Different species of turtles is equal.

Calculations:

1. Calculate the pooled variance:

   Since the sample sizes are relatively large (n1 = 40, n2 = 38) and the sample standard deviations are not significantly different, we can assume equal variances and use the pooled variance:

   s_pooled^2 = ((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2)
   

   s_pooled^2 = ((40 - 1) * 18.5^2 + (38 - 1) * 16.7^2) / (40 + 38 - 2) 
   s_pooled^2 ≈ 303.21
   

2. Calculate the t-statistic:

   t = (x̄1 - x̄2) / (s_pooled * sqrt(1/n1 + 1/n2))
   

   t = (300 - 305) / (sqrt(303.21) * sqrt(1/40 + 1/38))
   t ≈ -1.26
   

3. Determine degrees of freedom:

   degrees_of_freedom = n1 + n2 - 2 = 40 + 38 - 2 = 76
   

4. Find the critical value:

   Using a t-distribution table or a statistical software, find the critical value for a two-tailed test with α = 0.05 and 76 degrees of freedom. The critical values are approximately ±1.99.

5. Compare t-statistic to critical value:

   Since |-1.26| < 1.99, the calculated t-statistic does not fall in the rejection region.

Conclusion:

Since the calculated t-statistic does not fall in the rejection region, we fail to reject the null hypothesis. There is not enough evidence at the 0.05 significance level to conclude that the mean weight between the two species of turtles is different.

Two-Sample T-Test for Turtle Weights

Given:

• Sample 1:
  • Sample size (n1) = 40
  • Sample mean weight (x̄1) = 300
  • Sample standard deviation (s1) = 18.5
• Sample 2:
  • Sample size (n2) = 38
  • Sample mean weight (x̄2) = 305
  • Sample standard deviation (s2) = 16.7
• Significance level (α) = 0.05

Hypotheses:

• Null hypothesis (H0): μ1 = μ2 (The mean weight of the two turtle species is equal.)
• Alternative hypothesis (H1): μ1 ≠ μ2 (The mean weight of the two turtle species is not equal.)

Assumptions:

• The two samples are independent.
• The populations from which the samples are drawn are normally distributed.