Then, x bar, our sample mean, is distributed nearly normally with mean

equal to 100 the null value and standard area equal to s over square root of n.

S happens to be 6.5 here divided by square root of 36.

We get a standard error of 1.083.

So now we know the shape, the center, and the spread of our sampling distribution.

Which is going to come in handy when we're trying to

draw the sampling distribution, shade the p-value, and calculate the test statistic.

We have a nearly normal distribution, centered at 100.

And since the p-value is defined as the probability

of observed or more extreme outcome Given the null hypothesis

is true, we need to shape the tails of the curve beyond our observed mean of 118.2.

So, we want to shade here at the higher end.

However, since the alternative hypothesis is two-sided, we also need

to shade the symmetric tail area on the other side.

The cutoff for the lower tail will be equidistant

to the mean as the cutoff for the upper tail.

And since 118 minus 100 is 18.2, we subtract 18.2 from

100, and the lower lower tail, find the lower tail cutoff to be 81.8.

And we want to shade everything below that as well.

We can already see that these tail areas will be tiny.

But let's formally solve for the p-value for completeness anyway.

First,

we calculate a z score as the observed sample mean 118.2 minus the null

value 100 divided by the standard error. 1.083.

Our z-score comes out to be 16.8.

Such a high z-score means that the

observation is unusually far from the center, and

hence, the p-value, which is going to be the sum of the two tail areas here,

will be approximately 0.

Lastly we need to make a decision and interpret it in context of the data.

This smaller the p value, the stronger the evidence against the null.

And since we have a tiny p value, we have very strong evidence against the

null hypothesis, therefore, we reject the null hypothesis

and conclude that the data provide convincing evidence.

Of a difference between the average IQ score of mothers of

gifted children and the average IQ score for the population at large.