Question

1. The average score in the entrance examination in Mathematics at Sto.

Rosario National High School is 80 with a standard deviation of 10.A

random of 40 students was taken from this year's examinees and

was found to have a mean score of 84.

Is there a significant difference between the known mean and the

sample mean? Test at a= 0.05

Solution :

Step 1. H0: = 80 : There is no significant difference

hypothesized and the sample mean.

H1: $\neq$ 80 :

Step 2. Level of significance. a = 0.05.

Step 3. two tailed test, find the critical value. Zt =

4. Compute the test-statistic value:

5. Step :

6. Conclusion:

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=80$

$H_1:\mu\not=80$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{84-80}{10/\sqrt{40}}=2.5298$$

Since it is observed that $|z|=2.5298>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z>2.5298)=0.011413,$ and since $p= 0.011413<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 80, at the $\alpha = 0.05$ significance level.