Question: Comparing 25 and 29: - Difference = 25 - 29 = 37.06 - 37.03 = 0.03 - SE = [(0.15/40) + (0.15/37)] [(0.0038) + (0.0041)]

Comparing 25 and 29: - Difference = 25 - 29 = 37.06 - 37.03 = 0.03 - SE = [(0.15/40) + (0.15/37)] [(0.0038) + (0.0041)] 0.094 - t-value = Difference / SE = 0.03 / 0.094 0.32 - df = min(40-1, 37-1) = 36 - Critical t-value for a two-tailed test at = 0.0083 with df = 36 is approximately 2.704. - Since the absolute value of t-value (0.32) < critical t-value (2.704), we fail to reject the null hypothesis H0: 25 = 29. (is the calculation correct

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