Question

A zoologist believes the mean size of CO rabbits is smaller than the mean size of FL rabbits.

CO rabbit: x̅1= 9.2 lb. S1 = 3.4 lb, n1 = 20

FL rabbit: x̅2 = 11.4 lb, S2 = 1.3 lb, n 2 = 15

A) Find the 99% confidence interval for the difference of the means

B) At a significance level of 5% ( α = .05) Can the researcher conclude that an average CO rabbits are smaller than FL rabbits?

Answer 1

(A) MINITAB Procedure: Step 1: Choose Stat > Basic Statistics > 2-Sample t. Step 2: Choose Summarized data. Step 3: In first, enter Sample size as 20, Mean as 9.2, Standard deviation as 3.4. Step 4: In second, enter Sample size as 15. Mean as 11.4. Standard deviation as 1.3. Step 5: Select Assume equal variances. Step 6: Choose Options. Step 7: In Confidence level, enter 99. Step 8: In Alternative, select less than. Step 9: Click OK in all the dialogue boxes. MINITAB output: Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 20 9.20 3.40 0.76 2 15 11.40 1.30 0.34 Difference - (1) - (2) Estimate for difference: -2.200 99% CI for difference: (-4.517, 0.117) T-Test of difference - 0 (V3 <): T-Value = -2.65 P-Value = 0.007 DF = 25

The 99% confidence interval for the difference of the means is (-4.517,0.117). (B) Null hypothesis: H:44 = 42 There is no difference between the averages of CO rabbits and FL rabbits. Alternative hypothesis: H:4 <42 The average CO rabbits are smaller than FL rabbits. The value of the test statistic is t= -2.65. P-value = 0.007 Conclusion:

Use a significance level, a = 0.05. Here, p-value is 0.007, which is less than the value of a = 0.05. That is, p-value (= 0.007)s a(=0.05). Therefore, by the rejection rule, reject the null hypothesis H. Interpretation: There is sufficient evidence at level of significance a = 0.05. The average CO rabbits are smaller than FL rabbits.