s the following statment true or false? Why? The p-value was obtained using the t-distribution table. The sample sizes of the two groups were 33 and 22, and the degree of freedom was 53. To find the p-value, it is necessary to look at the row corresponding to 53 at a significance level of 0.05 and two tails. The value of the critical test statistic is 2. The computed t-statistic, 0.2112, falls between 0 and 0.679 from the same row. The next step is to look at the alpha levels at the top of the table with two tails, corresponding to 0 and 0.679. These alpha levels are 1 and 0.5. Therefore, the p-value of the test, which is 0.75, will be between 1 and 0.5. The p-value was confirmed using the TI-84 obtaining a similar p-value of 0.7994. Traditional Students (G1) Online Students (G2) mean 2.97 2.95 standard deviation 0.34 0.35 sample size 33 22 significance level: 0.05
Is the following statment true or false? Why?
The p-value was obtained using the t-distribution table. The sample sizes of the two groups were 33 and 22, and the degree of freedom was 53. To find the p-value, it is necessary to look at the row corresponding to 53 at a significance level of 0.05 and two tails. The value of the critical test statistic is 2. The computed t-statistic, 0.2112, falls between 0 and 0.679 from the same row. The next step is to look at the alpha levels at the top of the table with two tails, corresponding to 0 and 0.679. These alpha levels are 1 and 0.5. Therefore, the p-value of the test, which is 0.75, will be between 1 and 0.5. The p-value was confirmed using the TI-84 obtaining a similar p-value of 0.7994.
| Traditional Students (G1) | Online Students (G2) | |
| mean | 2.97 | 2.95 |
| standard deviation | 0.34 | 0.35 |
| sample size | 33 | 22 |
significance level: 0.05
degree of freedom: 53
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