1) The 90% confidence interval of the mean of each group 2) construct a complete hypothesis test and determine if the two groups have significantly difference balances using a=0.05. ll Sprint ?*7:18 PM@ 25%( All InboxesDirections: In cell B2below, where it saysname, type your nameand hit enter. Then copyand paste the data into aWord document.Sandra SantiagoBalance Time (sec)Number of Days WorkoutFavorite WorkoutMen 11776.ВackMen 2184BackMen 3108BackMen 41287ВackMen 5566ArmsMen 6744ChestMen 7681ChestMen 823ChestMen 91846.ArmsMen 101634ArmsMen 111587ChestMen 12162ВackWomen11047ArmsWomen1692ChestWomen31333LegsWomen41407ChestWomen5327ChestWomen1127ChestWomen7873LegsWomen8791Вack

Note: Hi there! Thank you for posting the question. As you have posted multiple questions, as per the policy, we have solved only one question for you. If you need any specific question to be answered, please re-submit the question separately or by specifying the question number or name.Given information: Given data represents the values of the variables balance time, number of days workout and favorite workout for men and women.State the hypotheses: Denote µ1 as the population mean balance time for men and µ2 as the population mean balance time for women. The null and alternate hypotheses are stated below: Null hypothesis H0: H0: µ1 = µ2 That is, there is no significant difference in the mean balance time between the two groups men women. Alternative hypothesis H1: H1: µ1 ≠ µ2 (Two tailed test) That is, there is a significant difference in the mean balance time between the two groups men women.Two sample t test: EXCEL software can be used to perform a two-sample t test for the given data. Software Procedure: The step-by-step procedure to perform a two-sample t test using EXCEL software is given below: Open an EXCEL file. Enter the data of Men in column A and name it as Men. Enter the data of Women in column B and name it as Women. Select Data > Data Analysis > Two-Sample Assuming Equal Variances > OK. In Variable 1 Range enter $A$1:$A$13 and in Variable 2 Range enter $B$1:$B$9. Click on Labels and enter Alpha as 0.05. Click OK. The obtained output is shown below: From the above obtained output below values are reported: The degrees of freedom is d.f = 18. The test statistic value is t = 0.1182. The P-value is P-value = 0.9072. Decision rule: The level of significance is α = 0.05. Decision rule based on P-value approach: If P-value ≤ α, then reject the null hypothesis H0. If P-value > α, then fail to reject the null hypothesis H0. Conclusion: Conclusion based on P-value approach: The P-value is 0.9072 and α value is 0.05. Here, P-value is greater than the α value. That is, 0.9072 (=P-value) > 0.05 (=α). By the rejection rule, fail to reject the null hypothesis. Thus, there is no significant difference in the mean balance time between the two groups men women.

Answered: construct a complete hypothesis test…

Math

Statistics

construct a complete hypothesis test and determine if the two groups have significantly difference balances using a=0.05.

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

Make a 95% confidence interval for p1 - p2 given that p-hat, is .38, p-hat2 is.35, n1 = 100 and n2 = 144.
A sample of 153 was selected from a large population. 69% of the sample had a characteristic of interest. Calculate a 99% confidence interval estimate for the proportion of the population with the characteristic.
11)
Assume that both populations are normally distributed. a) Test whether u >H2 at the a = 0.01 level of significance for the given sample data. b) Construct a 99% confidence interval about u - 42- Sample 1 28 52.6 Sample 2 23 44.1 9.4 O A. Ho: H1 H2 O B. Ho: H1 = H2, H:Hq H2, HHy H2 Determine the test statistic. t= (Round to two decimal places as needed.) Approximate the P-value. Choose the correct answer below. O A. P-value 20.10 O B. P-value <0.01 O C. 0.05 s P-value <0.10 O D. 0.01 SP-value <0.05 Should the hypothesis be rejected at the a = 0.01 level of significance? V the null hypothesis because the P-value is V the level of significance. b) The confidence interval is the range from to (Round to twwo decimal places as needed. Use ascending order
A sample of 164 was selected from a large population. 66% of the sample had a characteristic of interest. Calculate a 90% confidence interval estimate for the proportion of the population with the characteristic.
Two different simple random samples are drawn from two different populations. The first sample consists of 40 people with 22 having a common attribute. The second sample consists of 1900 people with 1375 of them having the same common attribute.compare results from a hypothesis test of p1-=p2 with a 0.05 significance level and a 95% confidence interval estimate of p1-p2 .
Researchers want to test a new hand exercise plan for arthritis patients to see if they have improved flexibility. They recruited 110 patients with arthritis in their hands and randomly assigned them to one of two treatment groups: group i took their regular medicine and did the new hand exercises; and group 2 only took medicine. After one month, 42 of the 55 in group 1 improved in their flexibility, and 37 of the 55 in group 2 improved their flexibility. Based on the 95% confidence interval (-0.076, 0.258), is there convincing evidence of a difference in the proportion of patients like these who would show improvement in flexibility if they did the new hand exercises along with taking the medicine versus just taking the medicine? v to believe there va difference in the proportion of patients like these who would show improvement in flexibility if they did the new hand exercises along with taking the medicine versus just taking the medicine. Because v contained in the interval, there…
A survey of 200 people was taken to determine whether distance from voting center was related to the likelihood of voting. Based on these results, does it appear that these two variables are related at the 10% level of significance?
Suppose college football scores are normally distributed with a population standard deviation of 6pts.
A study was done to provide data on sleep deprivation rates of Californians and Oregonians. The percent of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this percent is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. a) Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. Use California as Group A and Oregon as Group B. Round your test statistic to 2 decimal places and your p-value to 4 decimal places. Test statistic: p-value: Conclusion:
Assume that both populations are normally distributed. a) Test whether u > H2 at the a = 0.01 level of significance for the given sample data. b) Construct a 99% confidence interval about u - 42. 177 Sample 1 24 51.7 9.5 Sample 2 43.1 99 O A. Ho: H1 = H2, H:Hq>H2 O B. Ho: H1 = H2, HH1H2, H1: H1 H2 Determine the test statistic. %3D (Round to two decimal places as needed.) Approximate the P-value. Choose the correct answer below. O A. P-value 20.10 O B. P-value <0.01 O C. 0.01 s P-value <0.05 O D. 0.05 s P-value <0.10 Should the hypothesis be rejected at the a = 0.01 level of significance? V the null hypothesis because the P-value is the level of significance. b) The confidence interval is the range from to (Round to two decimal places as needed. Use ascending order.)
A graduate Statistic Student wants to estimate the true proportion of county college ("c.c.") students in here state who go on to receive an Associate Degree (AD ), She is able to take a random sample of n = 100 of those c.c. stuents, and 64 of those students received an AD. Find her 90% confidence interval for the true proportion of c.c. students in her state that receive an AD. Be sure to either i)Use full calculator syntax including proper label on your output or i)write down the formula, plug in the appropriate nun bers, and then give the calculated interval including a proper label. 1-prop-Z Int x: 64,n: 100, Clevel.90, interval (.56105,.71895 b. 1-prop-Z Int x: 64,n: 100, Clevel .90, interval (.66105,.71896 1-prop-Z Int x: 64,n: 100, Clevel .90, interval (.51105,.70950 1-prop-Z Int x: 64,n: 100, Clevel .90, interval (.61105,.81105 by C.