Uday M. answered 07/30/19
M.S. Engineering, 5+ years of teaching experience
Here, we are dealing with means, and since normality has not been implied it can be assumed that we are doing a t-test (as opposed to a z-test). Remember the meaning of the p-value: if the mean was actually 30.6, we would observe a sample empathy score mean more extreme than 39.5 about 100p% of the time. So the p-value is simply the area under the curve at the tail end.
The implication is that we need to do a one-sided t-test, although this is not very clear. Meaning, we would want to see the p-value for getting an empathy score *greater* than 39.5. (The alternative approach would be a two-sided t-test, where we would find the p-value for getting an empathy score *more extreme* than 39.5, meaning that it could either be greater than 39.5 or less than 21.7).
H0: mu = 30.6
Ha: mu > 30.6
First, let's calculate the test statistic.
t = (sample mean - null hypothesis mean)/(stdev/sqrt(sample size))
t = (39.5 - 30.6)/(6.6/sqrt(25)) = 6.74
Now to find the p-value.
p-value = probability that t > test statistic
p-value = P(t > 6.74)
You can compute this probability using the tcdf function on a calculator or by looking in a table of values for the t-distribution. You will need the degrees of freedom, which is n - 1 (where n is the sample size). A sample calculator input and output would be
p-value = tcdf(6.74, inf, 24) = 0.00001
This is a very low p-value, which makes sense, because the sample mean is quite far from the population mean. We are not likely to get values more extreme than this if we were to take several additional samples. Specifically compared to the alpha level (significance level),
p-value < alpha
0.00001 < 0.05
Therefore, we reject H0, and state that the sample mean does provide good evidence that the bullied individuals' empathy score is greater than that of the general population.