Answer to Question #157980 in Statistics and Probability for Ratul Das

Question

Consider the following ten IQ scores. IQ test scores are scaled to have

a mean of 100 and a standard deviation of 15.

65 98 103 77 93 102 102 113 80 94

Test the hypothesis that the mean is 100.

Answer 1

$$\bar{x}=\dfrac{65+ 98 +103+ 77 +93 +102 +102+ 113+ 80 +94}{10}$$

$$=\dfrac{927}{10}=92.7$$

$\sigma=15$

The provided sample mean is $\bar{x}=92.7$ and the known population standard deviation is $\sigma=15,$ and the sample size is $n=10.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=100$

$H_1:\mu\not=100$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{92.7-100}{15/\sqrt{10}}=-1.539$$

Since it is observed that $|z|=1.539<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 100, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=2P(z<-1.539)=0.123804,$ and since $p=0.123804>0.05=\alpha,$ 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 100, at the 0.05 significance level.