In hypothesis tests, the purpose was to make a decision about a parameter, in terms of it being greater than, less than, or not equal to a value. But what if you want to actually know what the parameter is. You need to do estimation. There are two types of estimation -- point estimator and confidence interval. The American Statistical Association (ASA) is recommending that confidence intervals are the process that should be followed when analyzing data.
A point estimator is just the statistic that you have calculated previously. As an example, when you wanted to estimate the population mean, \(\mu\), the point estimator is the sample mean, \(\bar{x}\). To estimate the population proportion, \(p\), you use the sample proportion, \(\hat{p}\). In general, if you want to estimate any population parameter, we will call it \(\theta\), you use the sample statistic, \(\hat{\theta}\).
Point estimators are really easy to find, but they have some drawbacks. First, if you have a large sample size, then the estimate is better. But with a point estimator, you don’t know what the sample size is. Also, you don’t know how accurate the estimate is. Both of these problems are solved with a confidence interval.
Confidence interval: This is where you have an interval surrounding your parameter, and the interval has a chance of being a true statement. In general, a confidence interval looks like: \(\hat{\theta}\pm E\), where \(\hat{\theta}\) is the point estimator and \(E\) is the margin of error term that is added and subtracted from the point estimator. Thus making an interval.
The statistical interpretation is that the confidence interval has a probability \(C=(1-\alpha)\) (where \(\alpha\) is the complement of the confidence level) of containing the population parameter. As an example, if you have a 95% confidence interval of $0.65 < p < 0.73$, then you would say, “you are 95% confident that the interval 0.65 to 0.73 contains the true population proportion.” This means that if you have 100 intervals, 95 of them will contain the true proportion, and 5 will not. The wrong interpretation is that there is a 95% confidence that the true value of \(p\) will fall between 0.65 and 0.73. The reason that this interpretation is wrong is that the true value is fixed out there somewhere. You are trying to capture it with this interval. So this is the chance that your interval captures it, and not that the true value falls in the interval.
There is also a real world interpretation that depends on the situation. It is where you are telling people what numbers you found the parameter to lie between. So your real world is where you tell what values your parameter is between. There is no probability attached to this statement. That probability is in the statistical interpretation.
The common probabilities used for confidence intervals are 90%, 95%, and 99%. These are known as the confidence level. The confidence level and the alpha level are related. If you are conducting a hypothesis test with \(H_a:\mu\ne \mu_o\), then the confidence level is \(C=1-\alpha\). This is because the \(\alpha\) is both tails and the confidence level is area between the two tails. As an example, for a hypothesis test \(H_a:\mu\ne \mu_o\) with \(\alpha\) equal to 0.05, the confidence level would be 0.95 or 95%. If you have a hypothesis test with \(H_a:\mu<\mu_o\), then your \(\alpha\) is only one tail of the curve. Because of symmetry the other tail is also \(\alpha\). You have \(2\alpha\) with both tails. So the confidence level, which is the area between the two tails, is \(C-2\alpha\).
Statistical Interpretation: You are 95% confident that the interval contains the mean age in 2013 that a woman gets married.
Real World Interpretation: The mean age that a woman married in 2013 is between 26 and 28 years of age.
Statistical Interpretation: You are 99% confident that the interval contains the proportion of Americans who have tried marijuana as of 2013.
Real World Interpretation: The proportion of Americans who have tried marijuana as of 2013 is between 0.35 and 0.41.
One last thing to know about confidence is how the sample size and confidence level affect how wide the interval is. The following discussion demonstrates what happens to the width of the interval as you get more confident.
Think about shooting an arrow into the target. Suppose you are really good at that and that you have a 90% chance of hitting the bull’s eye. Now the bull’s eye is very small. Since you hit the bull’s eye approximately 90% of the time, then you probably hit inside the next ring out 95% of the time. You have a better chance of doing this, but the circle is bigger. You probably have a 99% chance of hitting the target, but that is a much bigger circle to hit. You can see, as your confidence in hitting the target increases, the circle you hit gets bigger. The same is true for confidence intervals. This is demonstrated in Image \#8.1.1.
The higher level of confidence makes a wider interval. There’s a trade off between width and confidence level. You can be really confident about your answer but your answer will not be very precise. Or you can have a precise answer (small margin of error) but not be very confident about your answer.
Now look at how the sample size affects the size of the interval. Suppose Image \#8.1.2 represents confidence intervals calculated on a 95% interval. A larger sample size from a representative sample makes the width of the interval narrower. This makes sense. Large samples are closer to the true population so the point estimate is pretty close to the true value.
Now you know everything you need to know about confidence intervals except for the actual formula. The formula depends on which parameter you are trying to estimate. With different situations you will be given the confidence interval for that parameter.
Suppose you compute a confidence interval with a sample size of 25. What will happen to the confidence interval if the sample size increases to 50?
Suppose you compute a 95% confidence interval. What will happen to the confidence interval if you increase the confidence level to 99%?
Suppose you compute a 95% confidence interval. What will happen to the confidence interval if you decrease the confidence level to 90%?
Suppose you compute a confidence interval with a sample size of 100. What will happen to the confidence interval if the sample size decreases to 80?
A 95% confidence interval is \(6353km< \mu<6384km\), where \(\mu\) is the mean diameter of the Earth. State the statistical interpretation.
A 95% confidence interval is \(6353 km < \mu < 6384 km\), where \(\mu\) is the mean diameter of the Earth. State the real world interpretation.
In 2013, Gallup conducted a poll and found a 95% confidence interval of $0.52 < p < 0.60$, where p is the proportion of Americans who believe it is the government’s responsibility for health care. Give the real world interpretation.
In 2013, Gallup conducted a poll and found a 95% confidence interval of $0.52 < p < 0.60$, where p is the proportion of Americans who believe it is the government’s responsibility for health care. Give the statistical interpretation.
Suppose you want to estimate the population proportion, p. As an example you may be curious what proportion of students at your school smoke. Or you could wonder what is the proportion of accidents caused by teenage drivers who do not have a drivers’ education class.
\(x\) = number of successes
\(p\) = proportion of successes
State: A simple random sample of size \(n\) is taken. Check: describe how sample was taken.
State: The condition for the binomial distribution are satisfied. Check: argue that each condition has been met.
State: The sampling distribution of \(\hat{p}\) can be approximated by a normal distributed. check: To determine the sampling distribution of \(\hat{p}\) is normally distributed, you need to show that \(n*\hat{p}\ge5\) and , \(n*\hat{q}\ge5\) where \(\hat{q}=1-\hat{p}\). If this requirement is true, then the sampling distribution of \(\hat{p}\) is well approximated by a normal curve. (In reality this is not really true, since the correct condition deals with \(p\). However, in a confidence interval you do not know \(p\), so you must use \(\hat{p}\).)
This will be conducted using rStudio. The command is
prop.test(r, n, conf.level=C) #type C as a decimal
Statistical Interpretation: In general this looks like, “you are C% confident that \(\hat{p}\pm E\) contains the true proportion.”
Real World Interpretation: This is where you state what interval contains the true proportion.
A concern was raised in Australia that the percentage of deaths of Aboriginal prisoners was higher than the percent of deaths of non-Aboriginal prisoners, which is 0.27%. A sample of six years (1990-1995) of data was collected, and it was found that out of 14,495 Aboriginal prisoners, 51 died (\“Indigenous deaths in,\” 1996). Find a 95% confidence interval for the proportion of Aboriginal prisoners who died.
\(x\) = number of Aboriginal prisoners who die
\(p\) = proportion of Aboriginal prisoners who die
State: A simple random sample of 14,495 Aboriginal prisoners was taken. Check: The sample was not a random sample, since it was data from six years. It is the numbers for all prisoners in these six years, but the six years were not picked at random. Unless there was something special about the six years that were chosen, the sample is probably a representative sample. This condition is probably met.
State: The properties of the binomial experiment have been met. Check: There are 14,495 prisoners in this case. The prisoners are all Aboriginals, so you are not mixing Aboriginal with non-Aboriginal prisoners. There are only two outcomes, either the prisoner dies or doesn’t. The chance that one prisoner dies over another may not be constant, but if you consider all prisoners the same, then it may be close to the same probability. Thus the properties of the binomial experiment are satisfied
State: The sampling distribution of \(\hat{p}\) can be approximated with a normal distribution. Check: \(\hat{p}*n=\frac{51}{14495}*14495=51\ge5\) and \(\hat{q}*n=\frac{14495-51}{14495}*14495=14444\ge5\). The sampling distribution of \(\hat{p}\) can be approximated with a normal distribution.
The command in r Studio for a confidence interval for a proportion is
prop.test(51,14495, conf.level = 0.95)
1-sample proportions test with continuity correction
data: 51 out of 14495
X-squared = 14290, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.002647440 0.004661881
sample estimates:
p
0.003518455 the 95% confidence level is \(0.002647440<p<0.004661881\).
Statistical Interpretation: You are 95% confident that the interval \(0.0026<p<0.0047\) contains the proportion of Aboriginal prisoners who have died in prison.
Real World Interpretation: The proportion of Aboriginal prisoners who died in prison is between 0.26% and 0.47%.
A researcher who is studying the effects of income levels on breastfeeding of infants hypothesizes that countries with a low income level have a different rate of infant breastfeeding than higher income countries. It is known that in Germany, considered a high-income country by the World Bank, 22% of all babies are breastfeed. In Tajikistan, considered a low-income country by the World Bank, researchers found that in a random sample of 500 new mothers that 125 were breastfeeding their infant. Find a 90% confidence interval of the proportion of mothers in low-income countries who breastfeed their infants?
\(x\) = number of woman who breastfeed in a low-income country
\(p\) = proportion of woman who breastfeed in a low-income country
State: A simple random sample of 500 breastfeeding habits of woman in a low-income country was taken. Check: This was stated in the problem.
State: The properties of a Binomial Experiment have been met. Check: There were 500 women in the study. The women are considered identical, though they probably have some differences. There are only two outcomes, either the woman breastfeeds or she doesn’t. The probability of a woman breastfeeding is probably not the same for each woman, but it is probably not very different for each woman. The conditions for the binomial distribution are satisfied
State: The sampling distribution of \(\hat{p}\) can be approximated with a normal distributed. Check:\(n*\hat{p}= 500*\frac{125}{500}=125\ge5\) and \(n*\hat{q}=500*\frac{500-125}{500}=375\ge5\), so the sampling distribution of \(\hat{p}\) is well approximated by a normal distribution.
On rstudio, use the following command
prop.test(125, 500, conf.level = .90)
1-sample proportions test with continuity correction
data: 125 out of 500
X-squared = 124, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
90 percent confidence interval:
0.2185980 0.2841772
sample estimates:
p
0.25 90% confidence interval for \(p\) is \(0.2185980<p<0.2841772\).
Statistical Interpretation: You are 90% confident that \(0.2185980<p<0.2841772\) contains the proportion of women in low-income countries who breastfeed their infants.
Real World Interpretation: The proportion of women in low-income countries who breastfeed their infants is between 0.219 and 0.284.
In each problem show all steps of the confidence interval. If some of the conditions are not met, note that the results of the interval may not be correct and then continue the process of the confidence interval.
The Arizona Republic/Morrison/Cronkite News poll published on Monday, October 20, 2016, found 390 of the registered voters surveyed favor Proposition 205, which would legalize marijuana for adults. The statewide telephone poll surveyed 779 registered voters between Oct. 10 and Oct. 15. (Sanchez, 2016) Find a 99% confidence interval for the proportion of Arizona’s who supported legalizing marijuana for adults.
In November of 1997, Australians were asked if they thought unemployment would increase. At that time 284 out of 631 said that they thought unemployment would increase (\“Morgan gallup poll,\” 2013). Estimate the proportion of Australians in November 1997 who believed unemployment would increase using a 95% confidence interval?
According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, Arkansas had 1,601 complaints of identity theft out of 3,482 consumer complaints (\“Consumer fraud and,\” 2008). Calculate a 90% confidence interval for the proportion of identity theft in Arkansas.
According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints (\“Consumer fraud and,\” 2008). Calculate a 90% confidence interval for the proportion of identity theft in Alaska.
In 2013, the Gallup poll asked 1,039 American adults if they believe there was a conspiracy in the assassination of President Kennedy, and found that 634 believe there was a conspiracy (\“Gallup news service,\” 2013). Estimate the proportion of American’s who believe in this conspiracy using a 98% confidence interval.
In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) (\“Autism and developmental,\” 2008). Find the proportion of ASD in Arizona with a confidence level of 99%.
Suppose you want to estimate the mean height of Americans, or you want to estimate the mean salary of college graduates. A confidence interval for the mean would be the way to estimate these means.
\(x\) = random variable
\(\mu\) = mean of random variable
State: A random sample of size \(n\) is taken. Check: describe how the sample was collected.
State: The population of the random variable is normally distributed. Check: look at density plot and normal quantile plot. Note: though the t-test is fairly robust to the condition if the sample size is large. This means that if this condition isn’t met, but your sample size is quite large, then the results of the t-test are valid.
Use rStudio to find the confidence interval. The command is
t.test(~variable, data= Data_Frame, conf.level=C) #type C as a decimal
Statistical Interpretation: In general this looks like, “You are C% confident that the interval contains the true mean.”
Real World Interpretation: This is where you state what interval contains the true mean.
A random sample of 50 body mass index (BMI) were taken from the NHANES Data frame Table 8.1. Estimate the mean BMI of Americans at the 95% level.
sample_NHANES_50<- sample_n(NHANES, size=50)
knitr::kable(head(sample_NHANES_50))| ID | SurveyYr | Gender | Age | AgeDecade | AgeMonths | Race1 | Race3 | Education | MaritalStatus | HHIncome | HHIncomeMid | Poverty | HomeRooms | HomeOwn | Work | Weight | Length | HeadCirc | Height | BMI | BMICatUnder20yrs | BMI_WHO | Pulse | BPSysAve | BPDiaAve | BPSys1 | BPDia1 | BPSys2 | BPDia2 | BPSys3 | BPDia3 | Testosterone | DirectChol | TotChol | UrineVol1 | UrineFlow1 | UrineVol2 | UrineFlow2 | Diabetes | DiabetesAge | HealthGen | DaysPhysHlthBad | DaysMentHlthBad | LittleInterest | Depressed | nPregnancies | nBabies | Age1stBaby | SleepHrsNight | SleepTrouble | PhysActive | PhysActiveDays | TVHrsDay | CompHrsDay | TVHrsDayChild | CompHrsDayChild | Alcohol12PlusYr | AlcoholDay | AlcoholYear | SmokeNow | Smoke100 | Smoke100n | SmokeAge | Marijuana | AgeFirstMarij | RegularMarij | AgeRegMarij | HardDrugs | SexEver | SexAge | SexNumPartnLife | SexNumPartYear | SameSex | SexOrientation | PregnantNow |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 57763 | 2009_10 | female | 31 | 30-39 | 375 | White | NA | High School | Married | 35000-44999 | 40000 | 1.36 | 5 | Own | Working | 97.0 | NA | NA | 174.9 | 31.71 | NA | 30.0_plus | 92 | 125 | 74 | 118 | 76 | 126 | 80 | 124 | 68 | NA | 0.93 | 4.91 | 81 | 0.206 | NA | NA | No | NA | Good | 0 | 10 | None | None | 1 | 1 | NA | 9 | No | No | NA | NA | NA | NA | NA | No | NA | 0 | Yes | Yes | Smoker | 13 | Yes | 13 | No | NA | No | Yes | 17 | 6 | 1 | No | Heterosexual | No |
| 68588 | 2011_12 | female | 14 | 10-19 | NA | Mexican | Mexican | NA | NA | 20000-24999 | 22500 | 1.03 | 5 | Own | NA | 48.8 | NA | NA | 159.3 | 19.20 | NormWeight | 18.5_to_24.9 | 78 | 106 | 62 | 98 | 62 | 108 | 54 | 104 | 70 | 23.55 | 1.42 | 3.47 | 103 | 0.534 | NA | NA | No | NA | Good | 0 | 0 | NA | NA | NA | NA | NA | NA | NA | Yes | NA | 1_hr | 0_to_1_hr | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA |
| 65400 | 2011_12 | male | 16 | 10-19 | NA | Mexican | Mexican | NA | NA | 35000-44999 | 40000 | 1.41 | 7 | Rent | NotWorking | 94.3 | NA | NA | 152.7 | 40.40 | Obese | 30.0_plus | 72 | 114 | 70 | NA | NA | 114 | 70 | NA | NA | 307.00 | 0.98 | 4.86 | 369 | 1.277 | NA | NA | No | NA | Good | 5 | 0 | NA | NA | NA | NA | NA | 10 | No | Yes | 7 | 3_hr | 0_to_1_hr | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA |
| 66416 | 2011_12 | female | 63 | 60-69 | NA | White | White | High School | Married | 45000-54999 | 50000 | 3.30 | 6 | Own | NotWorking | 48.1 | NA | NA | 152.9 | 20.60 | NA | 18.5_to_24.9 | 66 | 117 | 76 | 118 | 74 | 118 | 76 | 116 | 76 | 15.04 | 2.09 | 6.62 | 211 | NA | NA | NA | No | NA | Good | 0 | 0 | None | None | 1 | 1 | NA | 8 | Yes | Yes | NA | 0_hrs | 0_hrs | NA | NA | Yes | 2 | 24 | NA | No | Non-Smoker | NA | NA | NA | NA | NA | Yes | Yes | 20 | 1 | NA | No | NA | NA |
| 61650 | 2009_10 | male | 80 | NA | NA | White | NA | High School | Married | 35000-44999 | 40000 | 2.75 | 7 | Own | NotWorking | 106.3 | NA | NA | 181.6 | 32.23 | NA | 30.0_plus | 58 | 119 | 58 | 116 | 62 | 122 | 58 | 116 | 58 | NA | 0.85 | 4.06 | 56 | 1.143 | NA | NA | No | NA | Good | 1 | 0 | Most | None | NA | NA | NA | 10 | No | No | NA | NA | NA | NA | NA | Yes | 1 | 364 | No | Yes | Smoker | 18 | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA |
| 67326 | 2011_12 | female | 47 | 40-49 | NA | Black | Black | Some College | NeverMarried | 20000-24999 | 22500 | 0.76 | 6 | Rent | NotWorking | 150.1 | NA | NA | 168.6 | 52.80 | NA | 30.0_plus | 96 | 116 | 86 | 124 | 82 | 118 | 86 | 114 | 86 | 24.08 | 1.14 | 6.44 | 82 | 0.812 | NA | NA | No | NA | Fair | 30 | 0 | Most | None | 4 | 2 | 26 | 6 | Yes | No | 3 | More_4_hr | More_4_hr | NA | NA | Yes | 4 | 20 | Yes | Yes | Smoker | 21 | Yes | 13 | Yes | 21 | Yes | Yes | 18 | 6 | 0 | No | Heterosexual | NA |
\(x\) = BMI of an American
\(\mu\) = mean BMI of Americans
A random sample of 50 BMI levels was taken. Check: A random sample was taken from the NHANES data frame using r Studio
The population of BMI levels is normally distributed. Check:
gf_density(~BMI, data=sample_NHANES_50, title="BMI of an American", xlab="Body Mass Index")
gf_qq(~BMI, data=sample_NHANES_50, title="BMI of an American") 
The density plot looks somewhat skewed right and the normal quantile plot looks somewhat linear. There doesn’t seem to be strong evidence that the sample comes from a population that is normally distributed. However, since the sample is moderate to large, the t-test is robust to this condition not being met. So the results of the test are probably valid.
On r Studio, the command would be
t.test(~BMI, data= sample_NHANES_50, conf.level=0.95)
One Sample t-test
data: BMI
t = 25.003, df = 46, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
24.55287 28.85224
sample estimates:
mean of x
26.70255 The sample statistic is the mean of \(x\) in the output, and confidence interval is under the words 95 percent confidence interval.
Statistical Interpretation: You are 95% confident that \(24.87190<\mu<28.71422\) contains the mean BMI of Americans.
Real World Interpretation: The mean BMI of Americans is between 24.87 and 28.71 \(kg/m^2\).
Notice that in example the chapter 7, you were asked if the mean BMI of Americans was different from Australians’ mean BMI of 27.2 \(kg/m^2\). The interval that Example: Confidence Interval for the Population Mean calculated does contain the value of 27.2. So you can’t say that Americans’ mean BMI and Australians’ mean BMI are different.This means that you can just use confidence intervals and not conduct hypothesis tests at all if you prefer.
Note: When creating this book, the random samples may change. So the answers may be different from what is said in the interpretations. This shows sampling variability, so it was not adjusted to show that this could happen.
The data in Table 7.5 are the life expectancies for all people in European countries (\“WHO life expectancy,\” 2013). The data in Table 7.6 filtered the data frame for just males and just year 2000. The year 2000 was randomly chosen as the year to use. Estimate the mean life expectancy for a man in Europe at the 99% level.
Code book for data frame Expectancy is below Table 7.5.
\(x\) = life expectancy for a European man
\(\mu\) = mean life expectancy for European men
State: A random sample of 53 life expectancies of European men in 2000 was taken.
Check: The data is actually all of the life expectancies for every country that is considered part of Europe by the World Health Organization in the year 2000. Since the year 2000 was picked at random, then the sample is a random sample.
State: The distribution of life expectancies of European men in 2000 is normally distributed.
Check:
gf_density(~expect, data=Expectancy_male, title="Life Expectancy of a male", xlab="Life Expectancy of a Male")
gf_qq(~expect, data=Expectancy_male, title="Male Life Expectancy")
This sample does not appear to come from a population that is normally distributed. This sample is moderate to large, so it is good that the t-test is robust.
On rStudio, the command would be
t.test(~expect, data=Expectancy_male, conf.level=0.99)
One Sample t-test
data: expect
t = 90.919, df = 52, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
68.60071 72.75778
sample estimates:
mean of x
70.67925 Sample statistic is 70.68 years, and the confidence interval is \(68.60071<\mu<72.75778\).
Statistical Interpretation: You are 99% confident that \(68.60071<\mu<72.75778\) contains the mean life expectancy of European men.
Real World Interpretation: The mean life expectancy of European men is between 68.60 and 72.76 years.
In each problem show all steps of the confidence interval. If some of the conditions are not met, note that the results of the interval may not be correct and then continue the process of the confidence interval.
Code book for data frame Emission is below Table 7.7.
Code book for data frame Sugar is below Table 7.8.
A new data frame will need to be created of just cereal for children. It is Table 7.9.
Code book for data frame Mercury is below Table 7.10.
Code book for data frame Pulse below Table 3.7.
A new data frame with just females who drink alcohol is Table 9.2 from chapter 7.
Code book for data frame Economics is below Table 7.12.
A data frame that contains only middle income countries was created in chapter 7 and is Table 7.13. Find a 95% confidence interval for the mean economic dynamism for middle income countries.
Code book for Data frame Fert_prenatal is below Table 2.18.
Code book for data frame Sway is below Table 7.14.