/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Presidents: Party Affiliation Fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Presidents: Party Affiliation For each successive presidential term from Teddy Roosevelt to George W. Bush (first term), the party affiliation controlling the White House is shown below, where R designates Republican and D designates Democrat. (Reference: The New York Times Almanac.) \(\begin{array}{llllllllllllllllllll}\mathrm{R} & \mathrm{R} & \mathrm{R} & \mathrm{D} & \mathrm{D} & \mathrm{R} & \mathrm{R} & \mathrm{D} & \mathrm{D} & \mathrm{D} & \mathrm{D} & \mathrm{D} & \mathrm{R} & \mathrm{D} & \mathrm{R} & \mathrm{R} & \mathrm{D} & \mathrm{R} & \mathrm{R} & \mathrm{R} & \mathrm{D} & \mathrm{D} & \mathrm{R}\end{array}\) Historical Note: In cases in which a president died in office or resigned, the period during which the vice president finished the term is not counted as a new term. Test the sequence for randomness. Use \(\alpha=0.05\).

Short Answer

Expert verified
The sequence does not significantly deviate from randomness at \( \alpha = 0.05 \).

Step by step solution

01

Define Expected Outcomes

In a random sequence, the transitions between parties should occur infrequently to demonstrate randomness. We are testing whether the observed number of such transitions significantly deviates from what is expected under randomness. Each party transition (R to D or D to R) is counted as one run.
02

Count Party Transitions

Analyze the sequence and count the total number of alternations, or 'runs', from one party to another. For the sequence given, we identify the transitions between parties: R to D or D to R. Upon examining the sequence of party affiliations, we find 10 transitions (runs): RR, R, DDD, R, DD, RRR, D, and R. Thus, there are 10 runs.
03

Calculate Expected Number of Runs

Calculate the expected number of runs (E) and variance (Var) for such a sequence of binary outcomes. The expected number of runs in a two-outcome sequence is given by: \[ E(R) = \frac{2n_1n_2}{n} + 1 \]Where \( n_1 \) is the number of R's, \( n_2 \) is the number of D's, and \( n \) is the total number of terms. In this case, \( n_1 = 11 \), \( n_2 = 12 \), and \( n = 23 \). Thus, \[ E(R) = \frac{2 \times 11 \times 12}{23} + 1 = 12.48 \]
04

Variance of the Run Count

The variance of the number of runs is calculated as:\[ \text{Var}(R) = \frac{2n_1n_2(2n_1n_2-n)}{n^2(n-1)} \]Substitute values: \[ \text{Var}(R) = \frac{2 \times 11 \times 12(2 \times 11 \times 12 - 23)}{23^2 \times (23-1)} = 5.89 \]
05

Calculate Test Statistic

The test statistic Z is calculated as:\[ Z = \frac{R - E(R)}{\sqrt{\text{Var}(R)}} \]Given \( R = 10 \), \( E(R) = 12.48 \), and \( \text{Var}(R) = 5.89 \), substitute these values:\[ Z = \frac{10 - 12.48}{\sqrt{5.89}} = -1.03 \]
06

Decision Rule and Conclusion

Compare the calculated Z value with the critical Z value from the standard normal distribution at \( \alpha = 0.05 \), which is approximately \( \pm 1.96 \). Since \(|Z| = 1.03\) is less than 1.96, we cannot reject the null hypothesis, indicating the sequence does not deviate significantly from randomness.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Randomness Test
A randomness test is performed to determine if a sequence behaves in a way that suggests random selection, rather than following a detectable pattern. It is important because this test helps assess the unpredictability of sequences, especially when trying to infer if the sequence occurs just by chance.
  • In the context of political party shifts in the White House, testing for randomness helps to check whether changes in party control happen by chance or follow a more systematic, predictable pattern.
  • This can assist in exploring whether historical patterns of political shifts have genuine stochastic behavior or are affected by underlying factors such as socio-political trends.
Other situational applications could involve similar tests in fields such as genetics, quality control in manufacturing, or finance, wherever random patterns are expected or assumed.
Party Affiliation Sequence
The party affiliation sequence refers to the historical listing of political parties in control, represented here by 'D' for Democrat and 'R' for Republican. This is a binary sequence because it consists of only two possible outcomes—either R or D.
  • The sequence can be significant as it portrays how often political control has shifted over time.
  • Some might analyze this pattern to predict future political landscapes or to understand the historical frequency of political shifts.
Understanding this sequence means observing the series of transitions from Democrat to Republican and vice versa within specified intervals, in this case, presidential terms.
In essence, it presents a concise history of political party dominance that can aid in evaluating political trends.
Statistical Significance
Statistical significance in the context of hypothesis testing refers to the probability of the observed result occurring under the null hypothesis. It helps determine whether the outcome can be attributed to chance or indicates an actual effect or pattern.
  • For this sequence of party affiliations, the statistical significance is assessed through the calculated test statistic and comparison to a critical value.
  • This step determines whether the observed number of transitions (observed in the sequence) significantly deviates from what would be expected by pure chance."
Bearing statistical significance implies that results are unlikely to have arisen randomly and possibly reflect real-world trends or effects, making it a core component of interpreting data in this randomness test.
Runs Test
The Runs Test is a non-parametric test used to determine whether a sequence is random. It focuses on identifying 'runs'—consecutive identical items (e.g., RRR or DD), and examines their number relative to what is statistically expected in a random sequence.
  • More formally, this involves comparing the observed number of runs with the expected number, considering the sequence's length and composition (number of R's and D's).
  • A run represents a crucial structural element in this context; too few or too many runs compared to the expected can signal non-randomness.
Runs Test calculates a Z-statistic to discern if the sequence deviates from randomness. The computed Z falls within a standard normal distribution reference range to make inferences, thereby supporting or refuting the null hypothesis of randomness. In this case, the calculated Z=-1.03 is within the acceptable range indicating randomness.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Horse Trainer: Jumps A horse trainer teaches horses to jump by using two methods of instruction. Horses being taught by method \(\mathrm{A}\) have a lead horse that accompanies each jump. Horses being taught by method \(\mathrm{B}\) have no lead horse. The table shows the number of training sessions required before each horse performed the jumps properly. \begin{tabular}{l|cccccccccccc} \hline Method A & 28 & 35 & 19 & 41 & 37 & 31 & 38 & 40 & 25 & 27 & 36 & 43 \\\ \hline Method B & 42 & 33 & 26 & 24 & 44 & 46 & 34 & 20 & 48 & 39 & 45 & \\ \hline \end{tabular} Use a \(5 \%\) level of significance to test the claim that there is no difference between the training session distributions.

Grain Yields: Feeding the World With an ever-increasing world population, grain yields are extremely important. A random sample of 16 large grainproducing regions in the world gave the following information about grain production (in kg/hectare). (Reference: Handbook of International Economic Statistics, U.S. Government Documents.) \begin{tabular}{l|cccccccc} \hline Region & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline Modern Production & 1610 & 2230 & 5270 & 6990 & 2010 & 4560 & 780 & 6510 \\ \hline Historic Production & 1590 & 2360 & 5161 & 7170 & 1920 & 4760 & 660 & 6320 \\ \hline \end{tabular} \begin{tabular}{l|cccccccc} \hline Region & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline Modern Production & 2850 & 3550 & 1710 & 2050 & 2750 & 2550 & 6750 & 3670 \\ \hline Historic Production & 2920 & 2440 & 1340 & 2180 & 3110 & 2070 & 7330 & 2980 \\ \hline \end{tabular} Does this information indicate that modern grain production is higher? Use a 5\% level of significance.

Economics: Stocks As an economics class project, Debbie studied a random sample of 14 stocks. For each of these stocks, she found the cost per share (in dollars) and ranked each of the stocks according to cost. After 3 months, she found the earnings per share for each stock (in dollars). Again, Debbie ranked each of the stocks according to earnings. The way Debbie ranked, higher ranks mean higher cost and higher earnings. The results follow, where \(x\) is the rank in cost and \(y\) is the rank in earnings. \begin{tabular}{l|rrrrrrrrrrrrrr} \hline tock & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline rank & 5 & 2 & 4 & 7 & 11 & 8 & 12 & 3 & 13 & 14 & 10 & 1 & 9 & 6 \\ rank & 5 & 13 & 1 & 10 & 7 & 3 & 14 & 6 & 4 & 12 & 8 & 2 & 11 & 9 \\ \hline \end{tabular} Using a \(0.01\) level of significance, test the claim that there is a monotone relation, either way, between the ranks of cost and earnings.

Statistical Literacy For data pairs \((x, y)\), if \(y\) always increases as \(x\) increases, is the relationship monotone increasing, monotone decreasing, or nonmonotone?

High School Dropouts: Male versus Female Is the high school dropout rate higher for males or females? A random sample of population regions gave the following information about percentage of 15 - to 19 -year-olds who are high school dropouts. (Reference: Statistical Abstract of the United States, 121 st edition.) \begin{tabular}{l|cccccccccc} \hline Region & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline Male & \(7.3\) & \(7.5\) & \(7.7\) & \(21.8\) & \(4.2\) & \(12.2\) & \(3.5\) & \(4.2\) & \(8.0\) & \(9.7\) \\ \hline Female & \(7.5\) & \(6.4\) & \(6.0\) & \(20.0\) & \(2.6\) & \(5.2\) & \(3.1\) & \(4.9\) & \(12.1\) & \(10.8\) \\ \hline \end{tabular} \begin{tabular}{l|cccccccccc} \hline Region & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\ \hline Male & \(14.1\) & \(3.6\) & \(3.6\) & \(4.0\) & \(5.2\) & \(6.9\) & \(15.6\) & \(6.3\) & \(8.0\) & \(6.5\) \\ \hline Female & \(15.6\) & \(6.3\) & \(4.0\) & \(3.9\) & \(9.8\) & \(9.8\) & \(12.0\) & \(3.3\) & \(7.1\) & \(8.2\) \\ \hline \end{tabular} Does this information indicate that the dropout rates for males and females are different (either way)? Use \(\alpha=0.01\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.