Is the evidence sufficient to conclude change in average IQ

Question

This seems like an easy statistics question. But, though, I wanted the complete details of it, as to the why's of the problem . The problem is:

The average adult population IQ is estimated to be about $100$ with a standard deviation of about $15$. Now, a new study was made on a sample of $70$ students(randomly selected from the adult population) and the resulting average was found to be $108$. Can we conclude that the average IQ of the population has really changed?

Well, according to me, the sample size is too less(intuitively, would like to know why in terms of confidence intervals or some similar stuff) and the new average being with the standard deviation, the evidence is insufficient to conclude anything. Any ideas. Thanks beforehand.

Answer 1

In the real world, you would have to care about how the sample was done, and whether the sample is representative. It has to be a very small population if a sample of 70 could ever be representative. In terms of a purely statistical question, you are looking for a hypothesis test.

We assume that IQ is approximately normally distributed, hence you can use a Z-test. First state null and alternative hypothesis.

Null hypothesis: $\mu \leq \mu_0 = 100$.

Alternative hypothesis: $\mu > \mu_0 = 100$.

Next, choose level of significance. Often with humans we use $\alpha = 0.05$ or sometimes $\alpha = 0.01$. Then

$$Z = \frac{\overline{X}-\mu_0}{\sigma/\sqrt{n}} = \frac{108-100}{15/\sqrt{70}}\approx 4.4622.$$

Then we reject null hypothesis if $Z> z_\alpha$, where $z_\alpha$ is such that the probability is $\alpha$ that it will be exceeded by a random variable having the standard normal distribution, i.e. $\alpha = P(Z > z_\alpha )$. This can be calculated as $$z_{0.01} = 2.326347874\quad\text{ and }\quad z_{0.05} = 1.644853627.$$

Since $Z = 4.4622 > z_\alpha$ we can reject the null hypothesis.

Answer 2

If there is no sample standard deviation, and the sample size is bigger than 50, so then we would tend to use the Z test rather than the T test.

$$\text{standard error}=SE=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{70}}=1.793$$

$$Z=\frac{\bar x-\mu}{SE}=\frac{108-100}{1.79}=4.462$$

Looking this up on a z score table, this gives a P value of about 99.75%.

Standard critical P values (one tailed tests) are 95% if we are being a bit sloppy, and 99% if we are being careful. 99.75% is above the 99% being careful value P value so someone could be pretty confident that this wasn't just a result of things randomly working out.

An ordinary person would say that was high enough to say that there was a change.

In statistics-land, however, they are super careful about false positives and so we reject the null hypothesis, which means the new sample is inconsistent with the previous result.