8.2: A Single Population Mean using the Normal Distribution
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A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of \(\bar{x} = 10\) and we have constructed the 90% confidence interval (5, 15) where \(EBM = 5\).
Calculating the Confidence Interval
To construct a confidence interval for a single unknown population mean \(\mu\), where the population standard deviation is known, we need \(\bar{x}\) as an estimate for \(\mu\) and we need the margin of error. Here, the margin of error (\(EBM\)) is called the error bound for a population mean (abbreviated EBM ). The sample mean \(\bar{x}\) is the point estimate of the unknown population mean \(\mu\).
The confidence interval estimate will have the form:
\[(\text{point estimate} - \text{error bound}, \text{point estimate} + \text{error bound})\nonumber \]
or, in symbols,
\[(\bar{x} - EBM, \bar{x} + EBM)\nonumber \]
The margin of error (\(EBM\)) depends on the confidence level (abbreviated \(CL\)). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percent of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of his or her conclusions.
There is another probability called alpha \((\alpha)\). \(\alpha\) is related to the confidence level, \(CL\). \(\alpha\) is the probability that the interval does not contain the unknown population parameter. Mathematically,
\[\alpha + CL = 1.\nonumber \]
Example \(\PageIndex{1}\)
Suppose we have collected data from a sample. We know the sample mean but we do not know the mean for the entire population. The sample mean is seven, and the error bound for the mean is 2.5: \(\bar{x} = 7\) and \(EBM = 2.5\)
The confidence interval is (7 – 2.5, 7 + 2.5) and calculating the values gives (4.5, 9.5). If the confidence level (\(CL\)) is 95%, then we say that, "We estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5."
Exercise \(\PageIndex{1}\)
Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2. What is the confidence interval estimate for the population mean?
- Answer
-
(11.8, 18.2)
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of \(\bar{x} = 10\), and we have constructed the 90% confidence interval (5, 15) where \(EBM = 5\). To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of \(\alpha = 10%\) in both tails, or 5% in each tail, of the normal distribution.
To capture the central 90%, we must go out 1.645 "standard deviations" on either side of the calculated sample mean. The value 1.645 is the z -score from a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail.
It is important that the "standard deviation" used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is
\[\dfrac{\sigma}{\sqrt{n}}\nonumber \]
This fraction is commonly called the "standard error of the mean" to distinguish clearly the standard deviation for a mean from the population standard deviation \(\sigma\).
In summary, as a result of the central limit theorem:
- \(\bar{X}\) is normally distributed, that is, \(\bar{X} \sim N(\mu_{x},\dfrac{\sigma}{\sqrt{n}})\).
- When the population standard deviation σ is known, we use a normal distribution to calculate the error bound.
Calculating the Confidence Interval
To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are:
- Calculate the sample mean \(\bar{x}\) from the sample data. Remember, in this section we already know the population standard deviation \(\sigma\).
- Find the z -score that corresponds to the confidence level.
- Calculate the error bound \(EBM\).
- Construct the confidence interval.
- Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.)
We will first examine each step in more detail, and then illustrate the process with some examples.
Finding the \(z\)-score for the Stated Confidence Level
When we know the population standard deviation \(\sigma\), we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of \(z\) that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution \(Z \sim N(0, 1)\).
The confidence level, \(CL\), is the area in the middle of the standard normal distribution. \(CL = 1 - \alpha\), so \(\alpha\) is the area that is split equally between the two tails. Each of the tails contains an area equal to \(\dfrac{\alpha}{2}\).
The \(z\)-score that has an area to the right of \(\dfrac{\alpha}{2}\) is denoted by \(z_{\dfrac{\alpha}{2}}\).
For example, when \(CL = 0.95, \alpha = 0.05\) and \(\dfrac{\alpha}{2} = 0.025\); we write \(z_{\dfrac{\alpha}{2}} = z_{0.025}\).
The area to the right of \(z_{0.025}\) is \(0.025\) and the area to the left of \(z_{0.025}\) is \(1 - 0.025 = 0.975\).
\[z_{\dfrac{\alpha}{2}} = z_{0.025} = 1.96\nonumber \]
using a calculator, computer or a standard normal probability table.
invNorm
\((0.975, 0, 1) = 1.96\)
Remember to use the area to the LEFT of \(z_{\dfrac{\alpha}{2}}\); in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution \(Z \sim N(0, 1)\).
Calculating the Error Bound
The error bound formula for an unknown population mean \(\mu\) when the population standard deviation \(\sigma\) is known is
\[EBM = z_{\alpha/2} \left(\dfrac{\sigma}{\sqrt{n}}\right)\nonumber \]
Constructing the Confidence Interval
The confidence interval estimate has the format \((\bar{x} - EBM, \bar{x} + EBM)\).
The graph gives a picture of the entire situation.
\[CL + \dfrac{\alpha}{2} + \dfrac{\alpha}{2} = CL + \alpha = 1.\nonumber \]
Writing the Interpretation
The interpretation should clearly state the confidence level (\(CL\)), explain what population parameter is being estimated (here, a population mean ), and state the confidence interval (both endpoints). "We estimate with ___% confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units)."
Example \(\PageIndex{2}\)
Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams).
Find a 90% confidence interval for the true (population) mean of statistics exam scores.
Answer
- You can use technology to calculate the confidence interval directly.
- The first solution is shown step-by-step (Solution A).
- The second solution uses the TI-83, 83+, and 84+ calculators (Solution B).
Solution A
To find the confidence interval, you need the sample mean, \(\bar{x}\), and the \(EBM\).
\[\bar{x} = 68\nonumber \]
\[EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)\nonumber \]
\[\sigma = 3; n = 36\nonumber \]
The confidence level is 90% ( CL = 0.90)
\[CL = 0.90\nonumber \]
so
\[\alpha = 1 – CL = 1 – 0.90 = 0.10\nonumber \]
\[\dfrac{\alpha}{2} = 0.05 z_{\dfrac{\alpha}{2}} = z_{0.05}\nonumber \]
The area to the right of \(z_{0.05}\) is \(0.05\) and the area to the left of \(z_{0.05}\) is \(1 - 0.05 = 0.95\).
\[z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645\nonumber \]
using \(\text{invNorm}(0.95, 0, 1)\) on the TI-83,83+, and 84+ calculators. This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.
\[EBM = (1.645)\left(\dfrac{3}{\sqrt{36}}\right) = 0.8225\nonumber \]
\[\bar{x} - EBM = 68 - 0.8225 = 67.1775\nonumber \]
\[\bar{x} + EBM = 68 + 0.8225 = 68.8225\nonumber \]
The 90% confidence interval is (67.1775, 68.8225).
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
7:ZInterval
.
Press
ENTER
.
Arrow to
Stats
and press
ENTER
.
Arrow down and enter three for
σ
, 68 for \(\bar{x}\), 36 for \(n\), and .90 for
C-level
.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (to three decimal places)(67.178, 68.822).
Interpretation
We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.
Explanation of 90% Confidence Level
Ninety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score.
Exercise \(\PageIndex{2}\)
Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time.
- Answer
-
(34.1347, 37.8653)
Example \(\PageIndex{3}\): Specific Absorption Rate
The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user’s body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Table shows the highest SAR level for a random selection of cell phone models as measured by the FCC.
| Phone Model | SAR | Phone Model | SAR | Phone Model | SAR |
|---|---|---|---|---|---|
| Apple iPhone 4S | 1.11 | LG Ally | 1.36 | Pantech Laser | 0.74 |
| BlackBerry Pearl 8120 | 1.48 | LG AX275 | 1.34 | Samsung Character | 0.5 |
| BlackBerry Tour 9630 | 1.43 | LG Cosmos | 1.18 | Samsung Epic 4G Touch | 0.4 |
| Cricket TXTM8 | 1.3 | LG CU515 | 1.3 | Samsung M240 | 0.867 |
| HP/Palm Centro | 1.09 | LG Trax CU575 | 1.26 | Samsung Messager III SCH-R750 | 0.68 |
| HTC One V | 0.455 | Motorola Q9h | 1.29 | Samsung Nexus S | 0.51 |
| HTC Touch Pro 2 | 1.41 | Motorola Razr2 V8 | 0.36 | Samsung SGH-A227 | 1.13 |
| Huawei M835 Ideos | 0.82 | Motorola Razr2 V9 | 0.52 | SGH-a107 GoPhone | 0.3 |
| Kyocera DuraPlus | 0.78 | Motorola V195s | 1.6 | Sony W350a | 1.48 |
| Kyocera K127 Marbl | 1.25 | Nokia 1680 | 1.39 | T-Mobile Concord | 1.38 |
Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates (SARs) for cell phones. Assume that the population standard deviation is \(\sigma = 0.337\).
Solution A
To find the confidence interval, start by finding the point estimate: the sample mean.
\[\bar{x} = 1.024\nonumber \]
Next, find the \(EBM\). Because you are creating a 98% confidence interval, \(CL = 0.98\).
You need to find \(z_{0.01}\) having the property that the area under the normal density curve to the right of \(z_{0.01}\) is \(0.01\) and the area to the left is 0.99. Use your calculator, a computer, or a probability table for the standard normal distribution to find \(z_{0.01} = 2.326\).
\(EBM = (z_{0.01})\dfrac{\sigma}{\sqrt{n}} = (2.326)\dfrac{0.337}{\sqrt{30}} =0.1431\)
To find the 98% confidence interval, find \(\bar{x} \pm EBM\).
\(\bar{x} - EBM = 1.024 – 0.1431 = 0.8809\)
\(\bar{x} - EBM = 1.024 – 0.1431 = 1.1671\)
We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram.
Solution B
- Press STAT and arrow over to TESTS.
- Arrow down to 7:Z Interval.
- Press ENTER.
- Arrow to Stats and press ENTER.
-
Arrow down and enter the following values:
- \(\sigma: 0.337\)
- \(\bar{x}: 1024\)
- \(n: 30\)
- C -level: 0.98
- Arrow down to Calculate and press ENTER.
- The confidence interval is (to three decimal places) (0.881, 1.167).
Exercise \(\PageIndex{3}\)
Table shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is \(\sigma = 0.337\).
| Phone Model | SAR | Phone Model | SAR |
|---|---|---|---|
| Blackberry Pearl 8120 | 1.48 | Nokia E71x | 1.53 |
| HTC Evo Design 4G | 0.8 | Nokia N75 | 0.68 |
| HTC Freestyle | 1.15 | Nokia N79 | 1.4 |
| LG Ally | 1.36 | Sagem Puma | 1.24 |
| LG Fathom | 0.77 | Samsung Fascinate | 0.57 |
| LG Optimus Vu | 0.462 | Samsung Infuse 4G | 0.2 |
| Motorola Cliq XT | 1.36 | Samsung Nexus S | 0.51 |
| Motorola Droid Pro | 1.39 | Samsung Replenish | 0.3 |
| Motorola Droid Razr M | 1.3 | Sony W518a Walkman | 0.73 |
| Nokia 7705 Twist | 0.7 | ZTE C79 | 0.869 |
Answer
\[\bar{x} = 0.940\nonumber \]
\[\dfrac{\alpha}{2} = \dfrac{1 - CL}{2} = \dfrac{1 - 0.93}{2} = 0.035\nonumber \]
\[z_{0.035} = 1.812\nonumber \]
\[EBM = (z_{0.035})\left(\dfrac{\sigma}{\sqrt{n}}\right) = (1.812)\left(\dfrac{0.337}{\sqrt{20}}\right) = 0.1365\nonumber \]
\[\bar{x} - EBM = 0.940 - 0.1365 = 0.8035\nonumber \]
\[\bar{x} + EBM = 0.940 + 0.1365 = 1.0765\nonumber \]
We estimate with 93% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8035 and 1.0765 watts per kilogram.
Notice the difference in the confidence intervals calculated in Example and the following Try It exercise. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter.
Changing the Confidence Level or Sample Size
Example \(\PageIndex{4}\)
Suppose we change the original problem in Example by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.
Answer
To find the confidence interval, you need the sample mean, \(\bar{x}\), and the \(EBM\).
\(\bar{x} = 68\)
\(EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)\)
\(\sigma = 3; n = 36\); The confidence level is 95% ( CL = 0.95).
\(CL = 0.95\) so \(\alpha = 1 – CL = 1 – 0.95 = 0.05\)
\(\dfrac{\alpha}{2} = 0.025 z_{\dfrac{\alpha}{2}} = z_{0.025}\)
The area to the right of \(z_{0.025}\) is 0.025 and the area to the left of \(z_{0.025}\) is \(1 – 0.025 = 0.975\).
\(z_{\dfrac{\alpha}{2}} = z_{0.025} = 1.96\)
when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.)
\(EBM = (1.96)\left(\dfrac{3}{\sqrt{36}}\right) = 0.98\)
\(\bar{x} – EBM = 68 – 0.98 = 67.02\)
\(\bar{x} + EBM = 68 + 0.98 = 68.98\)
Notice that the \(EBM\) is larger for a 95% confidence level in the original problem.
Interpretation
We estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98.
Explanation of 95% Confidence Level
Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score.
Comparing the results
The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider.
Summary: Effect of Changing the Confidence Level
- Increasing the confidence level increases the error bound, making the confidence interval wider.
- Decreasing the confidence level decreases the error bound, making the confidence interval narrower.
Exercise \(\PageIndex{4}\)
Refer back to the pizza-delivery Try It exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time.
- Answer
-
(33.37, 38.63)
Example \(\PageIndex{5}\)
Suppose we change the original problem in Example to see what happens to the error bound if the sample size is changed.
Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size and use \(n = 100\) instead of \(n = 36\)? What happens if we decrease the sample size to \(n = 25\) instead of \(n = 36\)?
- \(\bar{x} = 68\)
- \(EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)\)
- \(\sigma = 3\); The confidence level is 90% ( CL =0.90); \(z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645\).
Answer
Solution A
If we increase the sample size \(n\) to 100, we decrease the error bound.
When \(n = 100: EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right) = (1.645)\left(\dfrac{3}{\sqrt{100}}\right) = 0.4935\).
Solution B
If we decrease the sample size \(n\) to 25, we increase the error bound.
When \(n = 25: EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right) = (1.645)\left(\dfrac{3}{\sqrt{25}}\right) = 0.987\).
Summary: Effect of Changing the Sample Size
- Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.
- Decreasing the sample size causes the error bound to increase, making the confidence interval wider.
Exercise \(\PageIndex{5}\)
Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time.
- Answer
-
(34.6041, 37.3958)
Working Backwards to Find the Error Bound or Sample Mean
When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.
Finding the Error Bound
- From the upper value for the interval, subtract the sample mean,
- OR, from the upper value for the interval, subtract the lower value. Then divide the difference by two.
Finding the Sample Mean
- Subtract the error bound from the upper value of the confidence interval,
- OR, average the upper and lower endpoints of the confidence interval.
Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.
Example \(\PageIndex{6}\)
Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean.
Calculate the Error Bound:
- If we know that the sample mean is \(68: EBM = 68.82 – 68 = 0.82\).
- If we don't know the sample mean: \(EBM = \dfrac{(68.82−67.18)}{2} = 0.82\).
Calculate the Sample Mean:
- If we know the error bound: \(\bar{x} = 68.82 – 0.82 = 68\)
- If we don't know the error bound: \(\bar{x} = \dfrac{(67.18+68.82)}{2} = 68\).
Exercise \(\PageIndex{6}\)
Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean.
- Answer
-
Sample mean is 45, error bound is 2.88
Calculating the Sample Size \(n\)
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population mean when the population standard deviation is known is
\[EBM = \left(z_{\dfrac{a}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right) \label{samplesize}\nonumber \]
The formula for sample size is \(n = \dfrac{z^{2}\sigma^{2}}{EBM^{2}}\), found by solving the error bound formula for \(n\). In Equation \ref{samplesize}, \(z\) is \(z_{\dfrac{a}{2}}\), corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.
Example \(\PageIndex{7}\)
The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within two years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed?
Solution
- From the problem, we know that \(\sigma = 15\) and \(EBM = 2\).
- \(z = z_{0.025} = 1.96\), because the confidence level is 95%.
- \(n = \dfrac{z^{2}\sigma^{2}}{EBM^{2}} = \dfrac{(1.96)^{2}(15)^{2}}{2^{2}}\) using the sample size equation.
- Use \(n = 217\): Always round the answer UP to the next higher integer to ensure that the sample size is large enough.
Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within two years of the true population mean age of Foothill College students.
Exercise \(\PageIndex{7}\)
The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed?
- Answer
-
35 students
References
- “American Fact Finder.” U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/...html?refresh=t (accessed July 2, 2013).
- “Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. Available online at www.fec.gov/data/index.jsp (accessed July 2, 2013).
- “Headcount Enrollment Trends by Student Demographics Ten-Year Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. Available online at research.fhda.edu/factbook/FH...phicTrends.htm (accessed September 30,2013).
- Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at www.cdc.gov/growthcharts/2000...thchart-us.pdf (accessed July 2, 2013).
- La, Lynn, Kent German. "Cell Phone Radiation Levels." c|net part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013).
- “Mean Income in the Past 12 Months (in 2011 Inflaction-Adjusted Dollars): 2011 American Community Survey 1-Year Estimates.” American Fact Finder, U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/...prodType=table (accessed July 2, 2013).
- “Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at www.fec.gov/finance/disclosur...esummary.shtml (accessed July 2, 2013).
- “National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013).
Glossary
- Confidence Level (\(CL\))
- the percent expression for the probability that the confidence interval contains the true population parameter; for example, if the \(CL = 90%\), then in 90 out of 100 samples the interval estimate will enclose the true population parameter.
- Error Bound for a Population Mean (\(EBM\))
- the margin of error; depends on the confidence level, sample size, and known or estimated population standard deviation.


11.10: The Khanate of the Ilkhans (1265-1335)
Hülegü Khan (1256 – 1265), grandson of Genghis Khan and son of Tolui, served his brother Möngke (1251 – 1259), the great khan, and campaigned through the Middle East, where he whipped out the Assassins, a secret order of schismatic Shia entrenched in the mountains of Gilan province in 1256. He also destroyed the Abbasid capital of Baghdad in 1258, putting an end to the Caliphate. By 1260 Hülegü controlled parts of Armenia, Iraq, Anatolia, all of Azerbaijan, and all of Iran. Kublai Khan (1250 – 1294) had awarded his brother Hülegü the title of Ilkhan, a secondary khan who remained subordinate only to the great khan in Mongolia. This portion of the empire became known as the Khanate of the Ilkhans.
The Ilkhans were a Mongol minority ruling over a Muslim majority; religious problems plagued the Ilkhanate for much of its existence. To begin with, Hülegü, a Nestorian Christian, who later converted to Buddhism on his deathbed, had sacked Baghdad, one of the most politically important cities in the Islamic world, an act that alienated him from his Muslim cousin Berke Khan, ruler of the Golden Horde. The conversion of the Golden Horde to Islam had presented a real problem, for the Ilkhans had initially championed Buddhism in Iraq and Iran. As animosity continued to mount between the two parts of the Mongol Empire over religious differences, we see growing ties of alliance between the Muslim Golden Horde and their coreligionists, the Mamluk Sultanate of Egypt, against the Ilkhans. Belief transcended blood, as one part of Mongol Empire allied against another with an outside source. Faith-based civil wars consumed much of the reign of Abaga Khan (1265 – 1282). These wars were rooted in the Ilkhanate’s inappropriate treatment of their Muslim population. The Golden Horde’s alliance with the Mamluks threatened the Ilkhanate, and yet no longer could Abaga rely on the full might of centralized Mongol power; he was forced to appeal to Kublai Khan to assuage the hostilities between the Ilkhans and the Golden Horde.
Much of the religious conflict during the early Ilkhanate related to doctrinal differences between Islam and the traditional Mongol way of life. The most stubborn problem for the two was the contradiction between the traditional Mongolian method of animal slaughter, which required that no blood be spilled, and the Islamic code of cleanliness, which necessitated that all blood be drained. Each side was appalled by the other’s customs. Furthermore, as practitioners of rigid monotheism, Muslims found the Mongol worship of religious images repulsive, a ritual strictly forbidden in Islam.
While spiritual troubles remained a persistent problem for the Ilkhans, the economic situation deteriorated too. Gaykhatu Khan (1291 – 1295) practically emptied the royal treasury with profligate spending. He experimented with paper money recently adopted from China to compensate for his wasteful expenditures, but overprinting resulted in massive inflation. The Ilkhans also tried to extract the maximum amount of tribute from the countryside to offset declining revenues. This led to an abuse in tax gathering, known as tax farming, in which rulers sold contracts for the collection of revenues to the highest bidder. This method of tax collection provided a strong incentive to despoil peasants.
It was Mahmud Ghazan (1295 – 1304) who solved the Ilkhanate’s continued religious and economic problems. He was the first Ilkhan to convert to Islam, thus rehabilitating their image in the eyes of their Muslim subjects and making their rule much more acceptable. Their new public stance towards Islam moderated persistent conflict and paved the way for cultural flourishing. Ghazan patronized Ilkhanid art, scholarship, and science. Ilkhanid art reflected Chinese influence and helped contribute to Persian artistic development. In terms of scholarship, the first true history of the world was completed under the sponsorship of Mahmud Ghazan. Written by Rashid al-Din Hamadani (1247 – 1318), the book was richly illustrated with watercolors and portraiture in the Chinese style. Through his travels in the service of the Mongols, Rashid al-Din had become perceptively aware of Ilkhanid Persia’s cosmopolitan culture. It was Rashid al-Din, a Jewish convert to Islam, who had convinced Mahmud Ghazan to adopt the faith in order to be more attuned to the beliefs of his peoples. Regarding science, the Ilkhands attempted to amass large amounts of astronomical data from China to Europe. With unprecedented accuracy, they became very good at predicting lunar eclipses. Their data was used throughout Eurasia.
Despite the early looting and plundering indicative of a Mongol conquest, the Ilkhans eventually reactivated the Silk Road and promoted transcontinental trade. The newfound safety of the route throughout Eurasia stimulated trade and encouraged many different kinds of cultures to come together. Ghazan attempted to reform the tax policies that had led to the maximization of taxation. Cities did revive, but the long term negative consequence of conquest continued to be felt by the peasants who suffered from prolonged violence.
Much like the Mongols in Chaghatai Central Asia, eventually Ilkhans went native too. Here we see a Persian-Mongol fusion, as they began to identify with Persian culture and speak the Persian language. As they bonded with Persia, they adopted Islam and began to promote Persian as the written language of their land.


3.3: Clearances
Agencies and organizations must ensure that clearances are obtained in accordance with the Child Protective Services Law (CPSL). The required clearances may include:
- Pennsylvania Child Abuse History Clearance
- Pennsylvania State Police Criminal History Clearance
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Federal Bureau of Investigations (FBI) Criminal History Clearance
You will need to obtain updated clearances by the time the Practicum starts. Clearances can be obtained at: https://www.dhs.pa.gov/keepkidssafe/Pages/default.aspx
Click on “clearances.” You will need the clearances required by your practicum site. Check with your practicum site to see if they will allow submission of the free volunteer clearances, or if they require the clearances for employees. If you are doing volunteer clearances, use this code for the FBI clearance: 1KG6ZJ


6.13: Index Notation
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Overview
Index notation
is used to specify the elements of an array.
[1]
Most current programming languages use square brackets
[]
as the array index operator. Older programming languages, such as FORTRAN, COBOL, and BASIC, often use parentheses
()
as the array index operator.
Discussion
Example:
| Language | Example |
|---|---|
| C++ |
int ages[] = {49, 48, 26, 19, 16};
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| C# |
int[] ages = {49, 48, 26, 19, 16};
|
| Java |
int[] ages = {49, 48, 26, 19, 16};
|
| JavaScript |
var ages = [49, 48, 26, 19, 16];
|
| Python |
ages = [49, 48, 26, 19, 16]
|
| Swift |
var ages:[Int] = [49, 48, 26, 19, 16]
|
As an operator, square brackets either provide the value held by the member of the array (Rvalue) or change the value of member (Lvalue). In the above example, the member that is two offsets from the front of the array (the value 26) is assigned to the variable named myAge. The dereference operator of [2] means to go the 2 nd offset from the front of the ages array and get the value stored there. In this case, the value would be 26. In most current programming languages, the array members (or elements) are referenced starting at zero . The more common way for people to reference a list is by starting with position one . Consider:
| Position | Index | Miss America | Other Contests |
| zero offsets from the front |
ages[0]
|
Winner | 1 st Place |
| one offset from the front |
ages[1]
|
1 st Runner Up | 2 nd Place |
| two offsets from the front |
ages[2]
|
2 nd Runner Up | 3 rd Place |
| three offsets from the front |
ages[3]
|
3 rd Runner Up | 4 th Place |
| four offsets from the front |
ages[4]
|
4 th Runner Up | 5 th Place |
Saying that my cousin is the 2 nd Runner-Up in the Miss America contest sounds so much better than saying that she was in 3 rd Place. We would be talking about the same position in the array of the five finalists.
ages[3] = 20;
This is an example of changing an array’s value by assigning 20 to the 4 th member of the array and replacing the value 19 with 20. This is an Lvalue context because the array is on the left side of the assignment operator.
Key Terms
- array member
- An element or value in an array.
- index
- An operator that allows us to reference a member of an array.
- offset
- The method of referencing array members by starting at zero.