A researcher wants to determine if the number of years of education for a parent correlates with the number of years of education for their child. The table shows the number of years of education of the parent and the number of years of education of their child for a random sample of 7.    Years of Education of Parent: 13, 9, 12, 10, 12, 10, 12 Years of Education of their Child: 15, 11, 10, 16, 17, 12, 17   Conduct a hypothesis test to determine if there is a significant linear correlation between the two variables. Use α = 0.05. If there is a significant linear correlation between the variables, determine the regression equation.  There is a significant correlation between the variables and the regression equation is y' = 5.333 + 0.777x  There is a significant correlation between the variables and the regression equation is y' = 0.777 + 5.333x  There is not a significant correlation between the variables, so the regression equation should not be determined.  There is a significant correlation between the variables and the regression equation is y' = 8.451 + 0.1923x

The midterm and final exam grades for a statistics course are provided in the data set below. Jaymes, a student in the class, scored 86 on both exams. Treat the given data sets as samples. Midterm: 80, 78, 85, 82, 79, 79, 78, 86, 80, 84, 78, 84, 80, 84, 81, 78, 81, 82, 78, 84 Final: 81, 88, 68, 69, 69 81, 82, 86, 76, 71, 72, 68, 77, 86, 68, 83, 84, 71, 81, 87 z-score for midterm: 1.857 z-score for final: 1.181 Based on the z-scores calculated above, which of Jaymes's grades is more unusual, the midterm grade or the final exam grade? Select the correct answer below: A. The absolute value of the z-score for the final exam grade is greater than for the midterm grade, so the final exam grade is more unusual. B. The absolute value of the z-score for the midterm exam grade is less than for the final grade, so the midterm grade is more unusual. C. The absolute value of the z-score for the midterm exam grade is greater than for the final grade, so the midterm grade is more…

4. Find the mode of this data set: 3, 3, 4, 8, 5, 8, 1, 6, 7. OA. no mode OB. 3 O c. 5 OD. 3 and 8

The 2008 general social survey asked a nationally representative sample of 1324 adult americans to report how many hours of television they watch in a typical day. The resulting data are reported in the following table counts. Median number of hours of tv watching per day

Treat the dataset as your bag of peanut M&M’s (a simple random sample). Count the number of orange M&M’s and the total number of M&M’s in your bag. Then determine the proportion of your M&M’s that are orange. Number of orange M&M’s: 13 Total number of M&M’s: 65 Proportion of orange M&M’s: 0.2   Question below is according to this.

About 3% of the population has a particular genetic mutation. 900 people are randomly selected.Find the mean for the number of people with the genetic mutation in such groups of 900.

How many samples or treatments are here?   490 undergraduate students at a state university in the southwestern region of the United States were surveyed. Each student in the sample was classified according to class standing (freshman, sophomore, junior, or senior) and body are category (body piercings only, tattoos only, both tattoos and body piercings, no body art). Use the data in the accompanying table to determine if there is an association between class standing and body art category. Assume that it is reasonable to regard the sample as representative of the students at this university. Use α = 0.01.                                                             Freshman  Sophmore   Junior    Senior Body Piercings                                          61              7              14          86 Only Tattoos                                             43              11             10         64 Only Mildly Piercings and Tattoos            20                9              7…

About 2% of the population has a particular genetic mutation. 800 people are randomly selected. Find the mean for the number of people with the genetic mutation in such groups of 800.

A television station wishes to study the relationship between viewership of its 11 p.m. news program and viewer age (18 years or less, 19 to 35, 36 to 54, 55 or older). A sample of 250 television viewers in each age group is randomly selected, and the number who watch the station’s 11 p.m. news is found for each sample. The results are given in the table below.      Age Group     Watch11 p.m. News? 18 or less 19 to 35 36 to 54 55 or Older Total Yes 37 46 59 83 225 No 213 204 191 167 775 Total 250 250 250 250 1,000     (a) Let p1, p2, p3, and p4 be the proportions of all viewers in each age group who watch the station’s 11 p.m. news. If these proportions are equal, then whether a viewer watches the station’s 11 p.m. news is independent of the viewer’s age group. Therefore, we can test the null hypothesis H0 that p1, p2, p3, and p4 are equal by carrying out a chi-square test for independence. Perform this test by setting α = .05. (Round your answer to 3 decimal places.)…

A television station wishes to study the relationship between viewership of its 11 p.m. news program and viewer age (18 years or less, 19 to 35, 36 to 54, 55 or older). A sample of 250 television viewers in each age group is randomly selected, and the number who watch the station’s 11 p.m. news is found for each sample. The results are given in the table below.      Age Group     Watch11 p.m. News? 18 or less 19 to 35 36 to 54 55 or Older Total Yes 49 59 61 84 253 No 201 191 189 166 747 Total 250 250 250 250 1,000     (a) Let p1, p2, p3, and p4 be the proportions of all viewers in each age group who watch the station’s 11 p.m. news. If these proportions are equal, then whether a viewer watches the station’s 11 p.m. news is independent of the viewer’s age group. Therefore, we can test the null hypothesis H0 that p1, p2, p3, and p4 are equal by carrying out a chi-square test for independence. Perform this test by setting α = .05. (Round your answer to 3 decimal places.)…