91Ó°ÊÓ

Consider the following two questions designed to assess quantitative literacy: a. What is \(15 \%\) of 1000 ? b. A store is offering a \(15 \%\) off sale on all TVs. The most popular television is normally priced at \(\$ 1000\). How much money would a customer save on the television during this sale? Suppose the first question is asked of 200 randomly selected college students, with 164 answering correctly; the second one is asked of a different random sample of 200 college students, resulting in 140 correct responses (the sample percentages agree with those given in the article "Using the Right Yardstick: Assessing Financial Literacy Measures by Way of Financial Well-Being," J. of Consumer Affairs, 2013: 243-262; the investigators found that those who answered such questions correctly, particularly questions with context, were significantly more successful in their investment decisions than those who did not answer correctly). Carry out a test of hypotheses at significance level . 05 to decide if the true proportion of correct responses to the question without context exceeds that for the one with context.

Step by step solution

01

Define Hypotheses

We define the null hypothesis and the alternative hypothesis. Let \( p_1 \) be the true proportion of correct responses to the question without context, and \( p_2 \) be the proportion for the question with context. Our hypotheses are:- Null Hypothesis \( (H_0) \): \( p_1 = p_2 \)- Alternative Hypothesis \( (H_a) \): \( p_1 > p_2 \)

02

Calculate Sample Proportions

Calculate the sample proportions of correct responses for both questions. For the first question without context: \( \hat{p}_1 = \frac{164}{200} = 0.82 \). For the second question with context: \( \hat{p}_2 = \frac{140}{200} = 0.70 \).

03

Compute the Pooled Proportion

The pooled proportion \( \hat{p} \) is used in the hypothesis test and is calculated as the total number of correct responses over the total number of students. So, \( \hat{p} = \frac{164 + 140}{200 + 200} = \frac{304}{400} = 0.76 \).

04

Compute the Standard Error

We compute the standard error for the difference in sample proportions using the pooled proportion: \[ SE = \sqrt{ \hat{p} (1 - \hat{p}) \left( \frac{1}{200} + \frac{1}{200} \right) } = \sqrt{ 0.76 \times 0.24 \times \frac{1}{100} } \approx 0.0436 \]

05

Calculate the Z-score

The Z-score quantifies the difference between the sample proportions under the null hypothesis: \[ Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.82 - 0.70}{0.0436} \approx 2.75 \]

06

Determine the P-value

Using standard normal distribution tables, calculate the P-value from the Z-score. A Z-score of 2.75 gives a P-value of approximately 0.003. This is the probability that the observed difference (or one more extreme) would occur if \( H_0 \) were true.

07

Compare P-value to Significance Level

Since our P-value (0.003) is less than the significance level of 0.05, we reject the null hypothesis. This suggests that there is sufficient evidence to suggest the proportion of correct responses to the question without context is greater than that for the question with context.

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Key Concepts

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

Quantitative Literacy

Quantitative literacy is an essential skill that involves understanding and using numerical data effectively. In today's world, numbers are everywhere, from figuring out discounts at a store to analyzing data trends. It goes beyond mere arithmetic to include understanding percentages, probabilities, and the ability to interpret statistical information presented in various formats.
Developing quantitative literacy is crucial for individuals to make informed decisions in personal finance, health, and civic life. For example, being able to accurately calculate percentages can help someone determine the savings during a sale, as seen in the exercise question about \( 15\%\) of \( 1000 \).
This skill is not only important for academic success but also for everyday life, making individuals better equipped to engage in society. Integrating context into quantitative problems, as with the TV discount scenario, often enhances understanding because it makes the math more relevant and relatable.

Proportions

Proportions are a way of expressing a part of a whole, which is crucial in statistics and everyday situations. In hypothesis testing, proportions help us understand how one part relates to the total. For example, in the exercise, the proportion of students who answered correctly is calculated for two different scenarios.
To find a proportion, divide the number of desired outcomes by the total possible outcomes. With the college students, for instance, we found \( \hat{p}_1 = 0.82 \) and \( \hat{p}_2 = 0.70 \). These are the proportions of students answering correctly without and with context, respectively.
Understanding proportions helps compare different groups or scenarios—aspects central to making data-driven decisions. In our example, the larger \( \hat{p}_1 \) suggests more students answered correctly without the context than with it.

Significance Level

Significance level is a critical threshold in hypothesis testing that indicates the probability of rejecting a true null hypothesis, often set at \( 0.05 \) or 5%. This means there's a 5% risk of concluding a difference exists when it doesn't.
In our exercise, we conducted a hypothesis test comparing the proportion of correct answers between two groups of students. By setting the significance level at 0.05, we decide whether the observed difference is not due to random chance.
A smaller p-value than the significance level indicates strong evidence against the null hypothesis. For example, our p-value of 0.003 is much smaller, suggesting the proportion of answers without context is significantly higher.
Choosing the right significance level is crucial: too high might lead to unnecessary worries and interventions, while too low might miss something important.

Z-score

The z-score is a statistical measure that indicates how many standard deviations an element is from the mean of a dataset. In hypothesis testing, it helps assess whether the observed data significantly deviates from what is expected under the null hypothesis.
In this exercise, we calculate the z-score to compare proportions from two different groups. The formula \[ Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \] helps us quantify this difference.
A higher z-score indicates a larger difference between proportions than expected by chance. In our assessment with students, a z-score of approximately 2.75 suggests a significant difference.
Thus, using z-scores allows us to make informed decisions regarding statistical significance in hypothesis testing. If the z-score is beyond certain critical values depending on the significance level, it leads to rejecting the null hypothesis.