Respuesta :
Using the z-distribution, it is found that since the p-value of the test is less than 0.05, the data provides sufficient evidence to conclude that there is a difference in average GPA between the population of sorority students and the population of unaffiliated female students on this university campus.
What are the hypotheses tested?
At the null hypotheses, it is tested if the means are equal, that is, their subtraction is of 0, hence:
[tex]H_0: \mu_1 - \mu_2 = 0[/tex]
At the alternative hypotheses, it is tested if they are different, hence:
[tex]H_1: \mu_1 - \mu_2 \neq 0[/tex]
What is the mean and the standard error for the distribution of differences?
For each sample, they are given as follows:
- [tex]\mu_1 = 3.12, s_1 = \frac{0.41}{\sqrt{330}} = 0.0226[/tex].
- [tex]\mu_2 = 3.18, s_2 = \frac{0.37}{\sqrt{550}} = 0.0158[/tex].
For the distribution of differences, they are given by:
- [tex]\overline{x} = \mu_1 - \mu_2 = 3.12 - 3.18 = -0.06[/tex].
- [tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0226^2 + 0.0158^2} = 0.0276[/tex].
What is the test statistic?
The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence:
[tex]z = \frac{\overline{x} - \mu}{s}[/tex]
z = -0.06/0.0276
z = -2.17.
What is the decision rule?
Using a z-distribution calculator, considering a two-tailed test, as we are testing if the means are different of value, with z = -2.17, it is found that since the p-value of the test is less than 0.05, the data provides sufficient evidence to conclude that there is a difference in average GPA between the population of sorority students and the population of unaffiliated female students on this university campus.
More can be learned about the z-distribution at https://brainly.com/question/13873630
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