Preliminary data analysis indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test using the critical-value approach.%3DX = 44.2and s = 7.3Use a significance level of a = 0.05 to test the claim that u 35.8. The sample data consists of 10 scores for whichH:µ±35.80Н и-35.8Test statistics: t=3.64Critical Value: t=+/-2.262H:µ=35.8RejectThere is sufficient evidence to support the claim that the mean is different from 35.8.П. : и 3 44.20.H:µ±44.2a.Test statistics: t=3.64Critical Value: t= +/- 2.262H:µ=44.2Н,There is sufficient evidence to support the claim that the mean is different from 44.2.Reject H,:µ+44.2H:u=44.2aTest statistics: t=3.64Critical Value: t=+/-2.262Reject H:u=There is sufficient evidence to support the claim that the mean is different from 44.2H:µ3D35.8Н и+35.8Test statistics: t= 3.64Critical Values t=+/-2.262H, : µ=D35.8RejectThere is sufficient evidence to support the claim that the mean is different from 35.8.

Name: Preliminary data analysis indicate that it is reasonable to use a t-t
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: Preliminary data analysis indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test using the critical-value approach.%3DX = 44.2and s = 7.3Use a significance level of a = 0.05 to test the claim th

Answered: Image /qna-images/answer/d0eb62d9-99f9-4aa7-98c7-24dfd6a1294a.jpg

Math

Statistics

Preliminary data analysis indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test using the critical-value approach. Use a significance level of a = 0.05 to test the claim that u + 35.8. The sample data consists of 10 scores for which Y - 44 2. X = and s = 7.3 H, : µ + 35.8 H:u=35.8 a Test statistics: t=3.64 Critical Value: t=+/-2.262 H. : µ= 35.8 Reject There is sufficient evidence to support the claim that the mean is different from 35.8 H,: µ= 44.2 H:µ+44.2 Test statistics: t=3.64 Critical Value: t= +/- 2.262 Reject H.: u = 44.2 There is sufficient evidence to support the claim that the mean is different from 44.2.

Preliminary data analysis indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test using the critical-value approach. Use a significance level of a = 0.05 to test the claim that u + 35.8. The sample data consists of 10 scores for which Y - 44 2. X = and s = 7.3 H, : µ + 35.8 H:u=35.8 a Test statistics: t=3.64 Critical Value: t=+/-2.262 H. : µ= 35.8 Reject There is sufficient evidence to support the claim that the mean is different from 35.8 H,: µ= 44.2 H:µ+44.2 Test statistics: t=3.64 Critical Value: t= +/- 2.262 Reject H.: u = 44.2 There is sufficient evidence to support the claim that the mean is different from 44.2.

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

Use the random sample data to test the claim that the mean travel distance to work in California is less than 35 miles. Use 1% level of significance. Sample data: x¯=32.4 mis=8.3 min=35 Identify the tail of the test. Find the P-value Will the null hypothesis be rejected? Is the initial claim supported?
In the population of senior students at Shermer High School, the proportion who plan to attend college is 0.64. In a random sample of 100 students from this population, the proportion who plan to attend college is 0.75. We would call the value 0.64 a P-value. z-score. parameter statistic significance level.
One company produces premium grade carbon steel for Katana Sword manufacturers. Two different analytical methods were used to detect the contamination level of the carbon steel in 8 random specimensFrom the table provided, Is there sufficient evidence to conclude that tests differ in the mean contamination level use? α = 0.01?
Determine the Test statistic. Determine the P-Value. State the final conclusion that addresses the original claim.
The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n=761 and the test statistic is t=55.668. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.10. State the null and alternative hypotheses. H₁> 2.84 2.84 (Type integers or decimals. Do not round.) The test statistic is (Round to two decimal places as needed.)
A sample mean, sample size, and sample standard deviation are provided below. Use the one-mean t-test to perform the required hypothesis test at the 10% significance level. x=25, s=10, n=24, H0: μ=24, Ha: μ>24 The test statistic is t= (round to two decimals as needed)
Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below. H0:μ=8.2 seconds; Ha:μ<8.2 seconds α=0.04 (significance level) z0=−1.75 p=0.0401 Select the correct answer below: a. Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. b. Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. c. Reject the null hypothesis because the value of z is negative. d. Reject the null hypothesis because |−1.75|>0.04. e. Do not reject the null hypothesis because |−1.75|>0.04.
Q: Using the data (provided below this question) apply ANOVA to determine if the mean response for value for the dollar is the same for different types of cell phones. Use a 5% significance level if the problem doesn’t give an alpha. Note: Please only provide the t stat, p-value, and critical value, no need to explain the whole process. Gender Carrier Type Usage Signal strength Value for the Dollar Customer Service M AT&T Smart High 5 4 4 M AT&T Smart High 5 4 2 M AT&T Smart Average 4 4 4 M AT&T Smart Very high 2 3 3 M AT&T Smart Very high 5 5 2 M AT&T Smart Very high 4 3 5 M AT&T Smart Very high 3 4 4 F AT&T Smart Very high 3 2 3 F AT&T Smart Very high 4 3 4 M AT&T Smart Very high 3 3 1 M Other Smart Average 1 2 4 M Sprint Smart Very high 3 5 4 M Sprint Smart Very high 3 5 3 F Sprint Smart Average 2 5 4 F Sprint Smart Average 3 5 4 M Verizon Smart Average 4 3 3 F Verizon Smart Very high 4 3 2 M Verizon…
identify the test statistic. Identify the P-value.
I need help with this problem Conduct a hypothesis test about the mean cost per square foot at a chosen significance level. Reference an internet source for the current average (or median) cost per square foot of homes in Orange County. Discuss/explain your choice of significance level for your hypothesis test in #1 as it relates to the power of a test and Type I/II errors.
A teacher would like to determine if quiz scores improve after completion of a worksheet. The students take a pre-quiz before the worksheet and then another quiz after the worksheet. (Pre-quiz score - Post-quiz score) Assume quiz scores are normally distributed. The grades for each quiz are given below. Use a significance level of a = 0.05. H₂: Hd = 0 Ha: Hd <0 pre-quiz 12 16 17 16 11 15 12 7 15 13 post-quiz 13 16 15 16 14 15 13 9 14 15 What is the test statistic? test statistic = (Report answer accurate to 2 decimal places.) What is the p-value for this sample? p-value= (Report answer accurate to 3 decimal places.) The correct decision is to Select an answer Conclusion: Select an answer that the worksheet improved quiz scores.
You have been assigned to test the hypothesis that the average mumber of hours per week that an American works is higher than the average number of hours per week that a Swede works. The following data summarizes the sample statistics for the number of hours worked per week for workers in each country. Assume normality is satisfied. Conduct an appropriate hypothesis test using a = significance. 0.05 level of