Answer:
Step-by-step explanation:
Since n < 30, we will find the t-score and compare that to the t-score of a significance level of 1%.
Since they are asking if the difference is significant, we will have a two tailed test, with degree of freedom being 22, so our critical values are
t < -2.704 and t > 2.704
Our t-value for this situation is
t = ([µ + 5.8] - µ)/(1.4/√23)
It's µ + 5.8 because the problem told us that their levels are 5.8 mm higher than the average, so it's the average, plus 5.8
Simplify the equation...
t = 5.8/(1.4/√23)
t = 19.868
19.868 > 2.704, the evidence supports that there is significant difference between the sample and the population
