There is not enough evidence to conclude that there is a significant difference on exam performance between males and females, as the p-value of the test is greater than the alpha level, using the t-distribution.
What are the hypothesis tested?At the null hypothesis, it is tested if there is not a difference between the means, hence:
[tex]H_0: \mu_M - \mu_F = 0[/tex]
At the alternative hypothesis, it is tested if there is enough evidence to conclude that there is a difference between the means, hence:
[tex]H_1: \mu_M - \mu_F \neq 0[/tex]
What are the mean and the standard error of the distribution of differences?For each sample, the mean and the standard error are obtained as follows:
Males: [tex]\mu_M = 87.69, s_M = \frac{5.22}{\sqrt{26}} = 1.0237[/tex]Females: [tex]\mu_F = 88.32, s_F = \frac{4.98}{\sqrt{27}} = 0.9584[/tex]Hence, for the distribution of differences, the mean and the standard error are given as follows:
Mean: [tex]\overline{x} = \mu_M - \mu_F = 87.69 - 88.32 = -0.63[/tex]Standard error: [tex]s = \sqrt{s_M^2 + s_F^2} = \sqrt{1.0237^2 + 0.9584^2} = 1.4023[/tex]What is the test statistic?The test statistic is calculated as follows:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the hypothesis.
Hence:
t = -0.63/1.4023
t = -0.45
What is the p-value and the conclusion?Considering a two-tailed test, as we are testing if the mean is different of a value, with t = -0.45 and 26 + 27 - 2 = 51 df, the p-value, using a t-distribution calculator, is found to be of 0.6546.
Since this p-value is greater than the significance level of 0.05, there is not enough evidence to conclude that there is a significant difference on exam performance between males and females.
More can be learned about the t-distribution at https://brainly.com/question/13873630
#SPJ1