Tests of Differences between Means, Independent Groups, Standard Deviation Estimated (4 of 8)

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  1. (continued) For the present example,

    MSE = (81.54 + 244.03)/2 = 162.78,


    The probability value for t can be determined using a t table. The degrees of freedom for t is equal to the degrees of freedom for MSE which is equal to df = n1 - 1 + n2 -1 = 18 or, df = N - 2 where N = n1 + n2 The probability is: p = 0.008.

  1. In step 5, the probability computed in Step 4 is compared to the significance level stated in Step 2. Since the probability value (0.008) is less than the significance level (0.05) the effect is significant.

  2. Since the effect is significant, the null hypothesis is rejected. It is concluded that the mean memory score for experts is higher than the mean memory score for novices.

  3. A report of this experimental result might be as follows:
    The mean number of pieces recalled by tournament players (MT = 63.89) was significantly higher than the mean number of pieces recalled by novices (MN = 46.79), t(18) = 2.99, p = 0.008.
The expression "t(18) = 2.99" means that a t test with 18 degrees of freedom was equal to 2.99. The probability value is given by "p = 0.008." Since it was not mentioned whether the test was one- or two-tailed, it is assumed the test was two tailed.
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