In this problem, we're testing for statistically significant difference in the mean waterfall. Heights between Europe and Asia were given a sample of size and nine for European waterfalls and nine Asian waterfalls as well without going through the tedium of showing you the actual calculations. The X bar for the European waterfalls is 883.2 feet. That's the X bar for the Asian waterfalls is 840 0.3. We are not given Population Sigma's, so we have to calculate those from the sample is well doing.
So we get for the European waterfalls. We get a sample standard deviation okay for 3 87.2 and for the Asian waterfalls for 77.7. Now, even though the sample size is less than 30 it's reasonably safe to assume that the underlying population of waterfalls in each continent is normally distributed, and so therefore, we can take our typical approach of developing a test statistic of the form sample. Me minus my hypothesized expected value of the mean standardized by dividing by the standard error right now are no hypothesis in this circumstance is simply that the population means are equal to each other and in this case are alternative hypothesis is that they are different. That is to say, we are not making any assumption about if they're different in what direction or which one would be greater or lesser.
We're simply saying they're either. Where are null hypothesis is that on average, waterfalls are justus high in Europe, as they are in Asia are alternative hypothesis is that they're not that there is a statistically significant difference between the two continents. That being the case are specific. Test statistic in this case turns into this right. The difference of the sample means minus the difference in the expected population means and divided by the standard error, which, based on sample standard deviations, looks like this.
But because our no hypothesis is that these two means air equal, this term is going to translate to zero will simply go away. So therefore, our test statistic is simply going to be. The difference in the sample means over this standard error formulation. Great now, because we're not taking a position on which direction which continent would be greater or lesser. This is a to tail test, and the book specifies that we use a 10% or Alfa equals 100.10 significance level right now for the student t distribution.
We need to know that the degrees of freedom of the distribution and in this case, degrees of freedom, is going to equal the lesser off the respective sample sizes each of minus one in each case. Now, in this case, both of sample sizes equal nine they're four degrees of freedom are nine minus one. In this case or eight, if I have a two tailed T distribution was an Alfa of 0.1 and eight degrees of freedom. This gives me my critical value. This could be looked up, a variety of ways to get this the critical values plus or minus one point feet +595 And therefore, if we were to visualize, this would have.
So any values of the test statistic falling in this range would result in are not rejecting the null hypothesis, which is to say we would not be able to say with any statistical significance that the mean waterfall heights are different between the two continents Now, as it turns out, if we took all of those values, we had up above and plugged it into the test statistic formula. We would come up with a T statistic in this case of 0.2093 which is approximately there and clearly in the range where we do not reject are no hypothesis..