To test the claim about the mean lengths of Bass and Trout, we can use a two-sample t-test since we are comparing the means of two independent samples.

Given Data

• Bass (Group 1):

  • Sample size ($n_1$) = 11
  • Mean ($\bar{x}_1$) = 59 cm
  • Standard deviation ($s_1$) = 21.7 cm

• Trout (Group 2):

  • Sample size ($n_2$) = 8
  • Mean ($\bar{x}_2$) = 41 cm
  • Standard deviation ($s_2$) = 10.2 cm

Hypotheses

• Null hypothesis: $H_0: \mu_1 = \mu_2$ (the mean lengths are equal)
• Alternative hypothesis: $H_a: \mu_1 \neq \mu_2$ (the mean lengths are not equal)

Test Statistic

The formula for the t-statistic when not pooling variances is: $t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Calculating the values:

• $s_1^2 = (21.7)^2 = 470.89$
• $s_2^2 = (10.2)^2 = 104.04$

Now substituting the values into the formula: $t = \frac{59 - 41}{\sqrt{\frac{470.89}{11} + \frac{104.04}{8}}}$ Calculating the denominator: $\frac{470.89}{11} \approx 42.80, \quad \frac{104.04}{8} \approx 13.005$ $\sqrt{42.80 + 13.005} = \sqrt{55.805} \approx 7.47$ Now calculating $t$: $t \approx \frac{18}{7.47} \approx 2.41$

Degrees of Freedom

Using the Welch-Satterthwaite equation for degrees of freedom: $df \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}$ Calculating this gives: $df \approx \frac{\left(42.80 + 13.005\right)^2}{\frac{\left(42.80\right)^2}{10} + \frac{\left(13.005\right)^2}{7}} \approx \frac{55.805^2}{\frac{1838.44}{10} + \frac{169.13}{7}} \approx \frac{3115.295}{183.84 + 24.02} \approx \frac{3115.295}{207.86} \approx 14.97 \text{ (use 14 for df)}$

p-value Calculation

Using a t-table or calculator, we find the p-value for $t = 2.41$ with $df \approx 14$ in a two-tailed test.

Looking up or using software, we find: $\text{p-value} \approx 0.030$

Conclusion

Since the p-value (0.030) is less than the significance level (0.10), we reject the null hypothesis.

Thus, the correct conclusion is: Larry has sufficient evidence to reject Alex's claim.

Summary

• p-value: 0.030
• Conclusion: Larry has sufficient evidence to reject Alex's claim.

If you have any further questions or need more details, feel free to ask! Here are some related questions you might consider:

1. What assumptions must be met to perform a t-test?
2. How would the conclusion change if the p-value were greater than 0.10?
3. What are the implications of rejecting the null hypothesis in this context?
4. How can you increase the power of the test?
5. What other methods could be used to compare these two groups?

Tip: Always check the assumptions of normality and variance when performing t-tests, especially with small sample sizes.