In: Statistics and Probability
An eel supply company has two large vats of adolescent eels (called ’elvers’). One vat contains American eels, and the other contains European eels. An eel researcher suspects that American eels have a larger number of scales than European eels. Let µAm be the population mean for American, and µEu be the population mean for European. The species of eels can be considered independent. The researcher visits the company and uses an ingenious method to get random samples from each vat (details omitted) and counts the number of scales on each selected elver. Suppose the data from each sample supports the assumption that both species scale counts are approximately normal. The sample statistics are summarized below:
Type Sample Size Sample Mean Sample Variance
American 8 220 21
European 5 204 23
(a) Explain why a two sample t test with an equal variance assumption is reasonable to use here. List the null and alternative hypothesis.
(b) At α = 0.1, find the rejection region in the scale of t and use it to make a reject or not reject decision based on the observed test statistic. Then make a conclusion in the context of the problem.
(c) Decide the same test based on computing the p-value (by hand) and comparing it to α =0.1.
(d) Suppose we instead wanted to test: H0 : µAm −µEu =10 vs. HA : µAm −µEu > 10
If the same data is used for both tests, would the p-value for this test be larger or smaller than the p-value that was computed for the test in part (a)? Explain your answer. You do not need to actually compute either p-value to answer this question. (e) Suppose instead we choose to perform a Welch’s t test. What does that mean for the assumptions we are making? Perform a Welch’s T test for the same data (also by hand). How much do the p values differ?
Solution
Let Xbar,Ybar; s1,s2 ; n1 ,n2 ; be sample averages; sample standard deviations and sample sizes for ‘American’ and ‘European’ respectively.
Part (a)
The ratio of (s22/s12) = 1.2 is too small to be significant suggesting that the population variances are equal. [Fcrit = 6.09]. And we are concerned with equality the two population means. These justify the selection of 2-sample t-test with an equal variance assumption. Answer 1
Hypotheses:
Null: H0: µAM = µEU Vs Alternative: HA: µAM > µEUAnswer 2
where as given, µAM = the population mean for American and µEU = the population mean for European. ‘An eel researcher suspects that American eels have a larger number of scales than European eels.’ justifies the Alternative.]
Part (b)
Test Statistic:
t = (Xbar - Ybar)/[s√{(1/n1) + (1/n2)}]
= 6.021 Answer 3
tcrit = 1.363 Answer 4
Decision: Since tcal > tcrit, H0 is rejected. Answer 5
Conclusion: There is sufficient evidence to suggest that the researcher’s claim that American eels have a larger number of scales than European eels is valid. Answer 6
Calculation Details
n1 |
8 |
|
n2 |
5 |
|
Xbar |
220 |
|
Ybar |
204 |
|
s1 |
4.5826 |
|
s2 |
4.7958 |
|
s^2 |
21.7273 |
|
s |
4.661256 |
|
tcal |
6.021094 |
|
α |
0.05 |
|
tcrit |
1.36343 |
|
p-value |
|
Part (c)
p-value = 0.00005433 which is much smaller than the given significance level of 0.1. H0 is rejected. Answer 7
Part (d)
Test Statistic is actually t ={(Xbar - Ybar) – (µAM - µEU)}/[s√{(1/n1) + (1/n2)}]
Since under the earlier null hypothesis, (µAM - µEU) = 0, it had been dropped.
With the revised hypothesis, (µAM - µEU) would be replaced by 10 and hence all other things remaining the same, t value will come down.
Now, p-value = P(t11 > tcal). Since revised tcal < initial tcal, this probability will be larger. Thus, all other things remaining the same, the p-value would increase. Answer 8
DONE