91Ó°ÊÓ

The website pundittracker.com keeps track of predictions made by individuals in finance, politics, sports, and entertainment. Jim Cramer is a famous TV financial personality and author. Pundittracker monitored 678 of his stock predictions (such as a recommendation to buy the stock) and found that 320 were correct predictions. Treat these 678 predictions as a random sample of all of Cramer's predictions. (a) Determine the sample proportion of predictions Cramer got correct. (b) Suppose that we want to know whether the evidence suggests Cramer is correct less than half the time. State the null and alternative hypotheses. (c) Verify the normal model may be used to determine the \(P\) -value for this hypothesis test. (d) Draw a normal model with area representing the \(P\) -value shaded for this hypothesis test. (e) Determine the \(P\) -value based on the model from part (d). (f) Interpret the \(P\) -value. (g) Based on the \(P\) -value, what does the sample evidence suggest? That is, what is the conclusion of the hypothesis test? Assume an \(\alpha=0.05\) level of significance.

Step by step solution

01

- Calculate Sample Proportion

First, calculate the sample proportion of correct predictions. The sample proportion is given by the formula \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of correct predictions and \( n \) is the total number of predictions.

02

- Calculation

Given \( x = 320 \) and \( n = 678 \), the sample proportion \( \hat{p} = \frac{320}{678} \approx 0.472 \).

03

- State Hypotheses

To determine if Cramer is correct less than 50% of the time, state the null and alternative hypotheses. The null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)) are: \( H_0: p = 0.5 \) (Cramer is correct 50% of the time) and \( H_a: p < 0.5 \) (Cramer is correct less than 50% of the time).

04

- Verify Normal Model

Check if the normal model can be used for the hypothesis test. The conditions are met if both \( np \) and \( n(1-p) \) are greater than 10. Using \( p = 0.5 \), calculate \( np = 678 \times 0.5 = 339 \) and \( n(1 - p) = 678 \times 0.5 = 339 \). Both values are greater than 10, so the normal model is applicable.

05

- Draw Normal Model

Draw a normal distribution curve centered at \( p = 0.5 \). Shade the area to the left of the sample proportion \( \hat{p} = 0.472 \), which represents the \( P \)-value.

06

- Calculate P-Value

Calculate the standard error: \( SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{678}} \approx 0.0192 \). Then compute the z-score: \( Z = \frac{\hat{p} - p}{SE} = \frac{0.472 - 0.5}{0.0192} \approx -1.46 \). Using a z-table or calculator, find the corresponding \( P \)-value for z = -1.46, which is approximately 0.072.

07

- Interpret P-Value

Interpret the \( P \)-value: A \( P \)-value of 0.072 indicates that there is a 7.2% chance of obtaining a sample proportion of 0.472 or less if Cramer is correct 50% of the time.

08

- Conclusion

Based on the \( P \)-value and the significance level \( \alpha = 0.05 \), compare \( P \)-value to \( \alpha \). Since 0.072 > 0.05, do not reject the null hypothesis. There is insufficient evidence to suggest that Cramer is correct less than 50% of the time.

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.

Sample Proportion

The sample proportion is a fundamental concept in statistics, especially when analyzing data from a sample group. In the context of Jim Cramer’s stock predictions, the sample proportion (\( \hat{p} \)) is calculated to determine the proportion of correct predictions out of the total predictions observed. The sample proportion is given by the formula:
\[ \hat{p} = \frac{\text{correct predictions}}{\text{total predictions}} \]
For instance, given 320 correct predictions out of 678, the sample proportion would be:
\[ \hat{p} = \frac{320}{678} \approx 0.472 \]
This means Jim Cramer got approximately 47.2% of his stock predictions correct. Calculating the sample proportion is the first step in hypothesis testing, providing a basis for further analysis.

Null and Alternative Hypotheses

In hypothesis testing, you start by stating the null and alternative hypotheses. These hypotheses represent the claims or beliefs you want to test.
The null hypothesis (\( H_0 \)) is a statement that there is no effect or no difference, and it is assumed to be true until evidence suggests otherwise. In our context, the null hypothesis is that Jim Cramer is correct 50% of the time:
\[ H_0: p = 0.5 \]
The alternative hypothesis (\( H_a \)) is what you want to prove; it suggests there is an effect or a difference. Here, we want to test if Cramer is correct less often than 50% of the time:
\[ H_a: p < 0.5 \]
Stating these hypotheses helps clarify the aim of the analysis and sets the stage for statistical testing.

P-value Calculation

The P-value is a crucial concept in hypothesis testing. It quantifies the probability of observing the sample data, or something more extreme, if the null hypothesis is true. To calculate the P-value, follow these steps:

• First, calculate the standard error (SE): \[ SE = \sqrt{\frac{p(1-p)}{n}} \] In this example, using a hypothesized proportion (\( p \)) of 0.5: \ \[ SE = \sqrt{\frac{0.5(1-0.5)}{678}} \approx 0.0192 \]
  
• Then, compute the z-score, which measures how many standard errors the sample proportion is away from the hypothesized proportion: \ \[ Z = \frac{\hat{p} - p}{SE} = \frac{0.472 - 0.5}{0.0192} \approx -1.46 \]
  
• Finally, use a z-table or calculator to find the P-value corresponding to the z-score of -1.46, which is approximately 0.072.
  Thus, the P-value of 0.072 indicates a 7.2% chance of getting a proportion of 47.2% or less if Cramer's prediction accuracy were truly 50%.

Normal Distribution

The normal distribution, or bell curve, is a key concept underlying many statistical tests, including hypothesis testing. To use it appropriately, certain conditions must be met, such as sample size and proportion.
In this case, we checked the normality condition by ensuring both \[ np \] and \[ n(1-p) \] are greater than 10. Using \[ p=0.5 \]:
\[ np = 678 \times 0.5 = 339 \]
\[ n(1 - p) = 678 \times 0.5 = 339 \]
Since both values are greater than 10, we can use the normal model to approximate the distribution of the sample proportion.
The normal distribution helps visualize the hypothesis test by showing the mean proportion in the center and shading areas representing the P-value.

Standard Error

The standard error (SE) measures the amount of variation or dispersion of sample proportions around the mean proportion of the population. It provides a gauge of how much the proportion from a single sample is expected to vary from the actual population proportion.
The standard error formula for a proportion is:
\[ SE = \sqrt{\frac{p(1-p)}{n}} \]
For our example, using the hypothesized proportion (p) of 0.5 and a sample size (n) of 678:
\[ SE = \sqrt{\frac{0.5 \times 0.5}{678}} \approx 0.0192 \]
A smaller standard error indicates the sample proportion is likely to be very close to the population proportion, while a larger standard error reflects more variability.

Significance Level

The significance level, often denoted as \( \alpha \), is a threshold set by the researcher to determine whether to reject the null hypothesis. Commonly used levels are 0.05 (5%), 0.01 (1%), and 0.10 (10%).
In the context of our hypothesis test, we use a significance level of 0.05. This means we are willing to reject the null hypothesis if the P-value is less than 0.05, implying there is less than a 5% probability that our sample results are due to random chance.
Comparing our calculated P-value (0.072) with the significance level (0.05), since 0.072 is greater than 0.05, we do not reject the null hypothesis. This indicates that there is not enough evidence to conclude that Jim Cramer’s prediction accuracy is less than 50%.