Answer :
Using the z-distribution, it is found that since the value of the test statistic is greater than the critical value for the right-tailed test, there is sufficient evidence to support the principal’s claim.
At the null hypothesis, it is tested if they are not above average, that is, their mean is of 500 or below, hence:
[tex]H_0: \mu \leq 500[/tex]
At the alternative hypothesis, it is tested if they are above average, that is, their mean is of more than 500, hence:
[tex]H_1: \mu > 500[/tex]
We have the standard deviation for the population, thus, the z-distribution is used. 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.
For this problem, the values of the parameters are: [tex]\overline{x} = 541, \mu = 500, \sigma = 100, n = 50[/tex]
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{541 - 500}{\frac{100}{\sqrt{50}}}[/tex]
[tex]z = 2.9[/tex]
The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with the standard 0.05 significance level, is of [tex]z^{\ast} = 1.645[/tex].
Since the value of the test statistic is greater than the critical value for the right-tailed test, there is sufficient evidence to support the principal’s claim.
A similar problem is given at https://brainly.com/question/25728144