Respuesta :
Using the z-distribution, as we have the standard deviation for the population, it is found that there is no significant evidence that the group of teachers that could choose supplementary curriculum had a higher level of job satisfaction.
What are the hypothesis tested?
At the null hypothesis, it is tested if the mean satisfaction level of the group of teachers that could choose supplementary curriculum is the same, of 3.3, that is:
[tex]H_0: \mu = 3.3[/tex]
At the alternative hypothesis, it is tested if the group of teachers that could choose supplementary curriculum had a higher level of job satisfaction, that is:
[tex]H_1: \mu > 3.3[/tex]
What is the test statistic?
The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] is the value tested at the null hypothesis.
- [tex]\sigma[/tex] is the standard deviation of the sample.
- n is the sample size.
In this problem, the values of the parameters are:
[tex]\overline{x} = 3.5, \mu = 3.3, \sigma = 0.6, n = 40[/tex]
Hence, the value of the test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{3.5 - 3.3}{\frac{0.6}{\sqrt{40}}}[/tex]
[tex]z = 2.11[/tex]
What is the decision?
Considering that we have a right-tailed test, as we are testing if the mean is greater than a value, and a significance level of 0.01, the critical value is of [tex]z^\ast = 2.327[/tex].
Since the test statistic is less than the critical value, it is found that there is no significant evidence that the group of teachers that could choose supplementary curriculum had a higher level of job satisfaction.
More can be learned about the z-distribution at https://brainly.com/question/26454209