Question

Spencer sampled 50 students of a private school who were questioned about their scores in Mathematics. 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. z equals fraction numerator x with bar on top minus mu over denominator begin display style fraction numerator sigma over denominator square root of n end fraction end style end fraction If the sample mean is 65 marks, what is the z-score

Answer 1

Answer:

The z-score is 3.03.

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 62$

The alternate hypotesis is:

$H_{1} > 62$

The z-score is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ 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 $n = 50$

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 $\mu = 62, \sigma = 7$

If the sample mean is 65 marks, what is the z-score

This is z when $X = 65$. So

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{65 - 62}{\frac{7}{\sqrt{50}}}$

$z = 3.03$

The z-score is 3.03.