Respuesta :
Testing the hypothesis, we can conclude that since the p-value of the test is 0.0114 < 0.05, which means that the sample provides enough evidence to conclude that self-esteem scores for these adolescents are significantly different from those of the general population.
At the null hypothesis, we test if the mean is the same of the general population, that is:
[tex]H_0: \mu = 50[/tex]
At the alternative hypothesis, we test if it is different, that is:
[tex]H_1: \mu \neq 50[/tex]
The test statistic is given by:
[tex]z = \frac{M - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
For this problem, the parameters are:
[tex]M = 53.8, \mu = 50, \sigma = 15, n = 100[/tex]
Then, the value of the test statistic is:
[tex]z = \frac{M - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{53.8 - 50}{\frac{15}{\sqrt{100}}}[/tex]
[tex]z = 2.53[/tex]
Two tailed test, so the p-value is P(|z| < 2.53), which is 2 multiplied by the p-value of z = -2.53.
Looking at the z-table, z = -2.53 has a p-value of 0.0057
2(0.0057) = 0.0114
The p-value of the test is 0.0114 < 0.05, which means that the sample provides enough evidence to conclude that self-esteem scores for these adolescents are significantly different from those of the general population.
A similar problem is given at https://brainly.com/question/25147864
