Using the z-distribution, the observed test statistic is given as follows:
z = 3.69.
At the null hypothesis, it is tested if older students have the same mean as the general population, that is:
[tex]H_0: \mu = 115[/tex]
At the alternative hypothesis, it is tested if they have a greater mean, that is:
[tex]H_0: \mu > 115[/tex]
The test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which:
For this problem, the parameters are:
[tex]\overline{x} = 135.2, \mu = 115, \sigma = 30, n = 30[/tex]
Hence the test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{135.2 - 115}{\frac{30}{\sqrt{30}}}[/tex]
z = 3.69.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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