Question

5. (8 points) Suppose you work for the Department of Natural Resources and you want to estimate, with 95% confidence, the mean (average) length of all walleye fingerlings in a fish hatchery pond. You take a random sample of 36 fingerlings and determine that the average length is 7.5 inches and the population standard deviation is known to be 2.1 inches. a. Construct a 95% confidence interval. b. Interpret the Confidence Interval: b. If the hatchery supervisor claimed that the average length of the walleye fingerlings fish is 7.6 inches, would you agree or disagree?

Show all work and calc functions. For each Hypothesis test you need to use the p-value method and your calculator, write (i) H₀, H_a, (ii) Identify the Original Claim (iv) Identify the Random Variable & Distribution, (v) Calculator function used with values inputted to find the P-value (vi) the p-value with ,Decision and Conclusion

Answer 1

a.z score for 95% confidence interval is 1.96 as P(-1.96<z<1.96)=0.95

$E=z*\frac{\sigma}{\sqrt{n}}=1.96*\frac{2.1}{\sqrt{36}}=0.686$

$CI=\overline{x}\pmE=7.5\pm0.686=(6.814,8.186)$

So for 95% confidence interval

$6.814<\mu<8.186$

b.So we can say with 95% confidence that population mean will between 6.184 and 8.186

c.As the 7.6 inches lies between the range 6.184 and 8.186 we agree with the claim that average length is 7.6 inches