The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Listed below are IQ scores of randomly selected professional pilots. It is claimed that because professional pilots are a more homogeneous group than the general population, they have IQ scores with a standard deviation less than 15. Test that claim using a 0.05 significance level.

121
116
115
121
116
107
127
98
116
101
130
114

Identify the test statistic, ﻿begin mathsize 18px style p end style﻿ -value or critical value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

To test the claim that the standard deviation of IQ scores for professional pilots is less than 15, we will perform a hypothesis test for the population standard deviation using the chi-square test. Here's the step-by-step process:

### Step 1: State the Hypotheses

- Null Hypothesis (\(H_0\)): The standard deviation of IQ scores for professional pilots is equal to 15 (\(\sigma = 15\)).
- Alternative Hypothesis (\(H_1\)): The standard deviation of IQ scores for professional pilots is less than 15 (\(\sigma < 15\)).

This is a left-tailed test because we are testing if the standard deviation is less than 15.

### Step 2: Calculate the Sample Standard Deviation

We have the IQ scores of 12 professional pilots:

\[
121, 116, 115, 121, 116, 107, 127, 98, 116, 101, 130, 114
\]

Let's calculate the sample mean, sample variance, and sample standard deviation.

### Step 3: Test Statistic Calculation

The test statistic for the chi-square test for standard deviation is given by:

\[
\chi^2 = \frac{(n-1) \cdot s^2}{\sigma^2}
\]

where:
- \( n = 12 \) (sample size),
- \( s^2 \) is the sample variance,
- \(\sigma^2 = 15^2 = 225\) (population variance under the null hypothesis).

Let's calculate the values:

\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}
\]

where \(\bar{x}\) is the sample mean. We'll compute these values to find \(\chi^2\).The calculations yield the following results:

- **Sample Mean (\(\bar{x}\))**: \(115.17\)
- **Sample Variance (\(s^2\))**: \(90.33\)
- **Chi-Square Test Statistic (\(\chi^2\))**: \(4.42\)

### Step 4: Determine the Critical Value and P-Value

For a left-tailed test at a significance level of \(\alpha = 0.05\) with \(n - 1 = 11\) degrees of freedom, we need to find the critical value from the chi-square distribution and calculate the p-value.

Let's find the critical value \(\chi^2_{\alpha}\) and the p-value.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can explain the steps you would need to take next:

### Step 4: Determine the Critical Value and P-Value

1. **Critical Value**: Use the chi-square distribution table to find the critical value for a left-tailed test with \(\alpha = 0.05\) and \(df = 11\) (degrees of freedom). You can look this up or use a statistical calculator/software.

2. **P-Value**: You would calculate the p-value by finding the probability that the chi-square statistic falls to the left of the calculated test statistic (\(\chi^2 = 4.42\)).

### Step 5: Conclusion

- **If the p-value is less than the significance level (0.05)**, reject the null hypothesis and conclude that the standard deviation of IQ scores for professional pilots is less than 15.
- **If the p-value is greater than the significance level**, fail to reject the null hypothesis, meaning there's not enough evidence to support the claim that the standard deviation is less than 15.

Would you like to proceed with any specific questions or need further clarification?

Here are some related questions that might help expand your understanding:

1. How do you interpret the p-value in hypothesis testing?
2. What are the assumptions behind using a chi-square test for variance?
3. How would you calculate the sample variance manually?
4. What are the implications of a lower standard deviation in a specific professional group?
5. How does the sample size affect the power of a hypothesis test?

**Tip:** Always check assumptions like normality when conducting tests for variance, as deviations can affect your results significantly.

The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Listed below are IQ scores of randomly selected professional pilots. It is claimed that because professional pilots are a more homogeneous group than the general population, they have IQ scores with a standard deviation less than 15. Test that claim using a 0.05 significance level. IQ scores: 121, 116, 115, 121, 116, 107, 127, 98, 116, 101, 130, 114.

Explore how to perform a chi-square test for the standard deviation of IQ scores for professional pilots. This problem involves hypothesis testing, chi-square distribution, and variance calculation, including a step-by-step solution. The claim is tested using a significance level of 0.05, with real IQ score data. Suitable for advanced statistics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation