Answer to Question #339695 in Statistics and Probability for Agl

Question

1. According to the norms established for a history test, grade eight students

should have an average 81.7 with a standard deviation of 8.5. If 100 randomly

selected grade eight students from a certain school district average 79.6 in this

test, can we conclude at the 0.05 level of significance that grade eight students

from this school district can be expected to average less than the norm of 81.7?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge 81.7$

$H_a:\mu<81.7$

This corresponds to a left-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 left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z: z < -1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{79.6-81.7}{8.5/\sqrt{100}}\approx-2.4706$$

Since it is observed that $z = -2.4706 <-1.6449= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<-2.4706)=0.0067443,$ and since $p=0.006744<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 81.7, at the $\alpha = 0.05$ significance level.