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91Ó°ÊÓ

A college senior who took the Graduate Record Examination exam scored 620 on the Verbal Reasoning section and 670 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 462 with a standard deviation of \(119,\) and the mean score for the Quantitative Reasoning was 584 with a standard deviation of 151. Suppose that both distributions are nearly normal. (a) Write down the short-hand for these two normal distributions. (b) What is her \(\mathrm{Z}\) score on the Verbal Reasoning section? On the Quantitative Reasoning section? Draw a standard normal distribution curve and mark these two Z scores. (c) What do these \(Z\) scores tell you? (d) Relative to others, which section did she do better on? (e) Find her percentile scores for the two exams. (f) What percent of the test takers did better than her on the Verbal Reasoning section? On the Quantitative Reasoning section? (g) Explain why simply comparing her raw scores from the two sections would lead to the incorrect conclusion that she did better on the Quantitative Reasoning section. (h) If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b) - (f) change? Explain your reasoning.

Short Answer

Expert verified
She performed better on the Verbal Reasoning section (Z-score: 1.33) than on Quantitative (Z-score: 0.57).

Step by step solution

01

Normal Distributions Shorthand

For the Verbal Reasoning section, the scores are normally distributed with a mean of 462 and a standard deviation of 119. This can be represented as: \( N(462, 119) \). Similarly, the Quantitative Reasoning scores are normally distributed with a mean of 584 and a standard deviation of 151, represented as \( N(584, 151) \).
02

Calculate Z-Scores

The \( Z \)-score is calculated using the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.For Verbal Reasoning: \[ Z_{Verbal} = \frac{620 - 462}{119} \approx 1.33 \]For Quantitative Reasoning:\[ Z_{Quantitative} = \frac{670 - 584}{151} \approx 0.57 \]
03

Interpret Z-Scores

A \( Z \)-score tells us how many standard deviations an individual score is from the mean. * For the Verbal Reasoning section, a Z-score of 1.33 means she scored 1.33 standard deviations above the mean.* For the Quantitative Reasoning section, a Z-score of 0.57 means she scored 0.57 standard deviations above the mean.
04

Determine Better Performance

Comparing the two Z-scores, her score on the Verbal Reasoning is 1.33 standard deviations above the mean, which is better than her 0.57 standard deviations above the mean in Quantitative Reasoning.
05

Calculate Percentile Scores

Percentile scores can be found using a standard normal distribution table. * Verbal Z-score of 1.33 corresponds to approximately the 91st percentile. * Quantitative Z-score of 0.57 corresponds to approximately the 72nd percentile.
06

Percent Doing Better Than Her

Calculate what percent scored better than her by subtracting the percentiles from 100%. * On Verbal Reasoning, approximately 9% scored better (100% - 91%). * On Quantitative Reasoning, approximately 28% scored better (100% - 72%).
07

Misleading Raw Score Comparison

Raw scores alone do not reflect how well scores perform relative to the distribution. The Z-scores account for mean and variance, showing that Verbal Reasoning performance (1.33) is better than Quantitative Reasoning (0.57) despite the higher raw score in Quantitative.
08

Impact of Non-Normality on Results

If the distributions are not normal, Z-scores and percentile ranks might not be accurate, as these are based on the assumption of normality. The accuracy of conclusions in parts (b) to (f) would decrease if distributions were skewed or had outliers.

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Key Concepts

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

Normal Distribution
The Graduate Record Examination (GRE) scores mentioned in the problem follow a normal distribution. This is a common model in statistics that helps to predict how values are dispersed across a range. A normal distribution, also known as a bell curve due to its shape, is symmetrical about the mean. For any normal distribution, a few critical aspects give it its defined shape. These are the mean and the standard deviation.
  • The mean is the central point of the distribution, showing where the data clusters.
  • The standard deviation measures the spread of data around the mean.
For example, the Verbal Reasoning section has a mean (\(\mu\)) of 462 and a standard deviation (\(\sigma\)) of 119. In shorthand notation, it is expressed as \(N(462, 119)\). Similarly, the Quantitative Reasoning section is \(N(584, 151)\). This notation helps succinctly describe the distribution each set of scores follows. Understanding this concept is vital, as it forms the foundation for further statistical analysis like z-scores and percentiles.
Z-Scores
Z-scores are a fundamental concept in statistics, particularly in understanding standardized relationships between individual scores and the overall distribution. The z-score indicates how far and in what direction, an individual's score is from the mean, measured in standard deviations. The formula for calculating a z-score is: \[Z = \frac{X - \mu}{\sigma}\]where \(X\) is the individual's score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • For the Verbal Reasoning score of 620, using the mean of 462 and standard deviation of 119, her z-score is approximately 1.33. This means her score is 1.33 standard deviations above the mean.
  • For the Quantitative Reasoning score of 670, with a mean of 584 and a standard deviation of 151, her z-score is about 0.57. This indicates her score is 0.57 standard deviations above the mean.
These z-scores allow us to make direct comparisons even if the underlying distributions have different means and standard deviations, highlighting performance relative to peers.
Percentile Rank
Percentile rank is a statistical measure used to evaluate the relative standing of a value within a data set. It shows the percentage of scores that fall below a particular score. Knowing someone's percentile rank can help contextualize their performance in terms of how they compare to the rest of the test-takers.
For example, if you are told that a student's score is at the 91st percentile, this means they have scored better than 91% of the participants.
  • For the Verbal Reasoning section, the calculated z-score of 1.33 translates to approximately the 91st percentile.
  • For the Quantitative Reasoning section, the z-score of 0.57 corresponds to the 72nd percentile.
By knowing where a student stands in terms of percentile, one can understand how competitive their score is. It's crucial for students aiming to improve their scores, as higher percentile ranks also relate to better potential opportunities.
Statistical Analysis
Statistical analysis plays a crucial role when examining scores and performances. It helps translate raw data into meaningful insights. For test scores like those from the GRE, statistical analysis provides context by considering the distribution, mean, and variability of the data.
  • Z-scores are a tool for statistical analysis, letting us transform a raw score into a standard score. This illustrates how far a score varies from the average.
  • Percentile ranks, derived from z-scores, give interpretable insight into how a score stacks up against others. This is particularly useful for competitive exams like the GRE, where relative performance matters.
  • Finally, comparing scores using statistical methods can prevent misleading conclusions. For example, even though a student's quantitative score is higher numerically, their performance relative to the mean is better in verbal reasoning.
Such analysis answers not just what the scores are but what they mean in a broader context. It provides a clearer picture and enables well-informed decisions regarding performance, growth areas, and potential outcomes.

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Most popular questions from this chapter

Rosiglitazone is the active ingredient in the controversial type 2 diabetes medicine Avandia and has been linked to an increased risk of serious cardiovascular problems such as stroke, heart failure, and death. A common alternative treatment is pioglitazone, the active ingredient in a diabetes medicine called Actos. In a nationwide retrospective observational study of 227,571 Medicare beneficiaries aged 65 years or older, it was found that 2,593 of the 67,593 patients using rosiglitazone and 5,386 of the 159,978 using pioglitazone had serious cardiovascular problems. These data are summarized in the contingency table below. $$ \begin{aligned} &\text { Treatment }\\\ &\begin{array}{lccc} & {\text { Cardiovascular problems }} & \\ { 2 - 3 } & \text { Yes } & \text { No } & \text { Total } \\ \hline \text { Rosiglitazone } & 2,593 & 65,000 & 67,593 \\ \text { Pioglitazone } & 5,386 & 154,592 & 159,978 \\ \hline \text { Total } & 7,979 & 219,592 & 227,571 \end{array} \end{aligned} $$ Determine if each of the following statements is true or false. If false, explain why. Be careful: The reasoning may be wrong even if the statement's conclusion is correct. In such cases, the statement should be considered false. (a) Since more patients on pioglitazone had cardiovascular problems \((5,386\) vs. 2,593\(),\) we can conclude that the rate of cardiovascular problems for those on a pioglitazone treatment is higher. (b) The data suggest that diabetic patients who are taking rosiglitazone are more likely to have cardiovascular problems since the rate of incidence was \((2,593 / 67,593=0.038) 3.8 \%\) for patients on this treatment, while it was only \((5,386 / 159,978=0.034) 3.4 \%\) for patients on pioglitazone. (c) The fact that the rate of incidence is higher for the rosiglitazone group proves that rosiglitazone causes serious cardiovascular problems. (d) Based on the information provided so far, we cannot tell if the difference between the rates of incidences is due to a relationship between the two variables or due to chance.

In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages 30 - 34 group while Mary competed in the Women, Ages 25 29 group. Leo completed the race in 1:22:28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups: • The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. • The finishing times of the Women, Ages \(25-29\) group has a mean of 5261 seconds with a standard deviation of 807 seconds. • The distributions of finishing times for both groups are approximately Normal. Remember: a better performance corresponds to a faster finish. (a) Write down the short-hand for these two normal distributions. (b) What are the Z scores for Leo's and Mary's finishing times? What do these \(Z\) scores tell you? (c) Did Leo or Mary rank better in their respective groups? Explain your reasoning. (d) What percent of the triathletes did Leo finish faster than in his group? (e) What percent of the triathletes did Mary finish faster than in her group? (f) If the distributions of finishing times are not nearly normal, would your answers to parts (b) \- (e) change? Explain your reasoning.

Exercise 2.1 introduces a study that compares the rates of serious cardiovascular problems for diabetic patients on rosiglitazone and pioglitazone treatments. The table below summarizes the results of the study. $$ \begin{aligned} &\text { Treatment }\\\ &\begin{array}{lccc} & {\text { Cardiovascular problems }} & \\ { 2 - 3 } & \text { Yes } & \text { No } & \text { Total } \\ \hline \text { Rosiglitazone } & 2,593 & 65,000 & 67,593 \\ \text { Pioglitazone } & 5,386 & 154,592 & 159,978 \\ \hline \text { Total } & 7,979 & 219,592 & 227,571 \end{array} \end{aligned} $$ (a) What proportion of all patients had cardiovascular problems? (b) If the type of treatment and having cardiovascular problems were independent (null hypothesis), about how many patients in the rosiglitazone group would we expect to have had cardiovascular problems? (c) We can investigate the relationship between outcome and treatment in this study using a randomization technique. While in reality we would carry out the simulations required for randomization using statistical software, suppose we actually simulate using index cards. In order to simulate from the null hypothesis, which states that the outcomes were independent of the treatment, we write whether or not each patient had a cardiovascular problem on cards, shuffled all the cards together, then deal them into two groups of size 67,593 and 159,978 . We repeat this simulation 10,000 times and each time record the number of people in the rosiglitazone group who had cardiovascular problems. Below is a relative frequency histogram of these counts. i. What are the claims being tested? ii. Compared to the number calculated in part (b), which would provide more support for the alternative hypothesis, more or fewer patients with cardiovascular problems in the rosiglitazone group? iii. What do the simulation results suggest about the relationship between taking rosiglitazone and having cardiovascular problems in diabetic patients?

Assisted Reproductive Technology (ART) is a collection of techniques that help facilitate pregnancy (e.g. in vitro fertilization). A 2008 report by the Centers for Disease Control and Prevention estimated that ART has been successful in leading to a live birth in \(31 \%\) of cases \(^{55}\). A new fertility clinic claims that their success rate is higher than average. A random sample of 30 of their patients yielded a success rate of \(40 \%\). A consumer watchdog group would like to determine if this provides strong evidence to support the company's claim. (a) Write the hypotheses to test if the success rate for ART at this clinic is significantly higher than the success rate reported by the CDC. (b) Describe a setup for a simulation that would be appropriate in this situation and how the p-value can be calculated using the simulation results. (c) Below is a histogram showing the distribution of \(\hat{p}_{s i m}\) in 10,000 simulations under the null hypothesis. Estimate the p-value using the plot and use it to evaluate the hypotheses. (d) After performing this analysis, the consumer group releases the following news headline: "Infertility clinic falsely advertises better success rates". Comment on the appropriateness of this statement.

What percent of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region? Be sure to draw a graph. (a) \(Z<-1.35\) (b) \(Z>1.48\) (c) \(-0.42\)

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