Answer:
The z-score is 3.03.
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 62[/tex]
The alternate hypotesis is:
[tex]H_{1} > 62[/tex]
The z-score is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
Spencer sampled 50 students of a private school who were questioned about their scores in Mathematics.
This means that [tex]n = 50[/tex]
Spencer wants to test the hypothesis that the private school students score better than the general public which has an average of 62 marks with a population standard deviation of 7 marks.
This means that [tex]\mu = 62, \sigma = 7[/tex]
If the sample mean is 65 marks, what is the z-score
This is z when [tex]X = 65[/tex]. So
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{65 - 62}{\frac{7}{\sqrt{50}}}[/tex]
[tex]z = 3.03[/tex]
The z-score is 3.03.
