Math Problem Statement

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Solution

To conduct a hypothesis test regarding the brown trout's IQ, we can follow these steps, assuming a one-sample tt-test is appropriate for this small sample size.

1. Set up the Hypotheses

Let:

  • μ\mu = the true mean IQ of brown trout.

Our hypotheses are:

  • Null Hypothesis H0H_0: μ=4\mu = 4 (The mean IQ of brown trout is 4)
  • Alternative Hypothesis H1H_1: μ>4\mu > 4 (The mean IQ of brown trout is greater than 4)

2. Collect Sample Data

The sample data for IQs of the 12 brown trout is: 5,4,7,3,6,4,5,3,6,3,8,55, 4, 7, 3, 6, 4, 5, 3, 6, 3, 8, 5

3. Calculate Sample Mean and Standard Deviation

Let's calculate the sample mean xˉ\bar{x} and sample standard deviation ss.

The sample mean xˉ\bar{x} is given by: xˉ=xn\bar{x} = \frac{\sum x}{n} where xx are the individual IQ scores, and n=12n = 12.

The sample standard deviation ss is: s=(xixˉ)2n1s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}

4. Calculate the Test Statistic

Using a one-sample tt-test, the test statistic tt is given by: t=xˉμ0snt = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} where:

  • μ0=4\mu_0 = 4 (the hypothesized mean),
  • xˉ\bar{x} is the sample mean, and
  • ss is the sample standard deviation.

5. Determine the Critical Value and Make a Decision

We will use a significance level of α=0.05\alpha = 0.05 for a one-tailed test. For n1=11n - 1 = 11 degrees of freedom, we can find the critical value t0.05,11t_{0.05, 11} from the tt-distribution table.

If tt exceeds the critical value, we reject H0H_0; otherwise, we do not reject H0H_0.


Let's proceed with these calculations.The calculations yield the following results:

  • Sample Mean (xˉ\bar{x}) = 4.917
  • Sample Standard Deviation (ss) = 1.621
  • Test Statistic (tt) = 1.959
  • Critical Value at α=0.05\alpha = 0.05 for a one-tailed test with 11 degrees of freedom = 1.796

Conclusion:

Since the test statistic t=1.959t = 1.959 is greater than the critical value 1.7961.796, we reject the null hypothesis H0H_0. This suggests that there is statistically significant evidence at the 5% significance level to support the belief that the mean IQ of brown trout is greater than 4.

Would you like further details on any specific step, or have additional questions?

Here are some related questions:

  1. What would be the conclusion if we used a different significance level, like 1%?
  2. How does increasing the sample size affect the reliability of the hypothesis test?
  3. What are Type I and Type II errors in hypothesis testing?
  4. How would this test change if the alternative hypothesis were μ4\mu \neq 4?
  5. Can hypothesis tests be used to make decisions in fields other than psychology or biology?

Tip: Always check assumptions of normality when using tt-tests, especially with small sample sizes.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
One-Sample T-Test

Formulas

Sample Mean: x̄ = (Σx) / n
Sample Standard Deviation: s = √((Σ(x_i - x̄)^2) / (n - 1))
Test Statistic: t = (x̄ - μ) / (s / √n)

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate level, Statistics Course