does a test preparation course improve scores on the act test? the instructor gives a practice test at the start of the class and again at the end. the average difference (after - before) for his 30 students was 6 points with a standard deviation of the differences being 10 points. what is the test statistic for this test?

The test statistic for this test is 3.09.

To calculate the test statistic for this test, we need to use the formula:

t = [tex](\bar x - \mu) / (s /\sqrt n)[/tex]

where:

[tex]\bar x[/tex] = the sample mean difference (after - before)

[tex]\mu[/tex] = the population mean difference (assumed to be 0 if the test preparation course has no effect)

s = the standard deviation of the differences

n = the sample size (in this case, 30)

Plugging in the values given in the problem, we get:

t = [tex](6 - 0) / (10 / \sqrt 30)[/tex] = 3.09

To determine whether a test preparation course improves scores on the ACT test, we can use a paired samples t-test.

This test compares the mean difference between two related groups (in this case, the pre- and post-test scores of the same students) to the expected difference under the null hypothesis that there is no change in scores.

The test statistic for a paired samples t-test is given by:

t = (mean difference - hypothesized difference) / (standard error of the difference)

The mean difference is the average difference between the two groups, the hypothesized difference is the expected difference under the null hypothesis (which is 0 in this case), and the standard error of the difference is the standard deviation of the differences divided by the square root of the sample size.

The mean difference is 6 points, the hypothesized difference is 0, and the standard deviation of the differences is 10 points.

Since there are 30 students in the sample, the standard error of the difference is:

SE =[tex]10 / \sqrt{(30)[/tex]

= 1.83

Substituting these values into the formula for the test statistic, we get:

t = (6 - 0) / 1.83 = 3.28

The test statistic for this test is therefore 3.28.

To determine whether this test statistic is statistically significant, we would need to compare it to the critical value of t for 29 degrees of freedom (since there are 30 students in the sample and we are estimating one parameter, the mean difference).

The critical value for a two-tailed test at a significance level of 0.05 is approximately 2.045.

The test statistic (3.28) is greater than the critical value (2.045), we can conclude that the difference in scores between the pre- and post-test is statistically significant at a significance level of 0.05.

The test preparation course did indeed improve scores on the ACT test. It's important to note that this is just a single study and that further research would be needed to confirm these results.

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