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Predicting Election Outcomes Researchers conducted an experiment in which 695 individuals were shown black and white photographs of individuals running for Congress (either the U.S. Senate or House of Representatives). In each instance, the individuals were exposed to the photograph of both the winner and runner-up (in random order) for 1 second. The individuals were then asked to decide who they believed was more competent (and, therefore, more likely to receive their vote). Of the 695 individuals exposed to the photos, 469 correctly predicted the winner of the race. Do the results suggest that a quick 1-second view of a black and white photo represents enough information to judge the winner of an election (based on perceived level of competence of the individual) more often than not? Use the \(\alpha=0.05\) level of significance.

Step by step solution

01

Define the Hypotheses

First, define the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). \(H_0\): The proportion of correct predictions is 0.5 (no better than random chance).\(H_a\): The proportion of correct predictions is greater than 0.5 (better than random chance).

02

Calculate the Sample Proportion

The sample proportion \(p\) is calculated by dividing the number of correct predictions by the total number of trials. \(p = \frac{469}{695} \approx 0.6755\)

03

Calculate the Standard Error

The standard error of the proportion is calculated using the formula: \(SE = \sqrt{\frac{p_0(1-p_0)}{n}}\)where \(p_0 = 0.5\) and \(n = 695\). \(SE = \sqrt{\frac{0.5(1-0.5)}{695}} \approx 0.01898\)

04

Compute the Test Statistic

Calculate the test statistic using the formula for a proportion's z-test: \(z = \frac{p - p_0}{SE}$$z = \frac{0.6755 - 0.5}{0.01898} \approx 9.24\)

05

Determine the P-Value

Using standard normal distribution tables or a calculator, find the p-value corresponding to the z-value of 9.24. The p-value is approximately 0.

06

Compare the P-Value to \(\alpha\)

Since we are using an \(\alpha\) level of 0.05, compare the p-value to \(\alpha\). If p-value < \(\alpha\), reject the null hypothesis. Here, 0 < 0.05, so we reject the null hypothesis.

07

State the Conclusion

Based on the results, we reject the null hypothesis. There is significant evidence at the \(\alpha = 0.05\) level to conclude that a quick 1-second view of a black and white photo represents enough information to judge the winner of an election (based on perceived level of competence) more often than not.

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Key Concepts

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

null hypothesis

In statistics, the null hypothesis, often denoted as \(H_0\), represents a statement of no effect or no difference. It is a default assumption that there is no relationship between two measured phenomena. In the context of our election prediction exercise, the null hypothesis is exactly what you would expect if individuals guessed randomly. Here, \(H_0\): The proportion of people accurately predicting the winner is 0.5 or 50%. This means we begin by assuming that people cannot discern the winner better than chance.

For hypothesis testing, rejecting the null hypothesis means we have found evidence that something other than random chance is at work. Conversely, failing to reject \(H_0\) suggests our results could reasonably be due to random guessing.

alternative hypothesis

Contrasting the null hypothesis, the alternative hypothesis, denoted as \(H_a\), represents what we aim to support through our test. If the null hypothesis indicates no effect, the alternative suggests a meaningful effect.

In our election prediction example, the alternative hypothesis is that individuals are better than random chance at predicting the winner based on perceived competence from a photo. Therefore, \(H_a\): The proportion of correct predictions is greater than 0.5. By testing this hypothesis, we seek to provide support for the idea that people can identify more competent candidates based on a short viewing of their photographs.

If the evidence from our sample significantly favors \(H_a\), we reject \(H_0\) and infer that good predictions are not just due to luck.

proportion test

A proportion test is commonly used in statistics to determine if a sample proportion \(p\) differs significantly from a hypothesized population proportion \(p_0\). In our case, the sample proportion is the fraction of correct predictions out of 695 individuals, which is found to be approximately 0.6755.

This test helps us compare this observed proportion to what we would expect if people were guessing randomly (i.e., \(p_0 = 0.5\)). By calculating a test statistic, often a z-score under certain conditions, we can decide whether our sample provides enough evidence to reject the null hypothesis in favor of the alternative.

standard error

The standard error (SE) measures the variability or spread of the sample proportion. It tells us how much the sample proportion would vary from the true population proportion if we repeated our sample multiple times.

For a proportion, the SE can be calculated using the formula: \[SE = \sqrt{\frac{p_0(1-p_0)}{n}} \] where \(p_0\) is our hypothesized proportion (0.5) and \(n\) is the sample size (695). Substituting these values, we get \[SE \approx 0.01898\]

This calculation helps us understand how much we might expect our sample proportion (0.6755) to vary just by random chance.

z-test

A z-test for proportions compares an observed sample proportion to a hypothesized population proportion to see if there is a significant difference. In our example, we use this test to compare 0.6755 (sample) to 0.5 (hypothesized).

The z-test statistic is calculated using: \[z = \frac{p - p_0}{SE}\text{, where \ p is the sample proportion, \text{p_0 is the hypothesized proportion, and \text{SE is the standard error}}\]\text. For our provided values, this equates to roughly 9.24.

Assuming normal distribution, a z-score this high corresponds to a very low p-value, essentially suggesting a near zero probability that our result is due to random chance at the \alpha=0.05 level. Consequently, we reject \(H_0\) and accept \(H_a\), supporting that people can predict the winner better than random chance based on a 1-second photo glimpse.