Alex and Larry are both fairly new to fishing. Alex thinks that Bass and Trout are similar lengths, but Larry suspects that they have different mean lengths.

In a recent small sample of 11 randomly-selected Bass (group 1), the mean was 59 and the standard deviation was 21.7 centimeters (cm) long.

Also, a small sample of 8 randomly-selected Trout (group 2), had a mean of 41 with a standard deviation of 10.2 cm long.

Assume that each population distribution is normal-shaped. Test the claim at the 10% significance level. The null hypothesis would be H0:μ1=μ2

and which would be the alternative hypothesis?


Ha:μ1=μ2


Ha:μ1≠μ2


Ha:μ1>μ2


Ha:μ1<μ2


Do not pool the variances. Find the p-value, rounded to 3 decimal places.
p-value = 

Choose the best conclusion.

Larry has sufficient evidence to reject Alex's claim.

No conclusion can be made since at least one condition (requirement) is not satisfied.

Larry does not have sufficient evidence to reject Alex's claim.

There is sufficient evidence to prove Larry's claim is true.

There is sufficient evidence to prove Alex's claim is true.

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.

Alex and Larry are both fairly new to fishing. Alex thinks that Bass and Trout are similar lengths, but Larry suspects that they have different mean lengths. In a recent small sample of 11 randomly-selected Bass, the mean was 59 cm and the standard deviation was 21.7 cm long. Also, a small sample of 8 randomly-selected Trout had a mean of 41 cm with a standard deviation of 10.2 cm long. Test the claim at the 10% significance level. The null hypothesis would be H0:μ1=μ2 and which would be the alternative hypothesis?

Explore the hypothesis testing of mean lengths between Bass and Trout based on sample data. This problem includes calculations of means, standard deviations, and p-values using a two-sample t-test. Perfect for Grades 11-12 students studying statistics.

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