mean, what was our 𝑥 ? That was our sample statistic; that was all we had to start with… 𝑥 is a point estimator… our point estimate was ______. A point estimator is a statistic that provides an estimate of a population parameter. The value of that statistic from a sample is called a point estimate. Point estimator/estimate is our single best guess for value of the parameter.

But statistics vary from sample to sample…
Do you think the value of the population mean µ that I entered into the calculator exactly _______ ? Do you think µ is likely ‘somewhere around’ _______? How close is our 𝑥 likely to be to µ? How would the sample mean 𝑥 vary in repeated samples? Let’s go back to Chapter 7 and before ….

Our Mystery Mean… Sampling distribution Normal (because population is Normal, so is sampling distribution) Spread = 𝜎 𝑛 = = 5 Independent (infinite population) So….

Moving beyond a point estimate…
The sample mean 𝑥 is our point estimate for µ. We don’t expect 𝑥 to be exactly equal to µ. In many samples, the values of 𝑥 follows a Normal distribution with mean µ and SD = 5. The 95 part of the rule for Normal distributions says that 𝑥 is within 2 (5) = 10 of the population mean µ in about 95% of all samples of size n = 16.

Moving beyond a point estimate …
Whenever 𝑥 is within 10 points of µ, then µ is within 10 points of 𝑥 . This happens in about 95% of all possible samples. So the interval from 𝑥 to 𝑥 “captures’ the population mean µ in about 95% of all samples of size 16. For our data/statistic, µ lies somewhere in the interval from ________ to ________... we calculated this interval using a method that captures the true µ in about 95% of all possible samples of this size.

Visual way to think of a confidence interval … can I have a volunteer?
Is the distance from me to you the same as the distance from you to me? Student represents µ, the population mean. Extend your arms representing 2 SD’s on either side I represent 𝑥 , the sample mean If many samples were taken, how often would I be within the student’s arm span? Unfortunately we (almost never) know the true value of µ (put your arms down) But if I am within 2 SD’s of µ in 95% of all samples, then µ will be within 2 SD’s of me in 95% of all samples. So, if I construct an interval of 𝑥 ± 2 SD’s, about 95% of the possible samples will product intervals that capture µ Extend arms.

Vocab… Confidence Interval… A C% CI gives an interval of plausible values for a parameter. The interval is calculated from the data and has the form point estimate ± margin of error The difference between the point estimate and the true parameter value will be less than the margin of error in C% of all samples The confidence level C gives the overall success rate of the method for calculating the confidence interval. In C% of all possible samples, the method would yield an interval that captures the true parameter value.

Trade off’s… Trade off’s between the confidence level and the length of the confidence interval; High CL means a wide CI; Low CL means a narrow CI High CL, wide CI; Low CL, narrow CI

Practice with confidence intervals...
The admissions director at a university has an idea. He proposes using the IQ scores of current students as a marketing tool. The university agrees to provide him with enough money to administer IQ tests to 50 students. So the director gives the IQ test to an SRS of 50 of the university’s 5000 freshmen. The mean IQ score for the sample is 𝑥 = Further, suppose we know σ = 15 (unrealistic to actually know this). What can the director say about the mean IQ score μ of the population of all 5000 freshmen?

Confidence intervals... Is μ = 112? Higher? Lower?
We will never know μ (as is often the case; we rarely know population parameters) But we can say that it is likely that μ is somewhere near 112 due to randomization (and other stuff we will learn shortly) Previous chapter... 𝑥 is a point estimate/estimator, an unbiased estimator of μ; and 𝜎 𝑥 = σ/√n = 15/√50 ≈ 2.1 Point estimator

Confidence intervals...

Confidence intervals... 𝜎 𝑥 = σ/√n = 15/√50 ≈ 2.1
CLT  sample distribution ≈ Normal (n = 50) Normal, rule 95% of all samples of 𝑥 are within 2 standard deviations (4.2) of the center (unknown μ)

Confidence intervals... 𝑥 = 112; 𝜎 𝑥 ≈ 2.1
𝑥 = 112; 𝜎 𝑥 ≈ 2.1 Distribution ≈ Normal (n = 50) In 95% of all samples of 50 the interval 112 ± 4.2 will contain true population mean μ; or in repeated samples, 95% of these intervals capture μ We are 95% confident that the interval from to captures the true, unknown mean of all 5,000 freshmen IQ’s

Confidence intervals...

We are 95% confident in our method
We are 95% confident in our method... Gives correct results 95% of the time...

𝑥 = 112; 𝜎 𝑥 ≈ 2.1 Distribution ≈ Normal (n = 50) In 99.7% of all samples of 50 the interval 112 ± 6.3 will contain true population mean μ; or in repeated samples, 99.7% of these intervals capture μ We are 99.7% confident that the interval from to captures the true, unknown mean of all 5,000 freshmen IQ’s

𝑥 = 112; 𝜎 𝑥 ≈ 2.1 Distribution ≈ Normal (n = 50) In 68% of all samples of 50 the interval 112 ± 2.1 will contain true population mean μ; or in repeated samples, 68% of these intervals capture μ We are 68% confident that the interval from to captures the true, unknown mean of all 5,000 freshmen IQ’s

Confidence interval widths & levels… how do they relate
68% confidence level gives us a confidence interval of (109.9 to 114.1) 95% confidence level gives us a confidence interval of (107.8 to 116.2) 99.7% confidence level gives us a confidence interval of (105.7 to 118.3) Moral of the story? The higher the CL, the wider the CI, in general. We can manipulate this by increasing n though… more on this later … The higher the CL, the wider the CI, in general. We can manipulate this by increasing n though

Confidence intervals... Will we ever know for sure if we captured the true unknown population parameter μ? No. The vast majority of the time, actual μ is unknown. Interpretation of a confidence interval; always use this wording when interpreting a CI: Memorize: “I am ___% confident that the interval from ____ to ____ captures the (parameter in context)”

Practice on wording for interpreting CIs: Correct or not?
There is a 95% probability that the interval from to contains μ, the true, unknown IQ for all 5,000 freshmen. Incorrect. The probability is either 0 or 1 (but we don’t know which)

There’s a 95% chance that the freshmen IQ interval (107.8, 116.2) contains 𝑥 . Incorrect. The general form of CIs is always 𝑥 ± moe, so 𝑥 will always be in the center of the CI

There’s a 95% probability that μ = 112. Incorrect. Never use this wording. Don’t use ‘probability;’ there’s no mention of the CI either

We are 95% confident that the interval from to captures the true proportion of the 50 freshmen from the university who took the IQ test. Incorrect. This wording makes a conclusion about the 50 freshmen who took the IQ test rather than the all 5,000 freshmen (our population)

Incorrectly worded… how do we correct the wording?
We are 95% confident that the interval from to captures the true proportion of the freshmen from the university who took the IQ test. We are 95% confident that the interval from to captures the true proportion of the freshmen from the university who would took the IQ test.

Practice: True or False?
We are 95% confident that the interval from to captures the true, unknown population parameter, μ, for 5,000 freshmen IQ scores from this particular university. Correct. Perfect wording for CI interpretations

Confidence Interval vs. Level… again…
Interval  interval of numbers in context of problem; 𝑥 ± 4.2; (107.8, 116.2) Confidence Intervals usually have the form of estimate ± margin of error (or 𝑥 ± (z*) σ/√n) Level  how certain are we that we have proceeded correctly; used the given method correctly; usually 90%, 95%, 99% levels Usually confidence levels are 90% or higher; want to be quite sure of conclusions

Choose a partner… Let’s go out to the lab for a few applet activities…

Debrief… on confidence level
Part 1: Capture rate of the intervals/percent hits … in the long run is very close to confidence level; sample size doesn’t effect the confidence level/capture rate/percent hits Confidence level is the overall capture rate if the method is used many times. To say we are C% confident is shorthand for “If we take many samples of the same size from this population, about C% of them will result in an interval that captures the actual parameter value.”

Interpreting a Confidence Level…
If many samples are taken and 95% confidence intervals are constructed based on these samples, then about 95% of the intervals will capture the true parameter being estimated. In real life we usually calculate just one interval. Below describes what would happen if we were to take many, many samples

What does the applet picture mean
What does the applet picture mean? The applet is selecting an SRS of size 20, computing the sample mean 𝑥 and constructing a 95% CI to estimate µ, the unknown population parameter. If the interval crosses the vertical line, the interval succeeded in capturing µ.

Why are the intervals decreasing in length when the CL is decreased?
The intervals don’t need to be as long because the desired capture rate/confidence level is smaller

Bottom line… price we pay, in general, for greater/higher confidence level is a wider interval
If we’re satisfied with 80% confidence level, then our interval of plausible values for the unknown population parameter will be much narrower than if we insist on 90%, 95%, or 99% confidence level (and we will be much less confident in our estimate with an 80% confidence level). Taking the idea of confidence to an extreme, what if we want to estimate with 100% confidence the temperature of Santa Clarita today. I am 100% confident that the temperature in Santa Clarita today will be between -50 degrees F and degrees F! The activity also shows that we can get a more precise estimate of a parameter by increasing the sample size. Larger samples yield narrower confidence intervals. This result holds for any confidence level.

Calculating a Confidence Interval

Using Confidence Intervals Wisely…
Just an introduction; ‘nice’ situations so far We didn’t know µ (that’s the whole idea why are doing CI’s), but somehow we know σ … not very realistic at all. In reality, if we don’t know µ, then we don’t know σ More realistic CI’s for population means in Section 8.3 But first, we will practice CI’s for population proportions

Using Confidence Intervals Wisely…
Our method of calculation assumes that the data come from an SRS (or random assignment) of size n from the population of interest The margin of error in a confidence interval covers only chance variation due to random sampling or random assignment. It does not explain any other problems such as non-response, bias wording, under-coverage, etc.

Homework Page 489, # 1, 3, 5, 9, 11, 15, 17, 19, 21, 22, 23, 24 And FRQ’s (as needed) Section Quiz on 8-1 tomorrow…

8.2 Estimating a Population Proportion…
The beads/beans activity

Conditions for Estimating p, unknown population parameter
Random/Randomization: Data should come from a well-designed random sample or randomizes experiment (random assignment) If we can’t draw conclusions beyond the data at hand, there’s not much point in constructing a confidence interval

10%/Independence: When sampling without replacement (which is almost always the case; we rarely sample with replacement from a population) from a finite population, be sure to check that the population is at least 10 times the sample size (or make sure it is safe to assume this). Reason? When using the formula for the SD of a sampling distribution for 𝑝 , σ 𝑝 = 𝑝(1−𝑝) 𝑛 , acts as if we are sampling with replacement.

Large Count/Normality: Method we use to calculate a CI for p depends on the fact that the sampling distribution of 𝑝 is approximately Normal. Can use the Normal approximation to the sampling distribution of 𝑝 if np ≥ 10 and n(1 – p) ≥ 10; But we don’t know p (if we did, we wouldn’t have to construct a CI!) But in a large sample, 𝑝 will tend to be close to p; so we replace p by 𝑝 in checking the large count/Normality condition; n 𝑝 ≥ 10 and n (1 - 𝑝 ) ≥ 10

SRS (or random assignment) Independence Large Count/Normality

Beads/beans… check conditions
SRS (or random assignment)? 10%; sampled without replacement; is the population at least 10 times the sample size? Is this safe to assume? Large Count/Normality; Is n 𝑝 ≥ 10 & n (1 - 𝑝 ) ≥ 10 ?

What happens if any conditions are violated?
We can’t trust that our results will be meaningful. In general, unless you are told specifically, don’t proceed with calculations Note: Be careful. College Board won’t waste an entire FRQ on conditions not being met.

Constructing a Confidence Interval for p, Unknown Population Parameter Proportion
Format for Confidence Intervals: estimate ± margin of error estimate ± (critical value) (standard deviation of statistic) 𝑝 ± (z*) 𝑝 (1− 𝑝 ) 𝑛 Note: some books refer to SD of statistic as “standard error”

We have our beads/beans estimate; we can calculate our standard deviation of statistic… how do we get our critical value? What is a critical value? estimate ± margin of error estimate ± (critical value) (standard deviation of statistic) 𝑝 ± (z*) 𝑝 (1− 𝑝 ) 𝑛

What is a critical value? A critical value is the number of standard deviations away from our sample center (our estimate). Critical value is directly based on our confidence level. estimate ± margin of error estimate ± (critical value) (standard deviation of statistic) 𝑝 ± (z*) 𝑝 (1− 𝑝 ) 𝑛

Critical Value Detour…
A critical value is the number of standard deviations away from our sample center (our estimate). Our z*. Critical value is directly based on our confidence level. Let’s practice with a 95% confidence level; central 95%; what area is left over for tails? Split in ½; symmetry Look up area on Table A; go over and up for z score; that’s our z*

A critical value is the number of standard deviations away from our sample center (our estimate). Our z*. Critical value is directly based on our confidence level.

A critical value is the number of standard deviations away from our sample center (our estimate). Our z*. Critical value is directly based on our confidence level. What is our z* for 90% confidence level? z* = What is our z* for 99% confidence level? z* = 2.575

Now let’s calculate a 95% CI for our beads/beans…
estimate ± margin of error estimate ± (critical value) (standard deviation of statistic) 𝑝 ± (z*) 𝑝 (1− 𝑝 ) 𝑛 … and let’s interpret it too… we are 95% confident that the interval from ___ to ___ captures the true unknown population proportion of ___ color beads/beans in the entire jar.

Putting it all together… the four-step process

Practice... Alcohol abuse is considered by some as the number one problem on a campus. How common is it? A 2011 SRS of 10,904 U.S. college students collected information on drinking behavior and alcohol-related problems. The researchers defined “frequent binge drinking” as having 5 or more drinks in a row 3 or more times in the past 2 weeks. According to this definition, 2,486 students were classified as frequent binge drinkers. Based on these data, construct and interpret a 99% confidence interval for the proportion of all U.S. college students who have engaged in frequent binge drinking.

n = 10,904; x = 2,486; confidence level: 99%; 𝑝 = 2,486/10,904 = 0.228
State: What parameter do you want to estimate, and at what confidence level? We want to estimate the true proportion, p, of all U.S. college students who would say that they engage in frequent binge drinking with 99% confidence.

n = 10,904; x = 2,486; confidence level: 99%; 𝑝 = 0.228
Plan: Identify appropriate inference method. Check conditions. We should use a one-sample z-interval for p if the conditions are met. Random: SRS stated in problem 10%: We are sampling without replacement, so we need to be able to reasonably assume that there are at least 10 (10,904) U.S. college students Large Count/Normality – n 𝑝 = (10,904) (0.228) = 2486 ≥ 10 n (1 – 𝑝 ) = (10,904) (0.772) = 8418 ≥ 10 Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the population from which the sample came.

Do: If the conditions are met, perform calculations 𝑝 ± (z*)(√( 𝑝 (1 – 𝑝 )/n) = ± (2.575) ( √((0.228)(0.772)/(10,904)))  (0.218, 0.238) OR stat – test – 1PropZInt (NOT ZINT) x = 2486; n = 10,904; c-level 99; calculate  ( , )

Conclude: Interpret your interval in the context of the problem. We are 99% confident that the interval from to captures the true proportion of U.S. college students who would say that they engage in frequent binge drinking.

n = 10,904; x = 2,486; 𝑝 = 0.228 How about if we wanted to calculate a 95% confidence interval? Look up z* and calculate by hand; then use technology. How about if we wanted to calculate a 90% confidence interval? Look up z* and calculate by hand; then use technology

More Practice... The same study also asked that same SRS of 10,904 if they would classify themselves as abstainers (non-drinkers). Of the 10,904 U.S. college students asked, 2105 classified themselves as abstainers (nondrinkers). Construct a 99% confidence interval following the 4-step process (State, Plan, Do, Conclude)

n = 10,904; x = 2,105; 99% CL; 𝑝 = State: What parameter do you want to estimate, and at what confidence level? We want to estimate the true proportion, p, of all U.S. college students who would say that they consider themselves abstainers with 99% confidence. Plan: Identify appropriate inference method. Check conditions. We should use a one-sample z-interval for p if the conditions are met. Random: SRS stated in problem 10%: We are sampling without replacement, so we need to be able to reasonably assume that there are at least 10 (10,904) U.S. college students Large Count/Normality n 𝑝 = (10,904) (0.1930) = 2104 ≥ 10 n (1 – 𝑝 ) = (10,904) (0.8070) = 8799 ≥ 10 Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the populatin from which the sample came.

n = 10,904; x = 2,105; 99% CL; 𝑝 = Do: If the conditions are met, perform calculations 𝑝 ± (z*)(√( 𝑝 (1 – 𝑝 )/n) = ± (2.575) ( √((0.193)(0.807)/(10,904)))  (0.183, 0.203) OR stat – test – 1PropZInt (NOT ZINT) x = 2105; n = 10,904; c-level 99; calculate  (0.1833, )

n = 10,904; x = 2,105; 99% CL; 𝑝 = Conclude: Interpret your interval in the context of the problem. We are 99% confident that the interval from to captures the true proportion of all U.S. college students who would say that they abstain from drinking.

One more practice problem…
Spinning the globe. In her first-grade social studies class, Jordan learned that 70% of Earth’s surface was covered in water. She wondered if this was really true and asked her dad for help. To investigate, he tossed an inflatable globe to her 50 times, being careful to spin the globe each time. When she caught it, he recorded where her right index finger was pointing. In 50 tosses, her finger was pointing to water 33 times. Construct and interpret a 95% confidence interval for the proportion of Earth’s surface that is covered in water.

State – Plan – Do - Conclude
We want to estimate p, the true proportion of Earth’s surface that is covered in water with 95% confidence. We should use a one-sample z interval for p if the conditions are met. Random: The 50 locations are a random sample of all possible locations on the globe. 10%: We do not need to check the 10% condition; locations were selected with replacement. Large count: n 𝑝 = 33 ≥ 10 and n (1 - 𝑝 ) = 17 ≥ 10 Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the populatin from which the sample came.

State – Plan – Do - Conclude
𝑝 ± (z*) 𝑝 (1− 𝑝 ) 𝑛 = 0.66 ± (1−0.66) 50 = 0.66 ± = (0.529, 0.791) We are 95% confident that the interval from to captures the true proportion of Earth’s surface that is covered in water. This is consistent with the claim that 70% of Earth’s surface is covered in water, because 0.70 is one of the plausible values in the interval.

AP Exam Tip… You may use your calculator to compute a confidence interval on the AP exam (or in here). But there’s a risk involved. If you just give the calculator answer with no work, you will get either full credit for the “Do” step (if the interval is correct) or no credit (if it is wrong). If you opt for the calculator-only method, be sure to name the procedure (i.e., one-proportion z interval) and to give the interval (i.e., to 0.607).

AP Exam Common Error… Many students use the 1-PropZInt feature to correctly calculate the confidence interval and then try to ‘show their work’ with an incorrect formula. This will result in a loss of credit because the two attempts are considered ‘parallel solutions’ and students are scored on the worse response. If students want to include a formula in their response, they should make sure it produces the same results as the calculator. If the results aren’t the same, students should choose one of the answers and cross the other out.

AP Exam Common Error… Also, if students choose to include a formula, they should skip the symbolic formula and start with numbers substituted. In some cases, students choose the right formula and use the correct values, but include an incorrect symbol (i.e., p instead of 𝑝 or µ instead of 𝑥 ) and lose credit. Be careful with your symbols! Say what you mean and mean what you say.

Choosing the sample size…
Before a study is done (when is it in the planning stages), we may want to choose a sample size that allows us to estimate a population parameter within a given margin of error, while still maintaining a high confidence level. But we haven’t actually done the study yet, so we don’t have a 𝑝 yet… ME = z∗ 𝑝 (1− 𝑝 ) 𝑛

Choosing the sample size before the study is done…
But we haven’t actually done the study yet, so we don’t have a 𝑝 yet… Two options… (1) use a guess for 𝑝 based on past experience OR (2) use 𝑝 = 0.5 as the guess. This is a good, conservative estimate for 𝑝 . ME = z∗ 𝑝 (1− 𝑝 ) 𝑛

Practice... Choosing a sample size...
A company has received complaints about its customer service. They intend to hire a consultant to carry out a survey of customers. Before contacting the consultant, the company president wants some idea of the sample size that she will be required to pay for. One critical question is the degree of satisfaction with the company's customer service. The president wants to estimate the proportion p of customers who are satisfied. She decides that she wants the estimate to be within 3% (0.03) at a 95% confidence level.

Choosing a sample size... No idea of the true proportion p of satisfied customers; so use p* = 0.5. The sample size required is given by

Tattoos… Suppose that you want to estimate p, the true proportion of students at our school who have a tattoo with 95% confidence and a margin of error of no more than 0.10. Determine how many students should be surveyed to estimate p within 0.10 with 95% confidence. Because we don’t have any previous knowledge about the proportion of students with a tattoo, we will use 𝑝 = 0.5 to estimate the sample size needed.

Tattoos… (1−0.5) 𝑛 ≤ 0.10 n ≥ We need to survey at least 97 students to estimate the true proportion of students with a tattoo with 95% confidence and a margin of error at most 0.10.

AP Exam Formula Sheet… The specific formula for the margin of error we have used is not included on the formula sheet provided to students on the AP exam. However, the general formula for a confidence interval and the formula for the standard deviation of the statistic 𝑝 are included, so you should be able to put it together.

Homework… Page 507 #27, 29, 31, 33, 35, 37, 39, 43, 45, 47, 49, 50, 51, 52 And FRQ’s as needed Section Quiz on 8-2 tomorrow…

8.3: Estimating a Population Mean
Simulating Confidence Intervals for a Population parameter applet So why t vs. z for means? See directions in text pg 510

Estimating a Population Mean…
If we don’t know µ (that’s the whole reason why we are creating a confidence interval), then is it not reasonable for us to assume that we know σ For us to use Table A, or 𝑧= 𝑥−µ σ we must know σ. But that’s not reasonable when doing inference (confidence intervals). So what do we do? We use the t-distribution

Estimating a Population Mean…if we do not know σ …
t-distribution has a different (but similar) shape than the standard Normal curve Still symmetric, Still single-peaked Still same interpretation as any standardized statistic… it says how far 𝑥 is from it’s mean µ in standard deviation units BUT, has much more area in the tails… why?

t-distribution... Remember…σ unknown

t-distribution... Remember, σ unknown
Not Normal; but is similar in shape to Normal curve; Spread is greater than Normal curve; Has degrees of freedom (# of values that are allowed to vary) for each n; df = n – 1 (due to s in calculation); As degrees of freedom increase, t-distribution approaches Normal distribution But what are degrees of freedom (df)? It is single peaked but tails are fatter; talk about equations with 1, 2, 3 variables; how many can vary?; also discuss how table works (table c)

And… let’s look at Table B ….

Finding t* critical values…
What critical value t* from Table B should be used in constructing a confidence interval for the population mean in each of the following settings? A 95% confidence interval based on a SRS of size n = 12? A 90% confidence interval from a random sample of 30 observations? A 99% confidence interval based on a random sample of 95 observations?

Conditions for Estimating µ …
Very similar to CI’s for estimating p Random, 10%, Normal/Large Sample Random: Data is from well-designed random sample or randomized experiment (random assignment) 10% (Independence): When sampling without replacement (which is almost always the case), the population must be at least 10 times the sample size

Conditions for Estimating µ …
Normal/Large Sample: Three options, in this order… 1. If the population is Normal, then the condition is automatically met. 2. If the sample size is ≥ 30, then CLT kicks in and we have an approximately Normal population distribution 3. If population has an unknown shape and n < 30, we must use a graph of the sample data to assess the Normality of the population. T-procedures are fairly robust; just don’t use t-procedure if graph shows strong skewness or outliers. Goal with #3 is to answer the question, “is it reasonable to believe that these data came from a Normal population?”

If population has unknown shape and n < 30 …
… must use a graph of the sample data to assess the Normality of the population. T-procedures are fairly robust; just don’t use t-procedure if graph shows strong skewness or outliers Let’s use the calculator to simulate choosing random samples of size n = 20 from a Normal distribution with µ = 100 & σ = 15, then plot the data. Goal with #3 is to answer the question, “is it reasonable to believe that these data came from a Normal population?”

If population has unknown shape and n < 30 … n = 20, N (100, 15)
… must use a graph of the sample data to assess the Normality of the population. T-procedures are fairly robust; just don’t use t-procedure if graph shows strong skewness or outliers Math, Prb, randNorm (100, 15, 20) ► L1 enter Catalog F3 alpha 2 (R) to jump to the r’s, and choose randNorm (100, 15, 20) ► list1 enter Make a histogram, a boxplot, and a NPP of the data in L1 Do you see any obvious departures from Normality in the graphs of the sample data? Repeat above several times. Do the graphs of the sample data always look approximately Normal when the population distribution is Normal?

If population has unknown shape and n < 30 … n = 20, N (100, 15)
Repeat several times. Do the graphs of the sample data always look approximately Normal when the population distribution is Normal? Compare the results with those of your classmates. How easy do you think it will be to use a graph of sample data to determine whether or not a population has a Normal distribution? Note: On the AP exam, students will not be asked to construct a CI when the conditions for inference are clearly violated. If a student thinks that there is a problem with one of the conditions when asked to construct a CI, the student should state the concern(s) and proceed with calculating and interpreting the interval. In many cases the student has misremembered the condition and would lose out on the remaining points if they didn’t complete the problem. Do you see any obvious departures from Normality in the graphs of the sample data? Repeat above several times. Do the graphs of the sample data always look approximately Normal when the population distribution is Normal?

What’s the difference between ‘strongly skewed’ and ‘moderately skewed’?
Look at the stem plot. Compare maximum – median vs. median – minimum Now look at box plot. Compare maximum – median vs. median - minimum Stem plot: max – median = 336 – = 16.5; median – minimum = – 230 = 89.5… more than 5 times as long. Box plot: max – median 775 – 525 = 250 and median – minimum = 525 – 375 = 150… not quite twice as long.

What’s the difference between ‘strongly skewed’ and ‘moderately skewed’?
No accepted rule of thumb for identifying strong skewness. AP in general avoids boarderline cases. Stem plot: max – median = 336 – = 16.5; median – minimum = – 230 = 89.5… more than 5 times as long. Box plot: max – median 775 – 525 = 250 and median – minimum = 525 – 375 = 150… not quite twice as long.

CI’s… if we don’t know σ …
Standard deviation/standard error of the sampling distribution of 𝑥 is s/√n and its mean is μ (unbiased estimator 𝑥 ) General form of CI: estimate ± margin of error OR estimate ± (critical value) ( standard error) 𝑥 – (t*)(s/√n) and 𝑥 + (t*)(s/√n)

Practice… A manufacturer of high-resolution video terminals must control the tension on the mesh of fine wires that lies behind the surface of the viewing screen. Too much tension will tear the mesh and too little will allow wrinkles. The tension is measured by an electrical device with output readings in milli-volts (mV). Some variation is inherent in the production process. Here are the tension readings from an SRS of 20 screens from a single day's production:

n = 20; SRS Construct and interpret a 90% confidence interval for the mean tension μ of all the screens produced on this day. State: We want to estimate μ, the mean tension for all of screens produced on this day at a 90% confidence level.

n = 20; SRS; construct 90% CI Plan: Random. Data comes from a SRS, as stated in problem. 10%/Independence. We are sampling without replacement, so we need to assume that there are at least 10 (20) = 200 screens produced on that given day

n = 20; SRS; construct 90% CI Normal/Large Sample. We don’t know whether the population distribution of screens manufactured that day is Normal (if we did, then the condition is met) We can’t use Central Limit Theorem (CLT); n = 20 (need n ≥ 30 for CLT to apply and distribution be ≈ Normal) So we need to check Normality by creating a box plot, histogram, or a NPP. If there is no strong skewness or outliers, then we can assume the condition is met Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the populatin from which the sample came.

Input data into list and create box plot, histogram, or Normal Probability Plot

Checking Normality Condition
Boxplot shows no outliers, fairly symmetric; therefore population distribution is ≈ Normal NPP is fairly linear; only one possible outlier; therefore population distribution is ≈ Normal You will need to actually sketch the graph if FRQ. Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the populatin from which the sample came.

Constructing the CI (90% confidence level); n = 20
Do: CI = estimate ± (margin of error) = estimate ±(t*)(standard error) = 𝑥 ± (t*) (s/√n) How do we get our center and our SD? Look up our t*

𝑥 = & s =36.2 through 1-var stats df = 19 CI = estimate ± (margin of error) = estimate ±(t*)(standard error) = 𝑥 ± (t*) (s/√n) = ± (1.729) ( 36.2/√20) = ± 14.0 = (292.5, 320.3)

Conclude: We are 90% confident that the interval from mV to mV captures the true unknown population parameter, μ, of tensions of all video terminals that would be/were produced that day.

One-Sample t Confidence Interval
Environmentalists, government officials, and vehicle manufacturers are all interested in studying the auto exhaust emissions produced by motor vehicles. The major pollutants in auto exhaust from gasoline engines are hydrocarbons, monoxide, and nitrogen oxides (NOX). The following slide gives the NOX levels (in grams per mile) for a random sample of light-duty engines of the same type. Construct and interpret a 95% confidence interval for the mean amount of NOX emitted by all light-duty engines of this type.

Sample mean 1.329 grams/mile; s= 0.484 grams/mile; n = 46

95% CI; 1-sample t CI (if conditions are satisfied); 𝑥 = 1. 329; s = 0
95% CI; 1-sample t CI (if conditions are satisfied); 𝑥 = 1.329; s = 0.484; n = 46 State, Plan, Do, Conclude State: We want to estimate the true mean amount µ of NOX emitted by all light-duty engines of this type at a 95% confidence level

95% CI; 1-sample t CI (if conditions are satisfied); 𝑥 = 1.329; s = 0.484; n = 46 State, Plan, Do, Conclude Plan: Random. The data come from a random sample of 46 light-duty engines of the same type. 10%/Independence. Sampling without replacement, so we need to assume that there are at least 10 (46) = 460 light-duty engines of this type.

95% CI; 1-sample t CI (if conditions are satisfied); 𝑥 = 1.329; s = 0.484; n = 46 State, Plan, Do, Conclude Plan: Normal/Large Sample. Don’t know whether the population distribution of NOX emissions is Normal. But sample size is large (n = 46, n ≥ 30; CLT); so should be safe using a t-distribution. Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the populatin from which the sample came.

95% CI; 1-sample t CI (if conditions are satisfied); 𝑥 = 1.329; s = 0.484; n = 46 State, Plan, Do, Conclude Do: 𝑥 ± (t*) (s/√n); with df = 46 – 1 = 45 = ± (2.021)(0.484/√46) = ± 0.144 (1.185, 1.473)

95% CI; 1-sample t CI (if conditions are satisfied); 𝑥 = 1.329; s = 0.484; n = 46 State, Plan, Do, Conclude Conclude: We are 95% confident that the interval from to grams/mile captures the true unknown mean, µ, level of nitrogen oxides emitted by all light-duty engines of this type.

More Practice … The composition of the earth’s atmosphere may have changed over time. In an effort to discover the nature of the atmosphere long ago, we can examine the gas in bubbles trapped inside ancient amber. Amber is a tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber from the late Cretaceous era (75 to 95 million years ago) give these percents of nitrogen. 62.4 65 63.4 53.3 54.8 49.1 51 58.8 64.5

62.4 65 63.4 53.3 54.8 49.1 51 58.8 64.5 Practice... Assume that these observations are an SRS from the late Cretaceous atmosphere. Construct and interpret a 95% confidence interval for the mean percent of nitrogen in ancient air. State, Plan, Do, Conclude

State, Plan, Do, Conclude…
State: Want to estimate μ, the true, unknown population parameter of the percentage of nitrogen in ancient air Plan: Random. SRS (stated in problem) 10%/Independence. We are sampling without replacement, so we must assume that there exist at least 10 ( 9) hardened amber resin bubbles of this type Normal/Large Sample. Population Normal? Unknown. n ≥ 30? No. So we need to look at sample data; create boxplot, histogram, or NPP, sketch, comment on it; make sure there are no outliers or strong skewness in sample Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the populatin from which the sample came.

State, Plan, Do, Conclude…
𝑥 = s = df = 8 𝑥 ± (t*)(s/√n) ► ± (2.306)(6.1395/√9) ► (53.9, 62.7) Conclude: We are 95% confident that the interval from 53.9% to 62.7% captures the true, unknown population parameter mean percent of nitrogen in the atmosphere during the late Cretaceous era.

Milk’s favorite cookie…
For their second-semester project in AP Statistics, Ann & Tori wanted to estimate the average weight of an Oreo cookie to determine if the average weight was less than advertised. They selected a random sample of 36 cookies and found the weight of each cookie (in grams). The mean weight was 𝑥 = grams with a standard deviation of s = grams. Construct and interpret a 95% confidence interval for the mean weight of an Oreo cookie.

𝑥 = grams; s = grams; n = 36 We are 95% confident that the interval from grams to grams captures the true mean weight of any/all Oreo cookies. On the packaging, the stated serving size is 3 cookies (34 grams). Does the interval we just calculated provide convincing evidence that the average weight of an Oreo cookie is less than advertise? Why or why not? Answers and all work on page 519 of TE. Many students incorrectly state the normal/large sample condition by saying that the sample must have a normal distribution. This isn’t true (or even possible) in stead we use the sample data to make sure an inference about the shape of the populatin from which the sample came.

𝟗𝟓% 𝐂𝐈: (𝟏𝟏.𝟑𝟔𝟒𝟓, 𝟏𝟏.𝟒𝟏𝟗𝟕) On the packaging, the stated serving size is 3 cookies (34 grams). Does the interval we just calculated provide convincing evidence that the average weight of an Oreo cookie is less than advertise? Why or why not? The stated serving size is 34 grams for 3 cookies, or 34/3 = grams/cookie. Because all of the plausible values in the interval are greater than grams, there is no evidence that the average weight of an Oreo cookie is less than advertised. In fact, there is convincing evidence that the average weight is greater than advertised! Answers and all work on page 519 of TE

Tip… When do we use ‘z’ and when do we use ‘t’ ??
z is for proportions; t is for means ‘zap ta 𝑥 ’ z is for a proportion (ap); t is for a mean (a 𝑥 )

Tip… Many students incorrectly state the Normal/Large sample condition by saying that the sample must have a Normal distribution. This isn’t true (or even possible) instead we use the sample data to make an inference about the shape of the population from which the sample came.

Can you spare a square? As part of their final project in AP Statistics, Christina & Rachel randomly selected 18 rolls of a generic brand of toilet paper to measure how well this brand could absorb water. To do this, they poured ¼ cup of water onto a hard surface and counted how many squares of toilet paper it took to completely absorb the water. Here are the results from their 18 rolls: 29 20 25 21 24 27 28 26 22 23

99% CI; n = 18 Construct & interpret a 99% CI for µ = true mean number of squares of generic toilet paper needed to absorb ¼ cup of water. State, Plan, Do, Conclude 29 20 25 21 24 27 28 26 22 23

99% CI; n = 18 We are 99% confident that the interval from squares to squares captures the true mean number of squares of generic toilet paper needed to absorb ¼ cup of water. TIP: It is not enough just to make a graph of the data on your calculator when assessing Normality. You must sketch the graph on your paper & comment on it to receive credit. You do not have to draw multiple graphs, any appropriate graph will do. Full process on page 520 of TE

Should we use a t-procedure? Why or why not?
Discuss whether it would be appropriate to construct a t interval to estimate the population mean for: We want to estimate the average age at which U.S. presidents have died. So we obtain a list of all U.S. presidents who have died and their ages at death. Census; we have all the data;

Discuss whether it would be appropriate to construct a t interval to estimate the population mean for: We want to estimate the average age at which U.S. presidents have died. So we obtain a list of all U.S. presidents who have died and their ages at death. With data on all U.S. presidents, formal inference makes no sense. Census; we have all the data;

How much time do students spend on the Internet? We collect data from the 32 members of our AP Statistics class and calculate the mean amount of time that each student spent on the internet yesterday. If we had a SRS, n ≥ 30, so ≈ Normal, reasonable to assume independence, then ok to use.

How much time do students spend on the Internet? We collect data from the 32 members of our AP Statistics class and calculate the mean amount of time that each student spent on the internet yesterday. The 32 students in an AP Statistics class are not an SRS of all students, so the t-interval should not be used to make an inference about the mean amount of time all students spend on the internet. If we had a SRS, n ≥ 30, so ≈ Normal, reasonable to assume independence, then ok to use.

The stem plot below shows the force required to pull apart 20 pieces of randomly selected Douglas fir. If n ≥ 30, we could use t-procedure

CI vs. CL reminders … Many students confuse interpreting Confidence Levels with interpreting Confidence Intervals. Every time you construct a Confidence Interval, you are expected to interpret the interval you calculated. You are expected to interpret the Confidence Level only when you are specifically asked to do so. If you interpret the Confidence Level when you aren’t ask to, you won’t gain any credit and will lose credit if the interpretation is incorrect.

CI vs. CL reminders … A good interpretation of a Confidence Level will not include the endpoints of a specific interval. In fact, we can interpret a Confidence Level before the data are obtained because our confidence is in the method, not a particular interval. Interpretation of a Confidence Level: If we take many samples of the same size from this population, about C% of them will result in an interval that captures the actual parameter value.

Choosing the Sample Size…before we do the survey…
We can have both high confidence and a small margin of error at the same time… by taking enough observations… large enough n ME = t*( 𝑠 𝑛 ) But we don’t know the sample SD (s) because we haven’t done the survey yet/haven’t produced the data yet t* (our critical value) depends on n, our sample size

MOE = t*( 𝑠 𝑛 ) One alternative is to come up with a reasonable estimate for the population standard deviation, σ. If we use σ, then we use z* for our critical value, so… z* ( σ 𝑛 )≤ MOE and then we can solve for n

Example... High CI & Small MOE
Researchers would like to estimate the mean cholesterol level μ of a particular variety of monkey that is often used in laboratory experiments. They would like their estimate to be within 1 milligram per deciliter of blood (mg/dl) of the true value of μ at a 95% confidence level. It is (unrealistically) known that the standard deviation of cholesterol level is about σ = 5 mg/dl. Obtaining monkeys is time-consuming and expensive, so the researchers want to know the minimum number of monkeys they will need to generate a satisfactory estimate.

95% Confidence Level; MOE ≤ 1 mg/dl; σ = 5 mg/dl Need to determine minimum number of monkeys needed to achieve above Want high confidence level & small MOE (or interval) (z*)(σ/√n) = MOE So willing to increase n, can have both high confidence level and small interval; but expensive, realistic?

95% Confidence Level; MOE ≤ 1 mg/dl; σ = 5 mg/dl Need to determine minimum number of monkeys needed to achieve above Want high confidence level & small MOE (or interval) (z*)(σ/√n) = MOE (1.96)(5/√n) ≤ 1 n ≥  n ≥ 97 (why?) So willing to increase n, can have both high confidence level and small interval; but expensive, realistic?

So, if you want (need) high confidence level AND small(er) interval (margin of error), it is possible if you are willing to increase n Can be expensive, time-consuming Sometimes not realistic (why?) In reality, you may need to compromise on the confidence level (lower confidence level) So willing to increase n, can have both high confidence level and small interval; but expensive, realistic?

Sample Size for Desired MOE...
(z*)(σ/√n) ≤ m Note: sample size determines the MOE (not population size)

Practice... Texting at school… A counselor wants to estimate the average number of text messages sent by students at his school during school hours. He wants to estimate µ at the 99% confidence level with a margin of error of at most 2 texts. A pilot study indicated that the number of texts sent during school hours has a standard deviation of about 9 texts. How many students need to be surveyed to estimate the mean number of texts sent during school hours with 99% confidence and a margin of error of at most 2 texts? yes

99% CL; pilot study SD = 9 texts; MOE within (at most) 2 texts
2.576 ( 9 𝑛 ) ≤ 2 ► n ≥ The counselor needs to survey at least 135 students.

Case Closed, if time permits.. M&M’s activity if time permits

Homework... Page 527 # 55, 57, 59, 61, 63, 65, 67, 69, 71 All MC; #75 – 78 FRQs as needed, chapter review exercises, AP Stats practice test Our section quiz will be…. Our chapter test will be ….

Chapter 8: Estimating with Confidence

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