ponse. Question 3 An airline claims that the no-show rate for passengers is less than 8%. In a sample of 416 randomly selected reservations, 17 were no-shows. At a = 0.01, test the airline's claim. P-value 0.001 < 0.01; reject H0; There is not enough evidence to support the airline's claim, that is less than 8% O A. P-value = 0.002< 0.01; reject HO; There is not enough evidence to support the airline's claim, that is less than 8% O B. P-value = 0.123 > 0.01; do not reject H0; There is not enough evidence to support the airline's claim, that is less than 8% OC. P-value = 0.156>0.01; do not reject HO; There is not enough evidence to support the airline's claim, that is less than 8% OD.

ponse. Question 3 An airline claims that the no-show rate for passengers is less than 8%. In a sample of 416 randomly selected reservations, 17 were no-shows. At a = 0.01, test the airline's claim. P-value 0.001 < 0.01; reject H0; There is not enough evidence to support the airline's claim, that is less than 8% O A. P-value = 0.002< 0.01; reject HO; There is not enough evidence to support the airline's claim, that is less than 8% O B. P-value = 0.123 > 0.01; do not reject H0; There is not enough evidence to support the airline's claim, that is less than 8% OC. P-value = 0.156>0.01; do not reject HO; There is not enough evidence to support the airline's claim, that is less than 8% OD.

Name: A Moving to another question will save this response.Question 3An air
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: A Moving to another question will save this response.Question 3An airline claims that the no-show rate for passengers is less than 8%. In a sample of 416randomly selected reservations, 17 were no-shows. At a = 0.01, test the airline's claim.P-value

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A dietician read in a survey that 70.5% of adults in the U.S. do not eat breakfast at least 3 days a week. She believes that the proportion that skip breakfast 3 days a week is different than 0.705. To verify her claim, she selects a random sample of 71 adults and asks them how many days a week they skip breakfast.. 44 of them report that they skip breakfast at least 3 days a week. Test her claim at a = 0.10. The correct hypotheses would be: О Но:р 0.705 (claim) Но: р > 0.705 HA:p < 0.705 (claim) О Но: р — 0.705 HA:P + 0.705 (claim) Since the level of significance is 0.10 the critical value is 1.645 and -1.645 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.705. O There is not enough evidence to reject the claim that the proportion that skip…
In a fishing lodge brochure, the lodge advertises that 75% of its guests catch northern pike over 20 pounds. Suppose that last summer 71 out of a random sample of 86 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from 75% (either higher or lower)? Use ? = 0.05. (a) What is the level of significance? State the null and alternate hypotheses. H0: p = 0.75; H1: p > 0.75H0: p ≠ 0.75; H1: p = 0.75 H0: p < 0.75; H1: p = 0.75H0: p = 0.75; H1: p ≠ 0.75H0: p = 0.75; H1: p < 0.75 (b) What sampling distribution will you use? The Student's t, since np > 5 and nq > 5.The standard normal, since np > 5 and nq > 5. The Student's t, since np < 5 and nq < 5.The standard normal, since np < 5 and nq < 5. What is the value of the sample test statistic? (Round your answer to two decimal places.) (c) Find the P-value of the test…
a survey is conducted to compare the proportions of men and women whoaccess their bank statements online. Denoting the population propotions by Pm and Pw, a two-proportion z-test is carried out to test Ho: pm = pw against Ha: pm > pw. The values of the test statistic is found to be z = 0.784, and the p-value for the test is found to be 0.216. Which of the following is the correct interpretation of the p-value? a) given that ha is true, the probability of getting a value of z at least as large as 0.784 is 0.216. b) given that ho is true, the probability of getting a value of z at least as large as 0.784 is 0.216. c) given the results of the survey, the probability that Pm > Pw is 0.216. d) given the results of the survey, the probability that Pm = Pw is 0.216. e) none of the above
It has been claimed that at UCLA at least 40% of the students live on campus. In a random sample of 250 students, 90 were found to live on campus. Does the evidence support the claim at = .01?
It is bleieved that the percent of convicted felons who have a history of juvenile deliquency is 70%. Is there evidence to support the claim that the actual percentage is more than 70% if out of 200 convicted felons, we find that 154 have a history of juvenile deliquency? Set the level of significance at a = 0.05
It has been reported that 9% of Canadians have seen a doctor last month. A skeptical researcher believes this is not true and decides to test the claim. If you were to carry out a test for the claim, identify the following. (a) The Type I error in this context. O The error of failing to reject the claim that the true proportion is equal to 9% when it is actually not equal to 9%. O The error of failing to reject the claim that the true proportion is equal to 9% when it is actually equal to 9%. O The error of rejecting the claim that the true proportion is equal to 9% when it is actually equal to 9%. O The error of accepting the claim that the true proportion is equal to 9% when it is actually equal to 9%. O The error of rejecting the claim that the true proportion is equal to 9% when it is actually not equal to 9%. (b) The Type Il error in this context. O The error of rejecting the claim that the true proportion is equal to 9% when it is actually not equal to 9%. O The error of…
An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, 19 were no-shows. At =0.01, test the airlines claim.
In a sample of 47 adults randomly selected from one town, it is found that 9 of them have been exposed to a particular strain of the flu. Find the P-value for a test of the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is 8%. a. 0.005 O b. 0.05 OC. 0.002 O d. 0.08
Q1
Suppose "Naturally Gassy" magazine claims that 35% A hypothesis test is to be performed to determine if the natural gas is higher than the magazine's claim. A surve natural gas.
Suppose that the researchers wish to use a maximum error of 0.05 at the 90% confidence level, what is the minimum number of homes (round up to the nearest whole number) that they need to visit? Formula: E=zα/2√p¯(1−p¯)/n Hint: Isolate n on one side using algebra. Confidence Level 1−α Significance Level α α/2 zα/2 90% = 0.90 0.10 0.05 1.645 95% = 0.95 0.05 0.025 1.96 98% = 0.98 0.02 0.01 2.33 99% = 0.99 0.01 0.005 2.576
K A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, 14% chose chocolate pie, and the margin of error was given as ±4 percentage points. Describe what is meant by the statement that "the margin of error was given as ±4 percentage points." 身 Choose the correct answer below. OA. The statement indicates that the study is 100% -4% = 96% confident that the true population percentage of people that prefer chocolate pie is 14%. OB. The statement indicates that the true population percentage of people that prefer chocolate pie is in the interval 14% ±4%. OC. The statement indicates that the interval 14% ± 4% is likely to contain the true population percentage of people that prefer chocolate OD. The statement indicates that the study is only 4% confident that the true population percentage of people that prefer chocolate pie is exactly 14%.