Answer to Question #347311 in Statistics and Probability for rys

Question

the mathematics teacher claims that the mean iq of statistics students is 110 with standard deviation of 12. The mean IQ of 28 randomly selected statistics students is 112 . Test the difference of the population and sample means at 5%level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=110$

$H_1:\mu\not=110$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{112-110}{12/\sqrt{28}}=0.8919$$

Since it is observed that $|z|=0.8919<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(z>0.8919)=0.372447,$ and since $p= 0.372447>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 110, at the $\alpha = 0.05$ significance level.