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From the Independent sample t test results, the observed p-value (two-tailed) is p=0.00 at
alpha value of 0.05. We reject null hypothesis if and if the observed or calculated p-value is less
than the level of significance (Neuman & Neuman, 2006). Since p=0.00 <0.05, hence we have
sufficient evidence to reject null hypothesis. Therefore, the test results suggest that the mean
value of Group A prior training scores is not equal to the mean value of Group B revised training
scores.
One-way ANOVA results indicate that the observed p-value is p=0.00 and the test
statistic F; 𝐹
0.05,𝑑𝑓=79
=511.923 at alpha value of 0.05. We use either the p-value approach or the
test statistic approach to reject or fail to reject the null hypothesis. For one-way ANOVA, the
null hypothesis could be reject if the observed F value is greater than the critical value of F or
calculated p-value is less than the level of significance (Creswell, 2013). Since F> F-critical i.e
(11.923> 2.725), we reject the null hypothesis because F value is greater than the Critical value
of F at 5% level of significance. Therefore, we conclude that at least one of the means for A, B,
C, and D is different. that similarly, we reject the hull hypothesis because the observed p-value,
p=0.000<0.05.
From the ANOVA test results, a statistical significant difference was observed between
one of the means with a test value, 𝐹
0.05,𝑑𝑓=79
=511.923, p = 0.000. The test results suggest that
at least one of the means for A, B, C, and D is different

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From the Independent sample t test results, the observed p-value (two-tailed) is p=0.00 at alpha value of 0.05. We reject null hypothesis if and if the observed or calculated p-value is less than the level of significance (Neuman & Neuman, 2006). Since p=0.00 <0.05, hence we have sufficient evidence to reject null hypothesis. Therefore, the test results suggest that the mean value of Group A prior training scores is not equal to the mean value of Group B revised training scores. One-way ANOVA results indicate that the observed p-value is p=0.00 and the test statistic F; 𝐹0.05,𝑑𝑓=79 =511.923 at alpha value of 0.05. We use either the p-value approach or the test statistic approach to reject or fail to reject the null hypothesis. For one-way ANOVA, the null hypothesis could be reject if the observed F value is greater than the critical value of F or calculated p-value is less than the level of significance (Creswell, 2013). Since F> F-critical i.e (11.923> 2.725), we reject the null hypothesis because F value is greater than the Critical value of F at 5% level of significance. Therefore, we conclude that at least one of the means for A, B, C, and D is different. that similarly, we reject the hull hypothesis because the observed p-value, p=0.000<0.05. From the ANOVA test results, a statistical significant difference was observed between one of the means with a test value, 𝐹0.05,𝑑𝑓=79 =511.923, p = 0.000. The test results suggest that at least one of the means for A, B, C, and D is different Name: Description: ...
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