Question

6. in a certain population of fish, in order to determine an estimate of the lengths of individual fish, a random sample of 100 fish is taken. Based on the sample, the mean length is 54.1 mm and the standard deviation is 4.5 mm. What is the standard error of the mean length of the sampled 100 fish?

A 5.41

B 0.541

C 0.45

D 0.045

7. What would be the t-critical value for the 95% confidence interval in problem 6 above?

A 2.28

B 2.08

C 1.98

D 1.88

8. What would be the Margin of Error for the 95% confidence interval in problem 6 above?

A 1.0242

B 0.8929

C 0.08929

D 0.01024

9. What would be the 95% confidence interval for the true mean length of fish in problem 6 above?

A (54.00, 54.20)

B (53.08, 55.12)

C (53.21, 54.99)

D None of these

Answer 1

Question 6

Here standard deviation of sample = s = 4.5 mm

sample size = n = 100

standard error =s/sqrt(n) = 4.5/sqrt(100)= 0.45 mm

Option C is correct here.

Question 7

Here dF = n - 1 =100 -1=99

tcritical = TINV(0.05,99) = 1.984

Option C is correct here.

Question 8

Margin of error = critical test statistic * standard error = 0.45 * 1.984 = 0.893

Option B is correct here.

Question 9

95% confidence interval for the true mean length of fish = 54.1 +- 0.893 = (53.21, 54.99)

option C is correct here