To determine whether there has been an improvement in the students' scores after the new unit on factoring, we will start by computing the test statistic, which has already been given as -0.56. We then need to calculate the p-value to evaluate the significance of this test statistic.
The p-value tells us the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct. Given that our test statistic is -0.56, we use the cumulative distribution function (CDF) of the standard normal distribution.
For a two-tailed test, the p-value is calculated by finding the cumulative probability up to the given test statistic and then doubling that probability since it includes both tails of the distribution.
After calculating the p-value, we then categorize it based on its numerical value. Here are the categories we consider:
- p-value < 0.001
- p-value < 0.002
- p-value > 0.10
- p-value > 0.05
Using our calculations, we find that the p-value for a test statistic of -0.56 is approximately 0.575479. This p-value falls into the category of p-value > 0.10.
So, based on the given test statistic of -0.56, the p-value is approximately 0.575479. This places it in the range of:
[tex]\[ \text{p-value} > 0.10 \][/tex]
Therefore, the correct category for the p-value is:
[tex]\[ p\text{-value} > 0.10 \][/tex]
This p-value indicates that there is no statistically significant improvement in the test scores after the new unit on factoring.
