1 Chapter 7: Estimates & Sample Sizes 7 – 2: Estimating a Population Proportion Suppose you were asked to estimate the number of marbles in a 1 gallon jar. Suppose also that if you guessed the right amount you would win a substantial monetary prize. Suppose further that you were given the choice of guessing in two ways. Way #1 is to guess the exact number of marbles. Way #2 is to guess a range of numbers that you believe contains the exact number of marbles. The two methods of guessing represent the two kinds of estimates we could use when we are attempting to determine the value of a population parameter, but can’t take a census of the entire population. Way # 1 represents a____________ _. We take a well chosen sample from the population and calculate the mean, or proportion, or standard deviation, or variance or whatever statistic we are looking for from that sample and use that single number as an estimate for the population parameter. Way # 2 represents a ________________________. From a statistic calculated from a well chosen sample we construct an interval that we are reasonably confident contains the true population parameter we seek. A ________estimate is a single value (or point) used to approximate a population parameter A __________ interval, CI, is a range (or an interval) of values used to estimate the true value of a population parameter. Notation →
Population proportion: _________
Sample proportion :__________
n is the sample size and x is the number in the sample with the characteristic being studied. Point Estimate: The __________
_______________is the best point estimate of the population proportion.
Example 1: Find the point estimate for the population proportion in the following situations. a. In a telephone survey of 1,243 individuals regarding house siding preferences, 65 of them replied that they preferred aluminum siding. What is the point estimate for the population proportion of individuals who prefer aluminum siding?
b. An important issue facing Americans is the large number of medical malpractice lawsuits and the expenses that they generate. In a study of 1228 randomly selected medical malpractice lawsuits, it is found that 856 of them were later dropped or dismissed. What is the best point estimate of the proportion of medical malpractice lawsuits that are dropped or dismissed? p-hat = What about point estimate for proportion that are not dropped or dismissed?
2 Confidence Interval (as an estimate for the population proportion): A point estimate is not the only estimate we can make for a population proportion. We can also take the sample proportion and use it to build an interval, a range, in which we believe the true population proportion will lie. This is called a confidence interval or interval estimate or confidence interval estimate. We use 1) the sample proportion, 2) the sample size, 3) qɵ , 4) the 1-α confidence level, and 5) the critical value for α to build this range.
Confidence Interval for Estimating a population proportion, p Important Notation
p=
zα =
n=
pˆ =
2
E = zα / 2
=
p qɵ n
α is a value, like 0.05 or 0.10 and when we calculate 1 – α = _Confidence Level__ (Degree of Confidence) (or Confidence Coefficient) . *For example a confidence level of 0.95 (for α = 0.05) states that “we are 95% confident that the confidence interval ___________________________________ of the population proportion.” *Therefore the percentage 1 – α = How much confidence we have that this interval contains the true population parameter. Common values of C.L. are 90%, 95%, 99%. Once we set our confidence level, we can calculate the critical values (z-scores corresponding for our interval estimate. __Critical Values___: ± Z α / 2 A critical value is the number separating sample statistics that are likely to occur from those that are unlikely to occur. (Use invNorm function on the TI-84)
Confidence Level 95%
α 0.10 0.05 0.01
α/2
Critical Value, zα/2
0.025
1.96
Concept Check: Let us consider our marbles in the jar situation again. Is it easier to get the correct answer if I just guess the exact number of marbles or if I get to give a range of values in which I think the actual number of marbles occurs?
3
Finding the Confidence Interval The formula for the error estimate (also called the margin of error) is: Where zα is the critical value in the normal distribution and 2
E = zα / 2
pqɵ n
pqɵ is a form of the standard deviation n
for the proportion. E = the ___________________________ (with probability 1 – α) between the _________ proportion and the _________ of the population proportion p. Confidence Interval notations: OR
OR
( p – E , p + E)
Practice Find the 95% confidence interval for actual proportion of students who prefer jelly beans given that a sample of 14 students who are talking about their favorite candy and 5 of them prefer jelly beans. 5 ≈ .357 = 35.7% . (Any one of these a. The sample proportion, p , of students who like jelly beans is 14 three forms, fraction, decimal, or percent may be used to represent the proportion.) b. Find the 95% critical values for the normal distribution ( ± zα ) and multiply the values by 2
four decimal places. E = (1.96)
p-hat =
(.3571)(.6429) = 14
q-hat = 1 – _____ =
pqɵ to n
n =
Your answer for the error estimate should be ± . ☺
b. Add and subtract the error estimate of .2510 to our point estimate of 0.3571.
c. These two numbers represent the low value and high value of the 95% confidence interval of the population proportion of people who prefer jelly beans.
How do we interpret the confidence interval? We interpret this by saying… We are _______confident that our confidence interval from ______ to ______ contains the actual proportion of people who __ _____________________________.
4 Example 2: Find a confidence interval to estimate the population proportion in the following situations. In a telephone survey of 1,243 individuals regarding house siding preferences, 65 of them replied that they preferred aluminum siding. Find the 98% confident interval for the population proportion of individuals who prefer aluminum siding.
Write the interpretation:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ CA - A. An important issue facing Americans is the large number of medical malpractice lawsuits and the expenses that they generate. In a study of 1228 randomly selected medical malpractice lawsuits, it is found that 856 of them were later dropped or dismissed. Find the 95% Confidence interval of the proportion of medical malpractice lawsuits that are dropped or dismissed. Write the interpretation.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Finding point estimate for p, given a confidence interval: The point estimate for the proportion is in the middle of the interval, so you need to find the middle. Example 3) We are 98% confident that the true population proportion of weights that are above healthy is between (.35 , 0.41). What is the point estimate for p? What is the error?
****************************************************************************************************************** Requirements for Proportion Confidence Interval (CI) *** BE SURE TO CHECK BEFORE CALCULATING*** Otherwise your CI is USELESS 1) The sample is a simple random sample. (All samples of the same size have an equal chance of being selected.) 2) The conditions for the __________distribution are satisfied. 3) There are at least __________and ____________ Rounding: these are proportions, so you round as you would a proportion ….. ******************************************************************************************************************
5 Example 4: In the Chapter Problem we note that a Pew Research Center poll of 1501 randomly selected U.S. adults showed that 70% of the respondents believe in global warming. Find the 95% confidence interval estimate of the population proportion p and interpret the interval. Find population proportion of those who believe in global warming. Step 1) Describe the population parameter of interest. “P-hat represent the proportion of people who _____________________________________“ Step 2) Identify the probability distribution: _________ Step 3) Determine the level of confidence: _______ Step 4) Collect the sample evidence. x=
n=
=
qˆ =
α=
1–α=
5) with the TI graphing calculator. Press STAT TESTS, and down to “1-PropZInt” Where x, n are from p-hat = x/n . CL = 1-α 6) Find the lower and upper confidence limits.
7) Interpret the confidence interval: We are _____ confident that our confidence interval from ______ to _______ contains the _________________________ of people who believe in global warming. 8) Use limits to calculate E:
E = upper limit – p-hat
E=
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ CA - B: In July of 2008, a Quinnipiac University Poll asked 1783 registered voters nationwide whether they favored or opposed the death penalty for persons convicted of murder. 1123 were in favor. Obtain a 90% confidence interval for the proportion of registered voters nationwide who are in favor of the death penalty for persons convicted of murder. Step 1) Describe the population parameter of interest. Step 2) Identify the probability distribution:______________________ Step 3) Determine the level of confidence: ____ Step 4) Collect the sample evidence. qˆ = x= n= = α= 1–α= 7) Interpret the confidence interval: We are _________________ that the _________________________________________________ _______________________is between __________ and ________. 8) Use limits to calculate E (when needed):
6 CA - C: In a telephone survey of 1,243 individuals regarding house siding preferences, 65 of them replied that they preferred aluminum siding. Find the 95% confidence interval for the population proportion of individuals who prefer aluminum siding and interpret the interval. Perform same Steps 1 – 8 as above problems
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Sample size for estimating population proportion p Suppose we want to collect sample data in order to estimate some population proportion. question is how many sample items must be obtained?
If
2 z pˆ qˆ α2 is known: n = E2
If
is NOT known let
= 0.5
The
2 z (0.5)2 α 2 n= E2
Important: If the computed sample size n is not a whole number, round the value of n up to the next larger whole number. Example 6: The Internet is affecting us all in many different ways, so there are many reasons for estimating the proportion of adults who use it. Assume that a manager for E-Bay wants to determine the current percentage of U.S. adults who now use the Internet. How many adults must be surveyed in order to be 95% confident that the sample percentage is in error by no more than three percentage points? a. Use 73% as the sample proportion.
Interpretation of calculation for sample needed: To be ______ confident that our sample _______________is within ______ __________points of the true ______________ for all adults, we should obtain a simple random sample of ______ adults. b. No known possible value of the proportion.
Interpretation of calculation for sample needed:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ CA D: A sociologist wanted to determine the percentage of residents of America that only speak English at home. What size sample should be obtained if she wishes her estimate to be within 3 percentage points with 90% confidence assuming she uses the estimate of 82.4% obtained from the Census 2000 Supplementary Survey? Interpret your solution.
7 7 – 3: Estimating a population Mean: σ known ** NOTE we are doing more CI’s but this time for a mean…..*** The _______________is the best point estimate of the population mean.
Requirements: 1. The sample is a simple random sample. (All samples of the same size have an equal chance of being selected.) 2. The value of σ is known. 3. The population is normal or n > 30. (Or both) Confidence Interval (C.I.) for
µ (population mean)
x − E < µ < x + E ,where E = zα
σ
2
n
Example: 7-3 problem #22
In real–world problems,
σ is almost never a known quantity. Therefore, the standard error will almost
s . The use of an estimated standard error of the mean will require the n use of the Student’s t – distribution. Almost all real-world inferences about the population mean will be completed using the Student’s t – statistics. always be estimated using
Watch video: https://www.youtube.com/watch?v=s4SRdaTycaw – CI for mean using formula ******************** 7–3: Estimating a population Mean:
σ Unknown
**************************
Requirements: 1. The sample is a simple random sample. (All samples of the same size have an equal chance of being selected.) 2. The value of σ is NOT known. 3. The population is normal or n > 30. (Or both) Confidence Interval (C.I.) for
µ (population mean)
x − E < µ < x + E ,where E = tα
2
s n
8
Student t distribution: If the distribution of a population is approximately normal, then the distribution of t is called a Student t Distribution. Similarities to Normal Distribution 1. Symmetric (bell shaped) about the mean 2. Tails extend along the horizontal axis to infinity. 3. Area under the curve =1 4. Mean: µ = 0
Differences 1. t has lower height and wider spread. 2. d f = ( n − 1 ) degrees of freedom.
The number of degrees of freedom: df, is defined as the number of observations, for a collection of sample data that can be chosen freely. If the value of the mean is known for n sample data, how many values can be assigned freely?
*** Some calculators will not have invT, so you will need to use the “A-3 t-distribution chart” By Calculator: To find tα/2 , use [2nd] [Dist] 4: invT ( area, df)
By Chart:
Finding Critical value t: Example 8: Find the critical value t, where the degrees of freedom is 17 and the middle area under the curve is 0.90.
Example 9: Find the critical value t where the degrees of freedom is 19 and the middle area under the curve is 0.95.
9 Finding Confidence Interval for an estimate of the mean (σ unknown): Example 10: A common claim is that garlic lowers cholesterol levels. In a test of the effectiveness of garlic, 49 subjects were treated with doses of raw garlic, and their cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 0.4 and a standard deviation of 21.0. Use the sample statistics of n = 49, = 0.4 and s = 21.0 to construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Step 1) Describe the population parameter of interest.
Step 2) Identify the probability distribution and the degrees of freedom. Step 3) Determine the level of confidence. Step 4) Collect the sample evidence. n=
x
=
s=
Step 5) * USe TI graphing calculator. Press STAT TESTS, and down to “TInterval” Highlight “Stats”
Step 6) Find the lower and upper confidence limits.
Step 7) Find the margin of errors (E).
Step 8) Interpret the confidence interval. We are 95% confident that the actual value of the average change in LDL’s of those who took raw doses of garlic is in the interval (-5.6, 6.4)
10 Example 11: A survey of 3000 randomly selected Minnesotans aged 65 and older revealed that, on average, they spent $85.00 per month on prescription drugs, with a standard deviation of $50.35 per month. Construct a 99% confidence interval for the true mean amount spent per month using the procedure detailed above. Step 1) Describe the population parameter of interest.
Step 2) Identify the probability distribution and the degrees of freedom. Step 3) Determine the level of confidence. Step 4) Collect the sample evidence. n=
x
=
s=
$
Step 5) * USe TI graphing calculator. Press STAT TESTS, and down to “TInterval” Highlight “Stats”
Step 6) Find the lower and upper confidence limits.
Step 8) Interpret the confidence interval:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ CA E: The heights of adult men are normally distributed. The heights (in inches) of 9 young adult males are given: 72, 71, 71, 68, 68.75, 70.25, 72, 70.25, 68.25. Construct a 95% confidence interval for the true mean height (in inches) of young adult men. Interpret the interval. (Perform same steps 1 – 8 as above problems.)
CA F: A random sample of size 20 is taken from the weights of babies born at Northside Hospital during the year 1994. A mean of 6.87 lb and a standard deviation of 1.76 lb were found for the sample. Construct a 95% confidence interval of the mean weight of all babies born in this hospital in 1994. Based on past information, it is assumed that weights of newborns are normally distributed.
CA G: Write down a summary of what you learned about confidence intervals. How can you tell which one to use? What are key words or concepts you are looking for when making the decision? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
