Problem 3. (Finding Fish Lengths) [13 pts] Marine biologists at the Morro Bay Ma
ID: 3297902 • Letter: P
Question
Problem 3. (Finding Fish Lengths) [13 pts] Marine biologists at the Morro Bay Marine Life Institute are interested in whether blue tang fish are longer on average than clownfish. They collect a random sample of 30 clownfish, as well as a random sample of 40 blue tang fish (independently of one another). They then measure the length (in cm) of each fish in their two samples. The results of their findings are summarized below:
(a) Should we assume pooled variances here? Why or why not?
(b) Conduct a hypothesis test to assess whether blue tang fish are longer on average than clownfish. Use = 0.01. (c) Are the assumptions met for your hypothesis test in part (b) to be considered valid? Explain.
Explanation / Answer
a) For nearly same sample sizes, the rule of thumb is to check if the ratio of sample standard deviations falls within 0.5 to 2. If the condition is met, use poooled variance. For 30 Clownfish and 40 Blue tang fish (sample sizes are not equal) and s1/s2=10.2/4.4=2.32>2, therefore, donot use pooled variance.
b) Hypotheses: The null and alternative hypotheses are as follows:
H0:muCF-muBTF=0 (there is no difference in mean length of Clownfish and Blue Tang Fish)
H1:muCF-muBTF<0 (mean length of Blue Tan Fish is longer than Clown Fish)
Assumptions: Independent group assumption- Randomizing the experiment gives independent groups.
Independence assumption-The data have been collected for suitable randomization, hence length of Clownfish and Blue tang fishes are independent of each other.
Randomization condition-The experiment was randomized.
Nearly normal condition: In absence of raw data, this condition cannot be verified.
The assumptions ar emet, so use Student's t model to perform a two sample t test.
Test statistic:
t=(xbarCF-xbarBTF)/sqrt[s^2CF/nCF+s^2BTF/nBTF], where, xbar denotes sample mean, s denote sample standard deviation, and n denote sample size.
=(29.3-29.7)/sqrt[10.2^2/30+4.4^2/40]
=-0.20
p value at 37 df is 0.421
computation of degrees of freedom, df=[{(s1^2/n1)+(s2^2/n2)}^2/{1/n1-1(s1^2/n1)^2+1/n2-1(s2^2/n2)^2}]
Substitute the values assuming 1 to denote Clownfish and 2 to denote Blue tang fish.
Decision: Per rejection rule based on p value, reject null hypothesis if p value is less than alpha=0.01. Here, p value is not less than 0.01, therefore, fail to reject null hypothesis. There is insufficient sample evidence to conclude that blue tang fishes are longer on average than clownfish.
c) The independent group assumption, independence assumption, randomization condition are reasonably met. In absence of raw data the nearly normal condition cannot be checked.
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