Question

Set A: 250.66, 250.22, 251.40, 252.78, 254.39, 254.71, 250.36, 249.17, 249.28, 248.61, 245.90 Set B (have...

Set A: 250.66, 250.22, 251.40, 252.78, 254.39, 254.71, 250.36, 249.17, 249.28, 248.61, 245.90 Set B (have not added 0.5): 247.90, 250.66, 253.85, 250.87, 246.90, 246.63, 247.99, 249.43

3. Take the mean and standard deviation of data set A calculated in problem 1 and assume that they are population parameters (μ and σ) known for the variable fish length in a population of rainbow trouts in the Coldwater River. Imagine that data set B is a sample obtained from a different population in Red River (Chapter 6 problem!).

a) Conduct a hypothesis test to see if the mean fish length in the Red River population is different from the population in Coldwater River.

b) Conduct a hypothesis test to see if the variance in fish length is different in the Red River population compared to the variance in the Coldwater population.

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Answer #1

For set A

\bar x1  = 250.68, s1= 2.581, n1 = 11

For set B

\bar x2 = 249.28, s2 = 2.437, n2 = 8

A) H0: \mu1 = \mu2

H1: \mu1 \neq \mu2

The test statistic t = (\bar x1 - \bar x2)/sqrt(s1^2/n1 + s2^2/n2)

= (250.68 - 249.28)/sqrt((2.581)^2/11 + (2.437)^2/8) = 1.21

Df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

= ((2.581)^2/11 + (2.437)^2/8)^2/(((2.581)^2/11)^2/10 + ((2.437)^2/8)^2/7) = 16

At 0.05 significance level the critical value is t* = +/- 2.120

Since the test statistic value is not greater than the upper critical value(1.21 < 2.120), so we should not reject the null hypothesis.

So at 5% significance level there is not sufficient evidence to conclude that the mean fish length in the Red River population is different from the population in Coldwater River.

B) H0: \sigma1^2 = \sigma2^2

H1: \sigma1^2 \neq \sigma2^2

The test statistic F = s1^2/s2^2

= (2.581)^2/(2.437)^2 = 1.122

At alpha = 0.05, the critical values are

F(0.025, 10, 7) = 0.2532

F(0.975, 10, 7) = 4.7611

Since the test statistic value lies between the critical values (0.2532 < 1.122 < 4.7611), so we should not reject the null hypothesis.

So at 5% significance level there is not sufficient evidence to conclude that the variance in fish length is different in the Red River population compared to the variance in the Coldwater population

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