Please help and explain in the easiest way to understand 4. Organizers of a fishing tournament have picked a large lake that they believe has a population ofthousands of fish with a population mean weight of 10 lbs. One of the competitors frequently fishes atthis lake. He believes the lake has been overused for tournaments and the larger fish have been removedbring down the average weight of fish still in the lake. The organizers agree to move the tournament ifhe can show the true mean weight is significantly less than 10 lbs.The fisherman takes a random sample of 40 fish caught in a recent tournament at this lake. The sampleaverage was 8.9 lbs. The population standard deviation of 3.5 lbs. Use a = 0.05 to test the claim.a. State the Null and Alternative Hypotheses.b. Calculate the Mean and Standard Error. Please show work.c. Use the calculator to find the P-value and Test Statistic. Please show work.d. What is your decision based on the p-value you found? Explain.e. State your conclusion in context to the problem.f. Find the 90% confidence interval for the population average weight of fish still in the lake.Does this interval support your conclusion from the hypothesis test? (Hint: Look to see if 10 lbsinside the confidence interval?)

Name: Please help and explain in the easiest way to understand 4. Organizers
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: Please help and explain in the easiest way to understand 4. Organizers of a fishing tournament have picked a large lake that they believe has a population ofthousands of fish with a population mean weight of 10 lbs. One of the competitors frequently

Note: Hi there! Thank you for posting the question. As there are multiple sub parts, according to our policy we have solved the first three sub parts. If you need help with other sub parts, please re post separately.a) Denote μ as the population mean weight of fish. It is needed to check, whether the population mean weight of fish is less than 10 lbs Null hypothesis: H0: μ = 10. Alternative hypothesis: H1: μ < 10. The test is one directional. Hence, it is one-tailed test (left tailed). b) The sample size is n = 40. The population standard deviation is σ = 3.5 lbs. The sample mean is an unbiased estimator of the population mean. The sample mean is, x̅ = 8.9 lbs. The standard error is, σ/√n = 3.5/√40 = 0.55339859. c) The z-test statistic is: z = (x̅ – μ)/ (σ/√n) = (8.9 – 10)/(3.5/√40) = –1.1/0.55339859 ≈ –1.988. The test statistic is –1.988. The P-value for the left-tailed test is: P-value = PHo (Z ≤ –1.988). Hence, PHo (Z ≤ –1.988) = 0.0234 [using the EXCEL formula: =NORM.S.DIST(-1.988,TRUE)]. Thus, the P-value is 0.023. Decision rule: If P-value is less than the level of significance, then reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Since p-value (= 0.0234) < α (= 0.05), reject the null hypothesis H0. Thus, at the α = 0.05 level, there is sufficient evidence to claim that the true mean weight is significantly less than 10 lbs.