The average growth of a certain kind of tree is 6.9 inches in three years. A biologist claims that a new kind will have greater three-year growth. A random sample of 25 of the new variety has an average three-year growth of 13.7 inches and a standard deviation of 2.05 inches. The appropriate alternative hypotheses to test the biologist's claim are: O a. Ha: p>6.9 O b. Ha: u>2.05 OC. Ha: p>41.10 O d. not given in the question O e. Ha: u>13.7

The average growth of a certain kind of tree is 6.9 inches in three years. A biologist claims that a new kind will have greater three-year growth. A random sample of 25 of the new variety has an average three-year growth of 13.7 inches and a standard deviation of 2.05 inches. The appropriate alternative hypotheses to test the biologist's claim are: O a. Ha: p>6.9 O b. Ha: u>2.05 OC. Ha: p>41.10 O d. not given in the question O e. Ha: u>13.7

Name: The average growth of a certain kind of tree is 6.9 inches in three y
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: The average growth of a certain kind of tree is 6.9 inches in three years. A biologist claims that a new kind will have greater three-yeargrowth. A random sample of 25 of the new variety has an average three-year growth of 13.7 inches and a standard

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Similar questions

The incubation time for a breed of chicks is normally distributed with a mean of 29 days and standard deviation of approximately 3 days. Look at the figure below and answer the following questions. If 1000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.) (a) in 23 to 35 days __ chicks(b) in 26 to 32 days __chicks
Solve question 1 last part b where the question states... Probability that the total amount of time required to make an oil change in all cars from the sample of 100 cars is more than 3500 minutes is________
Question 1 produced Experience has shown that the weight of avocados by an orchard in Hoedspruit is normally distributed, with an average weight (u) of 1609, and a standard deviation (a) of 45g. A sample of 120 of these avocados is randomly selected. Determine the standard error for this sample. a) 0.45 4.29 C) 4.1079 d) 0.4091 e) 4.50
Questions 20 - 21 refer to the problem below. The recovery time from COVID19 is normally distributed with a mean of 12 days and a standard deviation of 2 days. 20. What is the probability that a randomly selected COVID19 patient recovers within a week? A. 0.0049 B. 0.0051 C. 0.0062 D. 0.0581 21. What is the probability that a randomly selected COVID19 patient takes between 9 and 15 days to recover? A. 0.8152 B. 0.7831 C. 0.9275 D. 0.8664
The incubation time for a breed of chicks is normally distributed with a mean of 29 days and standard deviation of approximately 3 days. Look at the figure below and answer the following questions. If 1000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.) (a) in 23 to 35 days chicks(b) in 26 to 32 days chicks(c) in 29 days or fewer chicks(d) in 20 to 38 days chicks
Question 4 Suppose a random sample of size n- 100 has been selected and the sample mean in found to be x 67. The population standard deviation is assumed to be a 12. What is the 95% confidence interval estimate of u? O 164.658. 69 2521 O163 648 70.352) O 164.648, 69.352I C (64.748, 69 452)
Suppose that the weights of people who work in an office building are normally distributed with a mean of u = 165 Ib. and a standard deviation of o = 25 lb. What is the chance that the total weight of 5 people is more than 1000 Ibs., i.e., what is P(X1+ ... + X5 > 1000)? [Hint: try to express this as an X-style problem.
Question 7.6.9 The packaging on the light bulbs you ordered for your store says they have a mean time to burnout of 200 hours, with a standard deviation of 18 hours, but the shape of the distribution is unknown. Someone buys 36 bulbs. a) Assuming the 36 bulbs can be considered a random sample, what is the distribution of the sample mean? N i b) What is the probability that the average time to burnout of the 36 bulbs will be less than 195 hours? Round your answer to 4 decimal places. The probability is i
Question 4. a. On a random day, a store can sell a mean value of 32 pasta packages and a standard deviation of 4. What is the probability that the total number of pasta packages sold by the store for a random sample of 58 days is less than 1862? Explain. Let b. X₁, X2, X3, and X4 represent the weight of shipment packages at a certain shipment facility. Suppose they are independent normal random variables with means M₁ = 0.8, M₂ = 2.6, μ3 = 8.4, µ4 = 2.3 pounds and variances . Find o² = 0² = 0} = 0} = 1 P(3X₁ + 1X2 + 1X3 + 2X4 ≤ 16.8)