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The article "More Students Taking AP Tests" (San Luis Obispo Tribune, January 10,2003 ) provided the following information on the percentage of students in grades 11 and 12 taking one or more AP exams and the percentage of exams that earned credit in 1997 and 2002 for seven high schools on the central coast of California. $$ \begin{array}{cccccc} & {\begin{array}{c} \text { Percentage of } \\ \text { Students Taking } \\ \text { One or More } \\ \text { AP Exams } \end{array}} & & {\begin{array}{c} \text { Percentage of } \\ \text { Exams That } \\ \text { Earned College } \end{array}} \\ { 2 - 3 } { 5 - 6 } & & & & {\text { Credit }} \\ { 2 - 3 } { 5 - 6 } \text { School } & 1997 & 2002 & & 1997 & 2002 \\ & 1 & 13.6 & 18.4 & & 61.4 & 52.8 \\ 2 & 20.7 & 25.9 & & 65.3 & 74.5 \\ 3 & 8.9 & 13.7 & & 65.1 & 72.4 \\ 4 & 17.2 & 22.4 & & 65.9 & 61.9 \\ 5 & 18.3 & 43.5 & & 42.3 & 62.7 \\ 6 & 9.8 & 11.4 & & 60.4 & 53.5 \\ 7 & 15.7 & 17.2 & & 42.9 & 62.2 \\ \hline \end{array} $$ a. Assuming that it is reasonable to regard these seven schools as a random sample of high schools located on the central coast of California, carry out an appropriate test to determine if there is convincing evidence that the mean percentage of exams earning college credit at central coast high schools in 1997 and in 2002 were different. b. Do you think it is reasonable to generalize the conclusion of the test in Part (a) to all California high schools? Explain. c. Would it be appropriate to use the paired-samples \(t\) test with the data on percentage of students taking one or more AP exams? Explain.

Short Answer

Expert verified
a. The two-sample t-test should be used to determine if there's convincing evidence of a significant difference in the mean percentage of exams earning college credit between 1997 and 2002. b. Generalization of the conclusion for all California high schools might not be reasonable due to potential lack of representativeness of the sample, sample size, and possible biases. c. The paired-sample t-test is likely not appropriate here because each observation in one group doesn't correspond with an observation in the second group.

Step by step solution

01

Preparing the Data

Organize the data from the percentage of exams that earned college credit for the years 1997 and 2002 into two different arrays or lists.
02

Calculating the Means

Calculate the mean percentage of exams that earned college credit for both 1997 and 2002.
03

Applying the Formula for the t-test

Calculate the standard deviation and standard error for each sample (1997 and 2002). Then use these values to conduct a two-sample t-test. The formula for the t-score is \( t = \frac{\bar{x_{1}} - \bar{x_{2}} }{\sqrt{ s_{1}^{2}/n_{1} + s_{2}^{2}/n_{2} }} \) where \( \bar{x_{1}} \) and \( \bar{x_{2}} \) are the sample means, \( s_{1}^{2} \) and \( s_{2}^{2} \) are the sample variances and \( n_{1} \) and \( n_{2} \) are the sample sizes.
04

Understanding the Result

Determine whether there is significant difference between the means of the two groups by comparing the obtained t value with the critical t value for a given level of significance (typically 0.05). If the obtained t value is greater than the critical t value, then there is a significant difference between the groups.
05

Generalizing the Results

Think critically about whether these results can be generalized for all high schools in California, keeping in mind representativeness of the sample, sample size, and possible biases.
06

Evaluating the Appropriateness of t-test

Determine whether it would be appropriate to use paired-samples t-test with the data on percentage of students taking one or more AP exams. A paired-samples t-test would require each observation in one group to correspond with an observation in the second group. Since the data provided in the exercise doesn't appear to be based on paired observations, paired t-test may not be appropriate.

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Key Concepts

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

Two-Sample t-Test
When investigating the average differences between two independent groups, statisticians often employ the two-sample t-test. This statistical method compares two sample means to see whether there is enough evidence to suggest that the means of the corresponding populations are significantly different from each other.

The two-sample t-test is based on the assumption that the two groups are independent, have normally distributed data, and have similar variances. The formula to calculate the t-score is: \[ t = \frac{\bar{x_{1}} - \bar{x_{2}} }{\sqrt{ s_{1}^{2}/n_{1} + s_{2}^{2}/n_{2} }} \]where \( \bar{x_{1}} \) and \( \bar{x_{2}} \) are the sample means for each group, \( s_{1}^{2} \) and \( s_{2}^{2} \) are the variances, and \( n_{1} \) and \( n_{2} \) are the sample sizes.This calculation generates a t-score, which is then compared to a critical value from a t-distribution table. If the t-score exceeds the critical value at a defined level of significance (often 0.05), we conclude there's a significant difference between the groups.
In our case, the aim is to compare the mean percentage of exams that earned college credit at central coast high schools in 1997 and 2002. By applying the two-sample t-test to our AP exam data, we can statistically infer whether the performance in these two years was significantly different.
Sample Means Calculation
The sample mean is a critical value in many statistical analyses, including t-tests, as it serves as an estimate of the population mean. Calculating the sample mean is straightforward: sum all the measurements in the sample and then divide by the number of observations in the sample. The mathematical notation for the sample mean is \( \bar{x} \), and the calculation can be expressed as: \[ \bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_{i} \]where \( x_i \) represents each value in the sample and \( n \) represents the sample size.When analyzing AP exam data, as is the case in our example, calculating the sample mean for each year's percentage of exams that earned college credit gives us a clear idea of the average performance for that year. Understanding and accurately calculating this value is essential, as it's the cornerstone of the subsequent statistical test—the two-sample t-test.
Data Representativeness
Data representativeness is concerned with how accurately a sample reflects the population from which it's drawn. For the conclusions drawn from statistical tests to be valid for the broader population, the sample must be representative. It involves ensuring that a variety of factors, such as demographics or geographic location, which might influence the variables being studied, are proportionally included in the sample.

When considering AP exam data, we must engage with questions of representativeness. For instance, when we look at AP exam results from seven high schools on the central coast of California, is it fair to generalize our findings to all California high schools? It's important to acknowledge that schools across the state might differ in significant ways, such as in resources, teaching quality, or student socio-economic backgrounds.
All these aspects might impact AP exam outcomes, which might mean our sample's findings wouldn't hold true for every school. Because representativeness affects the validity of our conclusions, it must be carefully considered, and any potential limitations should be transparently discussed.

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Most popular questions from this chapter

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be for a difference in population means. If not, explain why not. Scenario 1: The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" (Pediatrics [2009]: e953-e958) studied independent random samples of 764 Dutch boys and 748 Dutch girls ages 12 to \(19 .\) Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. You would like to determine if there is convincing evidence that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls. Scenario 2: The report "Highest Paying Jobs for \(2009-10\) Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010 ) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 is \(\$ 48,722\). A random sample of 50 accounting graduates at a large university resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3,300\). You would like to determine if there is strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722\). Scenario 3: Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25,2006 ). The sample mean and standard deviation were 15.1 hours and 11.4 hours for males and 14.1 and 11.8 for females. You would like to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers.

Babies born extremely prematurely run the risk of various neurological problems and tend to have lower IQ and verbal ability scores than babies that are not premature. The article "Premature Babies May Recover Intelligence, Study Says" (San Luis Obispo Tribune, February 12,2003 ) summarized medical research that suggests that the deficits observed at an early age may decrease as children age. Children who were born prematurely were given a test of verbal ability at age 3 and again at age 8 . The test is scaled so that a score of 100 would be average for normal-birth-weight children. Data for 50 children who were born prematurely were used to generate the accompanying Minitab output, where Age 3 represents the verbal ability score at age 3 and Age8 represents the verbal ability score at age \(8 .\) Use the Minitab output to determine if there is convincing evidence that the mean verbal ability score for children born prematurely increases between age 3 and age 8 . You can assume that it is reasonable to regard the sample of 50 children as a random sample from the population of all children born prematurely. $$ \begin{aligned} &\text { Paired T-Test and Cl: Age8, Age3 }\\\ &\begin{array}{l} \text { Paired } T \text { for } \text { Age8 - Age3 } \\ \begin{array}{rrrrr} & \text { N } & \text { Mean } & \text { StDev } & \text { Se Mean } \\ \text { Age8 } & 50 & 97.21 & 16.97 & 2.40 \\ \text { Age3 } & 50 & 87.30 & 13.84 & 1.96 \\ \text { Difference } & 50 & 9.91 & 22.11 & 3.13 \\ \text { T-Test of mean difference } & =0(\mathrm{vs}>0): \text { T-Value }=3.17 \\ \text { P-Value }=0.001 & & & \end{array} \end{array} \end{aligned} $$

In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were reported for a sample of students who worked and for a sample of students who did not work (University of Central Florida Undergraduate Research Journal, Spring 2005\()\) : $$ \begin{array}{cccc} & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Mean } \\ \text { GPA } \end{array} & \begin{array}{c} \text { Sandard } \\ \text { Deviation } \end{array} \\ \begin{array}{c} \text { Students Who Are } \\ \text { Employed } \end{array} & 184 & 3.12 & 0.485 \\ \begin{array}{c} \text { Students Who Are } \\ \text { Not Employed } \end{array} & 114 & 3.23 & 0.524 \\ \hline \end{array} $$ The samples were selected at random from working and nonworking students at the University of Central Florida. Does this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed?

The paper referenced in the Preview Example of this chapter ("Mood Food: Chocolate and Depressive Symptoms in a Cross-Sectional Analysis," Archives of Internal Medicine [2010]: \(699-703\) ) describes a study that investigated the relationship between depression and chocolate consumption. Participants in the study were 931 adults who were not currently taking medication for depression. These participants were screened for depression using a widely used screening test. The participants were then divided into two samples based on their test score. One sample consisted of people who screened positive for depression, and the other sample consisted of people who did not screen positive for depression. Each of the study participants also completed a food frequency survey. The researchers believed that the two samples were representative of the two populations of interest-adults who would screen positive for depression and adults who would not screen positive. The paper reported that the mean number of servings per month of chocolate for the sample of people that screened positive for depression was 8.39 , and the sample standard deviation was \(14.83 .\) For the sample of people who did not screen positive for depression, the mean was \(5.39,\) and the standard deviation was \(8.76 .\) The paper did not say how many individuals were in each sample, but for the purposes of this exercise, you can assume that the 931 study participants included 311 who screened positive for depression and 620 who did not screen positive. Carry out a hypothesis test to confirm the researchers' conclusion that the mean number of servings of chocolate per month for people who would screen positive for depression is higher than the mean number of chocolate servings per month for people who would not screen positive.

Wayne Gretzky was one of ice hockey's most prolific scorers when he played for the Edmonton Oilers. During his last season with the Oilers, Gretzky played in 41 games and missed 17 games due to injury. The article "The Great Gretzky" (Chance [1991]: 16-21) looked at the number of goals scored by the Oilers in games with and without Gretzky, as shown in the accompanying table. If you view the 41 games with Gretzky as a random sample of all Oiler games in which Gretzky played and the 17 games without Gretzky as a random sample of all Oiler games in which Gretzky did not play, is there convincing evidence that the mean number of goals scored by the Oilers is higher for games when Gretzky plays? Use \(\alpha=0.01\). $$ \begin{array}{lccc} & & \text { Sample } & \text { Sample } \\ & n & \text { Mean } & \text { sd } \\ \text { Games with Gretzky } & 41 & 4.73 & 1.29 \\ \text { Games without Gretzky } & 17 & 3.88 & 1.18 \end{array} $$

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