To test the hypothesis that students who finish an exam first get better grades, Professor Hardtack

kept track of the order in which papers were handed in. The first 25 papers showed a mean score of
77.1 with a standard deviation of 19.6, while the last 24 papers handed in showed a mean score of
69.3 with a standard deviation of 24.9. Is this a significant difference at á = .05? (a) State the
hypotheses for a right-tailed test. (b) Obtain a test statistic and p-value assuming equal variances.
Interpret these results. (c) Is the difference in mean scores large enough to be important? (d) Is it reasonable
to assume equal variances? (e) Carry out a formal test for equal variances at á = .05, showing
all steps clearly.

(a) The hypotheses for a right-tailed test are as follows:

Null hypothesis (H0): The mean score for students who finish an exam first is equal to or greater than the mean score for students who finish last.
Alternative hypothesis (Ha): The mean score for students who finish an exam first is less than the mean score for students who finish last.

(b) To obtain a test statistic and p-value assuming equal variances, we can use the two-sample t-test. The formula for the test statistic is:

t = (mean1 - mean2) / sqrt((s1^2 / n1) + (s2^2 / n2))

where mean1 and mean2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

In this case:
mean1 = 77.1, mean2 = 69.3
s1 = 19.6, s2 = 24.9
n1 = 25, n2 = 24

Plugging in these values, we get:
t = (77.1 - 69.3) / sqrt((19.6^2 / 25) + (24.9^2 / 24))

Next, we need to calculate the degrees of freedom (df) using the formula:
df = ( (s1^2 / n1 + s2^2 / n2)^2 ) / ( ( (s1^2 / n1)^2 / (n1 - 1) ) + ( (s2^2 / n2)^2 / (n2 - 1) ) )

Plugging in the values for s1, s2, n1, and n2, we can calculate the degrees of freedom.

Once we have the t-value and degrees of freedom, we can use a t-distribution table or a statistical software to find the p-value associated with the test statistic.

Interpretation: If the p-value is less than the significance level (0.05 in this case), we reject the null hypothesis. If the p-value is greater than the significance level, we fail to reject the null hypothesis.

(c) To determine if the difference in mean scores is large enough to be important, we need to examine the effect size. Common effect size measures for comparing means are Cohen's d or Hedges' g. These effect size measures quantify the magnitude of the difference between two means.

(d) To determine if it is reasonable to assume equal variances, we can conduct a formal test for equal variances.

(e) To carry out a formal test for equal variances at á = 0.05, we can use the F-test. We compare the variances of the two samples by comparing the ratio of the larger sample variance to the smaller sample variance against the critical value from an F-distribution table. The null hypothesis for equal variances is that the ratio should be equal to 1.

The steps for conducting the F-test for equal variances are as follows:
1. Calculate the ratio of the larger sample variance to the smaller sample variance.
2. Determine the degrees of freedom for each sample using the formula (n1 - 1) and (n2 - 1).
3. Look up the critical value from the F-distribution table using the degrees of freedom and the significance level (0.05 in this case).
4. Compare the calculated ratio to the critical value.
5. If the calculated ratio is less than or equal to the critical value, we have evidence to support the assumption of equal variances. If the calculated ratio is greater than the critical value, we reject the assumption of equal variances.

By following these steps, we can determine if it is reasonable to assume equal variances in this particular scenario.