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.

Answer the above questions, show all work.

Solution

To conduct a hypothesis test on the claim that the mean I.Q. of brown trout is greater than 4, we need to set up our hypotheses, calculate the test statistic, and compare it to the critical value for a one-tailed test. Let’s go through each step:

Step 1: State the Hypotheses

  1. Null Hypothesis (H0)(H_0): The mean I.Q. of brown trout is 4, μ=4\mu = 4.
  2. Alternative Hypothesis (Ha)(H_a): The mean I.Q. of brown trout is greater than 4, μ>4\mu > 4.

Step 2: Collect Data

The I.Q. scores for the 12 brown trout are: 5,4,7,3,6,4,5,3,6,3,8,55, 4, 7, 3, 6, 4, 5, 3, 6, 3, 8, 5

Step 3: Calculate the Sample Mean and Sample Standard Deviation

To perform the test, we first calculate the sample mean xˉ\bar{x} and the sample standard deviation ss.

  1. Sample Mean xˉ\bar{x}: xˉ=5+4+7+3+6+4+5+3+6+3+8+512=59124.92\bar{x} = \frac{5 + 4 + 7 + 3 + 6 + 4 + 5 + 3 + 6 + 3 + 8 + 5}{12} = \frac{59}{12} \approx 4.92

  2. Sample Standard Deviation ss: s=(xixˉ)2n1s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} where xix_i are the individual I.Q. scores and n=12n = 12. Calculating each squared deviation and summing:

    (xixˉ)224.67\sum (x_i - \bar{x})^2 \approx 24.67 s24.67111.50s \approx \sqrt{\frac{24.67}{11}} \approx 1.50

Step 4: Calculate the Test Statistic

The test statistic for a one-sample t-test is given by: t=xˉμs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}} Substitute the values: t=4.9241.50/120.920.4332.13t = \frac{4.92 - 4}{1.50 / \sqrt{12}} \approx \frac{0.92}{0.433} \approx 2.13

Step 5: Determine the Critical Value and Compare

For a one-tailed test with n1=11n - 1 = 11 degrees of freedom and a typical significance level of α=0.05\alpha = 0.05, we can look up the critical t-value in a t-table: tcritical1.796t_{\text{critical}} \approx 1.796

Step 6: Make a Decision

Since our calculated tt-value (2.13) is greater than the critical tt-value (1.796), we reject the null hypothesis.

Conclusion

There is sufficient evidence to support the belief that the mean I.Q. of brown trout is greater than 4.


Would you like further explanation on any of these steps or details?

Further Questions

  1. What would happen if the sample size increased?
  2. How does the significance level affect the outcome of this test?
  3. Can a hypothesis test conclude absolute truth about the mean I.Q.?
  4. What are the implications of using a one-tailed versus a two-tailed test here?
  5. How does variability in the sample affect the reliability of our conclusion?

Tip

When conducting hypothesis tests, always check if your sample meets the assumptions for the test (e.g., normality in the case of small sample sizes for t-tests).

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
t-Test
Statistical Inference
Mean
Standard Deviation
Critical Value
One-Tailed Test

Formulas

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

Theorems

Central Limit Theorem
t-Distribution

Suitable Grade Level

Grades 11-12