df = 25-1 = 24
For P = .05, with 24 df, you need to have difference between means ± 2.064(SEdiff)
t = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1) or to make it easier, you can use n
SD = √(variance)
I'll leave the calculation and explanation to you.
A researcher is testing a null hypothesis that states: H: ì = 50. A sample of 25 scores is selected and the mean is M = 55.Assuming that the sample variance is s2 = 100, compute the estimated standard error and the t statistic. Is this sample sufficient to reject the null hypothesis using a two-tailed test with á = .05? Assuming that the sample variance is s2 = 400, compute the estimated standard error and the t statistic. Is this sample sufficient to reject the null hypothesis using a two-tailed test with á = .05? Explain how increasing variance affects the standard error and the likelihood of rejecting the null hypothesis.
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