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Chapter 15: Problem 30

Glucose testing, continued. Shelia's measured glucose level one hour after having a sugary drink varies according to the Normal distribution with \(\mu=122\) \(\mathrm{mg} / \mathrm{dL}\) and \(\sigma=12 \mathrm{mg} / \mathrm{dL}\). What is the level \(L\) such that there is probability only \(0.05\) that the mean glucose level of four test results falls above L? (Hint: This requires a backward Normal calculation. See page 91 in Chapter 3 if you need to review.)

Short Answer

Expert verified
L is approximately 132 mg/dL.

Step by step solution

01

Understand the Problem

We need to find the glucose level \( L \) such that only 5% of the mean glucose levels of four tests fall above this value. This involves using the Normal distribution and the properties of sampling distributions.
02

Identify the Sample Distribution

Sheila's glucose levels are normally distributed with \( \mu = 122 \) mg/dL and \( \sigma = 12 \) mg/dL. The sampling distribution of the sample mean \( \bar{x} \) from 4 tests will also be normal with \( \mu_{\bar{x}} = \mu = 122 \) mg/dL and standard deviation \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{4}} = 6 \) mg/dL.
03

Use the Z-value for Probability

The problem asks for \( L \) such that there is a 0.05 probability of exceeding \( L \). This corresponds to the 95th percentile of the sample mean distribution. We find the Z-value corresponding to the cumulative probability of 0.95 using a standard normal distribution table or calculator. The Z-value for 0.95 is approximately 1.645.
04

Calculate the Value of L

Using the Z-score formula for the sample mean, \( L = \mu_{\bar{x}} + Z \times \sigma_{\bar{x}} \). Substitute the known values: \( L = 122 + (1.645 \times 6) = 122 + 9.87 = 131.87 \). Therefore, \( L \approx 132 \) mg/dL when rounded to the nearest whole number.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sampling Distribution
When conducting a glucose test like Shelia's, we often deal with something called the sampling distribution. To understand this concept, think of a sample not as a single number, but as a collection of data points. For example, if Shelia takes four glucose tests, the sampling distribution is the distribution of the average results of these tests.
  • It follows a normal distribution if the individual data points are normally distributed as well.
  • The mean of the sampling distribution ( \( \ \\ \mu_{\bar{x}} \ \) ) is the same as the mean of the individual tests, which in Shelia's case, is \(\mu=122\text{ mg/dL}\).
  • The standard deviation of the sampling distribution, known as the standard error, becomes smaller as you increase the sample size. Specifically, it's the population standard deviation \(\sigma\) divided by the square root of the sample size \(n\), giving us \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\).
For Shelia's tests, the standard deviation of the sampling distribution is 6 (\(\sigma_{\bar{x}} = \frac{12}{\sqrt{4}} = 6\)). This tells us how much the average glucose results will fluctuate if she takes multiple sets of tests.
Z-score
The Z-score is a statistical measure that helps us understand how far away a particular data point is from the mean, in terms of standard deviations. It provides a way to compare results from a sample to a standard normal distribution.
  • The Z-score is calculated using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is a single data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • In the context of Shelia's test, we calculated a Z-score to find a threshold level \(L\) for which just 5% of the test averages would be expected to exceed \(L\).
Z-scores allow researchers to calculate probabilities and percentiles, thereby helping determine where a specific point or average lies within the overall data distribution. For Shelia, a Z-score of 1.645 corresponds to the 95th percentile, which means that 5% of the glucose levels would be higher than \(L\).
95th Percentile
The 95th percentile is a point in your data below which 95% of the data falls. It is commonly used in statistics to identify unusually high values in a set of data points. In Shelia鈥檚 case, we needed to find the glucose level \(L\) such that only 5% of sample mean glucose levels are higher.
  • This percentile tells us what values are considered extreme, or in this context, what is a high glucose level after drinking sugar.
  • To determine the 95th percentile, you use a Z-score table or calculator to find the Z-score that corresponds to a cumulative probability of 0.95, and then plug it into the normal distribution formula for the sample mean.
For Shelia鈥檚 glucose test, knowing the 95th percentile allows medical professionals to judge whether further testing is needed if her test results are routinely above this threshold.
Standard Deviation
Standard deviation is a fundamental concept in statistics representing the spread or dispersion of a set of data. It tells us how much the values in a dataset vary from the mean.
  • A smaller standard deviation indicates that the data points tend to be close to the mean.
  • A larger standard deviation indicates that the data points are spread out over a wider range of values.
In the context of Shelia鈥檚 glucose test, the standard deviation \(\sigma\) is 12 mg/dL for individual tests. This means that each single test result of glucose might deviate from the mean value (122 mg/dL) by, on average, 12 mg/dL.
For the sampling distribution of the average of four tests, the standard deviation (standard error) is reduced to 6 mg/dL, which reflects less variability in the average results of these grouped tests as compared to individual tests.

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Most popular questions from this chapter

Pollutants in auto exhausts, continued. The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life \((150,000\) miles of driving) of cars of a particular model varies Normally with mean \(80 \mathrm{mg} / \mathrm{mi}\) and standard deviation \(4 \mathrm{mg} / \mathrm{mi}\). A company has 25 cars of this model in its fleet. What is the level \(L\) such that the probability that the average \(\mathrm{NOX}+\mathrm{NMOG}\) level \(\mathrm{x}^{-} 1\) for the fleet is greater than \(L\) is only \(0.01\) ? (Hint: This requires a backward Normal calculation. See page 91 in Chapter 3 if you need to review.)

Glucose testing. Shelia's doctor is concerned that she may suffer from gestational diabetes (high blood ghucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. In a test to screen for gestational diabetes, a patient is classified as needing further testing for gestational diabetes if the glucose level is above 130 milligrams per deciliter (mg/dL) one hour after having a sugary drink. Shelia's measured glucose level one hour after the sugary drink varies according to the Normal distribution with \(\mu=122 \mathrm{mg} / \mathrm{dL}\) and \(\sigma=12 \mathrm{mg} / \mathrm{dL}\). (a) If a single glucose measurement is made, what is the probability that Shelia is diagnosed as needing further testing for gestational diabetes? (b) If measurements are made on four separate days and the mean result is compared with the criterion \(130 \mathrm{mg} / \mathrm{dL}\), what is the probability that Shelia is diagnosed as needing further testing for gestational diabetes?

Guns in School. Researchers surveyed 15,624 Amercian high school students (grades 9-12) and found that \(27.2 \%\) of those surveyed were in grade 9 . The percent of all American high school students who are in grade 9 is \(27.5 \%\). The percent of those surveyed who were in grade 9 and had carried a gun to school was \(4.5 \%\). Is each of the boldface numbers a parameter or a staristic?

Sampling students. To estimate the mean score of those who took the Medical College Admission Test on your campes, you will obtain the scores of an SRS of students. From published information, you know that the scores are approximately Normal with standard deviation about 10.4. How large an SRS must you take to reduce the standard deviation of the sample mean score to 1 ?

The Law of Large Numbers Made Visible. Roll two balanced dice and count the total spots on the up-faces. The probability model appears in Example \(12.5\) (page 283). You can see that this distribution is symmetric with 7 as its center, so it's no surprise that the mean is \(\mu=7\). This is the population mean for the idealized population that contains the results of rolling two dice forever. The law of large numbers says that the average \(x-\bar{x}\) from a finite number of rolls tends to get closer and closer to 7 as we do more and more rolls. (a) Click "More dice" once in the Law of Large Numbers applet to get two dice. Click "Show mean" to see the mean 7 on the graph. Leaving the number of rolls at 1 , click "Roll dice" three times. How many spots did each roll produce? What is the average for the three rolls? You see that the graph displays at each point the average number of spots for all rolls up to the last one. This is exactly like Figure 15.1. (b) Click "Reset" to start over. Set the number of rolls to 100 and click "Roll dice." "The applet rolls the two dice 100 times. The graph shows how the average count of spots changes as we make more rolls. That is, the graph shows \(\mathrm{x}^{-} \bar{x}\) as we continue to roll the dice. Sketch (or print out) the final graph. (c) Repeat your work from part (b). Click "Reset" to start over, then roll two dice 100 times. Make a sketch of the final graph of the mean \(x^{-}\)- against the number of rolls. Your two graphs will often look very different. What they have in common is that the average eventually gets close to the population mean \(\mu=7\). The law of large numbers says that this will always happen if you keep on rolling the dice.
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