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91Ó°ÊÓ

A multiple-choice exam consists of 50 questions. Each question has five choices, of which only one is correct. Suppose that the total score on the exam is computed as $$ y=x_{1}-\frac{1}{4} x_{2} $$ where \(x_{1}=\) number of correct responses and \(x_{2}=\) number of incorrect responses. (Calculating a total score by subtracting a term based on the number of incorrect responses is known as a correction for guessing and is designed to discourage test takers from choosing answers at random.) a. It can be shown that if a totally unprepared student answers all 50 questions by just selecting one of the five answers at random, then \(\mu_{x_{1}}=10\) and \(\mu_{x_{2}}=40\). What is the mean value of the total score, \(y\) ? Does this surprise you? Explain. b. Explain why it is unreasonable to use the formulas given in this section to compute the variance or standard deviation of \(y\).

Short Answer

Expert verified
The mean value of the total score is 0. This is expected since for a student answering by guessing, the incorrect responses cancel out the correct ones due to the correction for guessing. It's not reasonable to calculate the variance or standard deviation of the total score using the given formulas because the total score is a derived variable, and its variance or standard deviation may have a totally different distribution from the original variables.

Step by step solution

01

Understand the given formulas

The formula for the total score on the exam (\(y\)) is given as \(y=x_{1}-0.25x_{2}\), where \(x_{1}\) represents the number of correct responses and \(x_{2}\) represents the number of incorrect responses.
02

Find the mean total score

It is given that if a student answers all 50 questions just selecting one of the five answers at random, the mean number of correct responses (\(\mu_{x_{1}}\)) is 10 and the mean number of incorrect responses (\(\mu_{x_{2}}\)) is 40. We can substitute these values into the formula for \(y\) to find the mean total score (\(\mu_{y}\)):\n\[\mu_{y}=\mu_{x_{1}}-0.25\mu_{x_{2}}=10-0.25(40)=10-10=0\]
03

Discuss the results

A mean total score of 0 should not surprise us because, for a student who answers by guessing, we would expect that the penalty for incorrect answers would negate the points gained from correct answers. This is a consequence of the correction for guessing scheme, which decreases the likelihood of benefiting from random guessing.
04

Discuss the unreasonableness of calculating the variance or standard deviation of \(y\)

The variance or standard deviation (\(\sigma\)) of a variable is a measure of how spread out the numbers of a distribution are. But here, the variable is a 'total score' that is derived from a simple mathematical transformation of the original variables (number of correct and incorrect responses) so it is not possible to use the formulas given in the section for calculating \(\sigma\) of \(y\). The direct transformation of the variables means that the variance, or spread, of the original variables, may not be applicable or meaningful for the derived variable. The distribution of \(y\) could be very different from the distributions of \(x_{1}\) and \(x_{2}\).

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Key Concepts

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

Mean Value of Total Score
When students take a multiple-choice test, each question they answer correctly is a step towards a better score. But in some exams, to correct for guessing, points are subtracted for each incorrect answer. If we imagine a student who guesses answers randomly, the mean value of their total score often becomes an intriguing topic of discussion.

The formula for the total score, denoted as \(y\), is expressed as \(y=x_{1}-0.25x_{2}\), where \(x_{1}\) is the number of correct answers and \(x_{2}\) is the number of incorrect answers. Intriguingly, when a student guesses all answers, the expected mean for correct responses (\(\mu_{x_{1}}\)) will be 10, as with each question having a 1 in 5 chance of being correct, and the mean for incorrect responses (\(\mu_{x_{2}}\)) will be 40. Therefore, the mean total score (\(\mu_{y}\)) calculates to zero: \[\mu_{y} = 10 - 0.25 \times 40 = 0\].

This might come as a surprise at first, but it underscores the effectiveness of the correction for guessing strategy. It's purposely designed to neutralize the advantage that could be gained from random guessing, leveling the playing field by rewarding knowledge rather than luck.
Variance and Standard Deviation in Statistics
Diving deeper into statistics, two crucial concepts are variance and standard deviation, which respectively measure the spread of a set of data points and the average distance of each data point from the mean. However, when it comes to computed scores like our total score \(y\) from a multiple-choice test with a guessing correction, calculating these measures directly isn't quite correct.

Variance (\
Impact of Random Guessing on Test Scores
Many students might wonder about the impact of random guessing on their test scores. In exams without a guessing penalty, guessing could potentially increase a student's score since there's a chance of selecting the correct answer. However, when a correction for guessing is applied, as in our example with the formula \(y=x_{1}-0.25x_{2}\), the scenario changes significantly.

An unprepared student's strategy to answer randomly can result in a mean score of zero, purely due to the design of the scoring system. This underscores the testing strategy's aim to emphasize knowledge over chance: it essentially canciles out the 'noise' introduced by guessing, making it improbable for a student to score purely on the basis of luck. Consequently, this encourages students to study and learn the material, knowing that guesses won't artificially inflate their scores.

By understanding the statistical implications of test-scoring methods, students can better prepare for their assessments, focusing on material mastery rather than relying on the uncertainty of random guesses. An informed approach to test-taking can make the difference between guesswork and demonstrated knowledge.

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Most popular questions from this chapter

A grocery store has an express line for customers purchasing at most five items. Let \(x\) be the number of items purchased by a randomly selected customer using this line. Give examples of two different assignments of probabilities such that the resulting distributions have the same mean but quite different standard deviations.

Let \(x\) denote the amount of gravel sold (in tons) during a randomly selected week at a particular sales facility. Suppose that the density curve has height \(f(x)\) above the value \(x\), where $$ f(x)=\left\\{\begin{array}{ll} 2(1-x) & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ The density curve (the graph of \(f(x)\) ) is shown in the following figure: Use the fact that the area of a triangle \(=\frac{1}{2}\) (base)(height) to calculate each of the following probabilities: a. \(P\left(x<\frac{1}{2}\right)\) b. \(P\left(x \leq \frac{1}{2}\right)\) c. \(P\left(x<\frac{1}{4}\right)\) d. \(P\left(\frac{1}{4}

State whether each of the following random variables is discrete or continuous: a. The number of defective tires on a car b. The body temperature of a hospital patient c. The number of pages in a book d. The number of draws (with replacement) from a deck of cards until a heart is selected e. The lifetime of a lightbulb

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Suppose that only \(20 \%\) of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are ultimately replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the \(20 \%\) figure? Explain.

In a study of warp breakage during the weaving of fabric (Technometrics [1982]: 63), 100 pieces of yarn were tested. The number of cycles of strain to breakage was recorded for each yarn sample. The resulting data are given in the following table: $$ \begin{array}{rrrrrrrrrr} 86 & 146 & 251 & 653 & 98 & 249 & 400 & 292 & 131 & 176 \\ 76 & 264 & 15 & 364 & 195 & 262 & 88 & 264 & 42 & 321 \\ 180 & 198 & 38 & 20 & 61 & 121 & 282 & 180 & 325 & 250 \\ 196 & 90 & 229 & 166 & 38 & 337 & 341 & 40 & 40 & 135 \\ 597 & 246 & 211 & 180 & 93 & 571 & 124 & 279 & 81 & 186 \\ 497 & 182 & 423 & 185 & 338 & 290 & 398 & 71 & 246 & 185 \\ 188 & 568 & 55 & 244 & 20 & 284 & 93 & 396 & 203 & 829 \\ 239 & 236 & 277 & 143 & 198 & 264 & 105 & 203 & 124 & 137 \\ 135 & 169 & 157 & 224 & 65 & 315 & 229 & 55 & 286 & 350 \\ 193 & 175 & 220 & 149 & 151 & 353 & 400 & 61 & 194 & 188 \end{array} $$ a. Construct a frequency distribution using the class intervals 0 to \(<100,100\) to \(<200\), and so on. b. Draw the histogram corresponding to the frequency distribution in Part (a). How would you describe the shape of this histogram? c. Find a transformation for these data that results in a more symmetric histogram than what you obtained in Part (b).

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