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First of all, we need to determine the system of hypothesis, which must have a null and alternative hypothesis.
In all hypothesis proof procedure, we must work with the null one, rejecting it or accepting it. If we reject the null hypothesis, it will mean that our principal hypothesis is true.
According to the problem, the null hypothesis must be: ; because it says that 100 is the normal mean that is used.
The alternative hypothesis must be: ; because the question is asking if there's enough evidence to say that they children have a higher IQ than 100.
Then, we must calculate the P-value, which is the probability value that we are gonna use to accept or reject the null hypothesis. This P-value we can calculate it with the formula: ; where x is the sample mean, u is the hypothetical mean parameter, o is the sample deviation and n is the sample.
Extracting all values from the problem: x = 104.3; u = 100; o = 14 and n = 20.
Applying all values to the formula:
Then, this Z-value will give us the P-value by using the Z-table with the 0.05 level of significance. In table attached, we can see that the result is 0.9265. However, the P-value is 1 - 0.9265 = 0.735. We need to do this subtraction because, the result 0.9265 belong to the bigger area (as it's shown is the image in blue colour), and we need the smaller area, which belong to the 0.05 level of significance.
<em>In this problems, if the P-value is more than the level of significance, the null hypothesis is accepted. If the P-value is less than the level of significance the null hypothesis is rejected.</em>
Therefore, there is no enough evidence to say that the sample children have higher IQs, because the P-value is higher than the level of significance. (0.735>0.05).
