4. Fish of a certain species live in two separate rivers, X and. A scientist cla

Question

4. Fish of a certain species live in two separate rivers, X and. A scientist claims that the mean length of fish in X is greater than the mean length of fish inY. To test his claim, he catches a random sample of 9 fish from X and a random sample of 7 fish from Y. The lengths of the 9 fish from X, in appropriate units, are as follows 5.3 12.0 15.1 11.2 14.4 13.8 12.4 11.8 12.5 The lengths of the 7 fish fromY, in the same units, are as follows. 5.0 10.7 13.6 12.4 11.6 12.6 13.2 i) Stating any assumptions that you make, test at the 5% significance level whether the mean length of fish in X is greater than the mean length of fish inY. 6 Marks ii) Find a 90% confidence interval for the mean difference in the lengths of the fish from the two rivers 5 Marks

Explanation / Answer

1)
a) The two populations have the same variance. b) This assumption is called the assumption of homogeneity of variance.
c) The populations are normally distributed.
Each value is sampled independently from each other value.

Hypothesis:
H0 : mu1 = mu2
Ha : mu1 > mu2
Test statistics:

x1 = 13.1667 , s1 = 1.5141 , n1 = 9 , x2 = 12.7286 , s2 = 1.3937 , n2 = 7

t = (x1 - x2) /sqrt(s1^2/n1+s2^2/n2)
= ( 13.1667 - 12.7286)/sqrt(1.5141^2/9 + 1.3937^2/7)
= 0.6005

P value = .2788

we cannot reject the null hypothesis.

2)
t value at 90% = 1.7613

CI = (x1 - x2) +/- t *sqrt(s1^2/n1+s2^2/n2)
= ( 13.1667 - 12.7286)+/- 1.7613 *sqrt(1.5141^2/9 + 1.3937^2/7)
= (-0.8468,1.7230)

The 90% CI is (-0.8468,1.7230)

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