Week 7 Homework Problems

docx

School

University of Maryland, University College *

*We aren’t endorsed by this school

Course

200

Subject

Statistics

Date

Apr 3, 2024

Type

docx

Pages

7

Uploaded by jlsolomon1228

Report
Letticia Solomon 10.1.2 Table #10.1.6 contains the value of the house and the amount of rental income in a year that the house brings in. a) Create a scatter plot and find a regression equation between house value and rental income. b) Then use the regression equation to find the rental income for a house worth $230,000 and for a house worth $400,000. c) Find the correlation coefficient. Data of House Value versus Rental Value Rental Value Rental 67500 6864 170000 9568 77000 4576 174000 10400 85000 7072 190000 8320 94000 8736 200000 12272 104000 7904 200000 10400 115000 7904 208000 10400 125000 7904 225000 12480 130000 9776 240000 12064 135000 7488 244500 11232 140000 9152 270000 12896 148000 8320 300000 12480 165000 13312 310000 12480 a. STAT200 Week 7 Homework
b. For a house worth $230,000 y = 0.028*$230,000 + $4,941.4 = $6,440+ $4941.4 = $11,381.4 For a house worth $400,000 y = 0.028*$400,000 + $4,941.4 = $11,200 + $4941.4 = $16,141.4 c. Using Excel’s in-built correl function, the correlation coefficient is = 0.835075 The high value indicates that there is a strong positive correlation between house value and rental income. 10.1.4 The World Bank collected data on the percentage of GDP that a country spends on health expenditures and the percentage of women receiving prenatal care. The data is showed in table. a) Create a scatter plot of the data and find a regression equation between percentage spent on health expenditure and the percentage of women receiving prenatal care. b) Use the regression equation to find the percent of women receiving prenatal care for a country that spends 5.0% of GDP on health expenditure and for a country that spends 12.0% of GDP. c) Find the correlation coefficient Health Expenditure (% of GDP) Prenatal Care (%) 9.6 47.9 3.7 54.6 5.2 93.7 5.2 84.7 10.0 100.0 4.7 42.5 4.8 96.4 6.0 77.1 5.4 58.3 4.8 95.4 4.1 78.0 6.0 93.3 9.5 93.3 6.8 93.7 6.1 89.8
a. b. Using Excel’s in-built slope function, the slope of the regression equation is Thus; Slope = 1.6606 Similarly, using Excel’s in-built intercept function, the y-intercept is y-Intercept = = 69.739% The regression equation for the percentage of GDP spent on health expenditure, and the percentage of women receiving prenatal care is y = 1.6606x + 69.739% For a country that spends 5.0% of GDP on health expenditure y = 1.6606*5% + 69.739% = 8.303% + 69.739% = 78.04% Thus, percent of women receiving prenatal care is 78.04% For a country that spends 12.0% of GDP on health expenditure y = 1.6606*5% + 69.739% = 19.927% + 69.739% = 89.67% Thus, percent of women receiving prenatal care is 89.67% c. Using Excel’s in-built correl function, the correlation coefficient is
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
= 0.171505 The low value indicates that there is a weak positive correlation between the percentage of GDP spent on health expenditure and the percentage of women receiving prenatal. 11.1.2 Researchers watched groups of dolphins off the coast of Ireland in 1998 to determine what activities the dolphins partake in at certain times of the day. The numbers in table represent the number of groups of dolphins that were partaking in an activity at certain times of days. Is there enough evidence to show that the activity and the time period are independent for dolphins? Test at the 1% level. X 2 = ( 6 14.9 ) 2 14.9 + ( 6 3.1 ) 2 3.1 + + ( 10 25.9 ) 2 25.9 ¿ 68.46 Df= (3 – 1)(4 – 1) = 6 Significance level = 0.01 Using chi-square tables, Probability, P ( x 2 ( 6 ) ( 0.01 ) > 68.46 ) = 8.44 E 13 0 Since the p-value is less than the significance level (0.01), we reject the null hypothesis. We can, therefore, conclude that there is sufficient statistical evidence to indicate that a dolphin’s activity is dependent on the time of day. 11.1.4 Activit y Period Row Total Morning Noon Afternoo n Evening Travel 6 6 14 13 39 Feed 28 4 0 56 88 Social 38 5 9 10 62 Column Total 72 15 23 79 189
A person’s educational attainment and age group was collected by the U.S. Census Bureau in 1984 to see if age group and educational attainment are related. The counts in thousands are in the table. Do the data show that educational attainment and age are independent? Test at the 5% level. Education Age Group Row Total 25-34 35-44 45-54 55-64 >64 Did not complete HS 5416 5030 5777 7606 13746 37575 Competed HS 16431 1855 9435 8795 7558 44074 College 1-3 years 8555 5576 3124 2524 2503 22282 College 4 or more years 9771 7596 3904 3109 2483 26863 Column Total 40173 20057 22240 22034 26290 130794 x 2 = 27,603 Df= = (5 – 1)(4 – 1) = 12 Significance level = 0.05 Using chi-square tables, Probability, Pr ( x 2 ( 12 ) 0.05 > 27,603 ) 0 Since the p-value is less than the significance level (0.05), we reject the null hypothesis. We can, therefore, conclude that there is sufficient statistical evidence to indicate that educational attainment is dependent on age. 11.2.4 In Africa in 2011, the number of deaths of a female from cardiovascular disease for different age groups are in table #11.2.6 ("Global health observatory," 2013). In addition, the proportion of deaths of females from all causes for the same age groups are also in table #11.2.6. Do the data show that the death from cardiovascular disease are in the same proportion as all deaths for the different age groups? Test at the 5% level. Table #11.2.6: Deaths of Females for Different Age Groups Age 5-14 15-29 30-49 50-69 Total Cardiovascular Frequency 8 16 56 433 513 All Cause Proportion 0.10 0.12 0.26 0.52
x 2 = 1.034 Df= = (2 – 1)(4 – 1) = 3 Significance level = 0.05 Using chi-square tables, Probability, Pr ( x 2 ( 3 ) 0.05 > 1.034 ) 0.7930 Since the p-value is greater than the significance level (0.01), we fail to reject the null hypothesis. We can, therefore, conclude that there is sufficient statistical evidence to indicate that cardiovascular disease is in the same proportion as all deaths for the different age groups. 11.2.6 A project conducted by the Australian Federal Office of Road Safety asked people many questions about their cars. One question was the reason that a person chooses a given car, and that data is in the table. Table: Reason for Choosing a Car Safety Reliability Cost Performance Comfort Looks 84 62 46 34 47 27 Do the data show that the frequencies observed substantiate the claim that the reasons for choosing a car are equally likely? Test at the 5% level. P ( 1 ) = P ( 2 ) = P ( 3 ) = P ( 4 ) = P ( 5 ) = 1 6 Expected Frequency = n p = 300 1 6 = 50 x 2 = 42.2 Df- = (6 – 1) = 5 Significance level = 0.05 Using chi-square tables, Probability, Pr ( x 2 ( 5 ) 0.05 > 42.2 ) 5.366 E 8 Since the p-value is less than the significance level (0.01), we reject the null hypothesis. We can, therefore, conclude that there is sufficient statistical evidence to indicate that the reasons for choosing a car are equally likely among safety, reliability, cost, performance, comfort, and looks.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help