You may need to use the appropriate appendix table or technology to answer this question.What percentage of the population live in their state of birth? According to the U.S. Census Bureau's 2014 American Community Survey, the figure ranges from 25% in Nevada to 78.7% in Louisiana. The average percentage across all states and the District of Columbia is 57.7%. The datain the file HomeState are consistent with the findings in this American Community Survey. The data are for a random sample of 120 Arkansas residents and for a random sample of 180 Virginia residents.(a) Formulate hypotheses that can be used to determine whether the percentage of stay-at-home residents in the two states differs from the overall average of 57.7%.Ho: p > 0.577Ha: p ≤ 0.577Ho: p = 0.577Ha: P = 0.577Ho: p ≤ 0.577Ha: p > 0.577Ho: p≥ 0.577Ha: p < 0.577Ho: P < 0.577Ha: p≥ 0.577(b) Estimate the proportion of stay-at-home residents in Arkansas. (Round your answer to four decimal places.)0.6167Does this proportion differ significantly from the mean proportion for all states? Use a = 0.05.Find the value of the test statistic. (Round your answer to two decimal places.)0.88Find the p-value. (Round your answer to four decimal places.)p-value = 0.3788What is your conclusion?Reject Ho. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of 0.577.Do not reject Ho. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of0.577.Reject Ho. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of 0.577.Do not reject Ho. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Arkansas differs from the overall proportion of 0.577.(c) Estimate the proportion of stay-at-home residents in Virginia. (Round your answer to four decimal places.)0.4944Does this proportion differ significantly from the mean proportion for all states? Use a = 0.05.Find the value of the test statistic. (Round your answer to two decimal places.)-2.24 (c) Estimate the proportion of stay-at-home residents in Virginia. (Round your answer to four decimal places.)0.4944Does this proportion differ significantly from the mean proportion for all states? Use a = 0.05.Find the value of the test statistic. (Round your answer to two decimal places.)-2.24Find the p-value. (Round your answer to four decimal places.)p-value: = 0.0249What is your conclusion?Reject Ho. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of 0.577.Do not reject Ho. There is sufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of0.577.Reject Ho. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of 0.577.Do not reject Ho. There is insufficient evidence to conclude that the proportion of stay-at-home residents in Virginia differs from the overall proportion of0.577.(d) Would you expect the proportion of stay-at-home residents to be higher in Virginia than in Arkansas? Support your conclusion with the results obtained in parts (b) and (c).From the results obtained in parts (b) and (c), we would expect the number of stay-at-home residents to be higherthere is 2.08in Arkansas than Virginia. The sample results show that, rounded to two decimal places, the estimated percentage of Arkansas residents who were born%.%. The sample results also show that, rounded to two decimal places, the estimated percentage of Virginia residents who were born there is 0.0372

(D) Would you expect the proportion of stay-at-home residents to be higher in Virginia than in…

Answered: (d) Would you expect the proportion of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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This question will be answered in three parts. Use the following sample data of math test scores to answer all three parts: 43; 55; 54; 83; 95; 58; 85; 65; 72; 96; 71; 47; 71; 82 Part A) Calculate the mean for the data above: Round your answer to 2 decimal places.
Listed below are heights in inches of fathers and their first sons. Father Son Difference 1. Complete the table and find the differences 2. Use a 5% level of significance to test the claim there is no difference in heights between fathers and first sons. 72 73 66 68 Sample Data 69 68 70 71 70 70 70 70 70 71 75 71 68 70
A)In the two tables in the picture, in which examples are the samples representative of the statistical population? Which are not? B)For those samples that are not representative, identify the statistical population actually sampled.
I ONLY NEED THE ANSWER TO PART 7!!!!!!!!!!!!! 1. For the sample data use Excel to find the mean price, square footage, number of bedrooms, and number of bathrooms. Price Square Feet Beds Baths Average $507, 502.78 1876.56 ft. 3.11 1.88 2. Use Excel to estimate the following four models. Round the coefficients to the nearest whole number. (a) Model 1: 171332 + 179 * Sqft (b) Model 2: 154402 + 96 * Sqft + 91909 * baths (c) Model 3: 160656 + 27438 * beds + 139458 * baths (d) Model 4: 153348 + 96 * Sqft + 557 * beds + 92023 * baths 3. For Model 4 in (d) above, interpret the coefficients for SqFt, Bed, and Bath. 96 Sqft: This means that every additional square foot added is related to a $96 increase in price, as long as beds and baths are constant. 557 beds: This means that every additional bed is related to a $557 increase in price, as long as square feet and baths are constant. 92023 baths: This means that every additional bath is related…
What is the sample difference in means? Question 7 options: a. Lake Rick sample is non-normal, so no comparison is possible. b. 0.5125 kg c. observed p= 0.0167, reject Ho: mean(1) - mean(2) =0 at =0.05. d. +- 0.5 kg e. 0.6572 kg
Suppose a two-way Chi-square test is used to evaluate the relationship between two variables. If both variables have 4 categories or 'items', the number of degrees of freedom (df) for the Chi-square test would be:A. 8B. 7C. 4D. 9 In a two-way Chi-square situation, you are trying to determine if there is a significant relationship between two variables that represent the nominal and/or ordinal scale of measurement.A. TrueB. False What is stated by the null hypothesis in a two-way Chi-square test? A. The two variables have a curvilinear relationshipB. The two variables have very different frequency distributionsC. There is a significant relationship between the two variablesD. There is no significant relationship between the two variables
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1. Sample data about the listing price of a house is collected from Surrey, Guilford area in 2022 and the data is given below. Using the data answer the following questions: N 1 2 3 4 5 Listing Price 1049900 11 1229000 1049900 12 1230000 1050000 13 1075000 14 1298999 1299000 1299000 1075000 15 16 1398000 1078800 1179000 17 18 1398000 1399000 1188000 1188000 19 1449900 1225000 20 1450000 a. What is the mean listing price of a home? b. Generate appropriate numerical summaries for the listing price of a home, the sample standard deviation, etc. 6 7 8 9 10 Listing Price N Listing Price 21 1450000 22 1450000 23 1469900 24 1475000 25 1475000 26 1480000 1499000 27 28 1499000 29 1499233 30 1499900 N C. Use these summaries to develop a 95% confidence interval to estimate the mean listing price of a house in your selected city. That is, you will be using the sampling data you collected to estimate the population parameter (listing price). Make sure to show both your calculations and a final…
Part C. Compute the sample mean and sample deviation.

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