BUSI 2305 chapter 13 homework

A study of 20 worldwide financial institutions showed the correlation between their assets and pretax profit to be 0.86. 1. State the decision rule for 0.05 significance level: H0: ρ ≤ 0; H1: ρ > 0 (Round your answer to 3 decimal places.) 2. Compute the value of the test statistic. (Round your answer to 2 decimal places.) 3. Can we conclude that the correlation in the population is greater than zero? Use the 0.05 significance level.

1. reject H0 is t> 1.734 2. value of the test statistic 0.86√20−2/√1−(0.86)^2=7.15 3. Reject H0. It is Reasonable to conclude that there is positive association in the population between assets and pretax profit.

A regional commuter airline selected a random sample of 25 flights and found that the correlation between the number of passengers and the total weight, in pounds, of luggage stored in the luggage compartment is 0.94. State the decision rule for 0.05 significance level: H0: ρ ≤ 0; H1: ρ > 0. (Round your answer to 3 decimal places.) Compute the value of the test statistic. (Round your answer to 3 decimal places.) Can we conclude that there is a positive association between the two variables?

H0: ρ ≤ 0 H1: ρ > 0 Reject H0 if t > 1.714 r = 0.94 t=0.94√25−2/√1−(0.94)^2√=13.213 Reject H0, there is a positive correlation between passengers and weight of luggage.

Bloomberg Intelligence listed 50 companies to watch in 2019 (www.bloomberg.com/features/companies-to-watch-2019). Twelve of the companies are listed here with their total assets and 12-month sales. Company Total Assests ($ billions)12-Month Sales ($ billions)12-Month Sales ($ billions) Anthem 74.37 89.6 General Electric 65.36 23.2 Barclays 34.45 8.51 Kroger 36.96 124.18 CBS 20.39 14.32 Nike 22.5 36.4 Comcast 194.25 87.52 Petrobas 220.09 92.66 Dell Technologies 123.38 85.84 Tableau Software 1.47

Reject H0 and conclude the slope is different from zero

Waterbury Insurance Company wants to study the relationship between the amount of fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30 claims for the last year, the director of the actuarial department determined the distance from the fire station (x) and the amount of fire damage, in thousands of dollars (y). ANOVA table Source SS df MS F Regression 1,864.5782 1 1,864.5782 3

a-1. 12.360+4.796x a-2. the relationship between distance and damage is direct b. $36,338, found by 12.3601 + 4.7956(5) c-1. 0.581, found by 1,865/3,209. c-2. 0.581x100=58.1 d-1. .762 d-2 there is a strong association between the variables d-3. 0.762 which is √0.581. it is positive because the slope is positive. e-1. reject h0 if t< -2.763 or t> 2.763 e-2. t=0.762√30 − 2/√1 − (0.762)^2=6.23 e-3. reject h0there is a significant relationship between distance and fire damage.

The Airline Passenger Association studied the relationship between the number of passengers on a particular flight and the cost of the flight. It seems logical that more passengers on the flight will result in more weight and more luggage, which in turn will result in higher fuel costs. For a sample of 15 flights, the correlation between the number of passengers and total fuel cost was 0.667. a. State the decision rule for 0.01 significance level: H0: ρ ≤ 0; H1: ρ > 0. (Round your answer to

a. Reject H0 if t> 2.650 b. value of the test statistic 0.667√15−2/√1−(0.667)^2 =3.23 c. Reject H0. It is reasonable to conclude that there is positive association in the population between the two variables.

The following hypotheses are given. H0: ρ ≥ 0 H1: ρ < 0 A random sample of 15 paired observations has a correlation of −0.46. Can we conclude that the correlation in the population is less than zero? Use the 0.05 significance level. a. State the decision rule for 0.05 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) b. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your an

a. reject H0 if t< -1.771 b. value of the test statistic is = −0.46 √15 − 2/√1 − ( −0.46)^2 =-1.867 c. Reject H0, We can conclude that the correlation in the population is less than zero.

Bi-lo Appliance Super-Store has outlets in several large metropolitan areas in New England. The general sales manager aired a commercial for a digital camera on selected local TV stations prior to a sale starting on Saturday and ending Sunday. She obtained the information for Saturday-Sunday digital camera sales at the various outlets and paired it with the number of times the advertisement was shown on the local TV stations. The purpose is to find whether there is any relationship between the

a. sales is the dependent variable b. see image c. coefficient of correlation is =.93 36/(5 − 1)(1.58)(6.12)= 0.93 d. the statistical measures obtained here indicates a strong positive correlation between the variables

We are studying mutual bond funds for the purpose of investing in several funds. For this particular study, we want to focus on the assets of a fund and its 5-year performance. The question is: Can the 5-year rate of return be estimated based on the assets of the fund? Nine mutual funds were selected at random, and their assets and rates of return are as follows: Fund Assets ($ millions) Return (%) AARP High Quality Bond $622.2 10.8 MFS Bond A $494.5 11.6 Babson Bond L 160.411.3 Nichols Income 1

a. see image b-1. −116.13(9−1)(192.11)(1.6409)=-0.046 b-2. (−0.046)^2=.002 c. there is very little association between the variables d. b= −0.04605(1.6409/192.11) = -.0004 a= 87.9/9−(−0.0004)3,504.5/9 =9.9198 e. Y^=9.9118 − 0.0004(400.0) =9.7620.

The following sample observations were randomly selected. x 4 5 3 6 1 0 y 4 6 5 7 7 a.Fill in the blanks below:(Round your answers to 2 decimal places.) x¯(xbar) y¯(ybar) sx sy Coefficient of correlation (r) b. Choose the right option. the correlation coefficient obtained here indicates correlation between x and Y

a. xbar:28/5= 5.6 ybar: 29/5=5.8 sx: √29.2/4=2.70 sy: √6.8/4=1.30 coefficient of correlation= r = 10.6 / ((5 − 1)(2.7)(1.3)) =0.75 b. a strong positive

The owner of Maumee Ford-Volvo wants to study the relationship between the age of a car and its selling price. Listed below is a random sample of 12 used cars sold at the dealership during the last year. Car Age (years) Selling Price ($000) 1 9 8.1 2 7 6.0 3 11 3.6 4 12 4.0 5 8 5.0 6 7 10.0 7 8 7.6

a. Age is the independent variable and selling price is the dependent variable. b. see image xbar: 107/12= 8.917 ybar: 82.9/12= 6.908 sx: √54.917/11 =2.234 sy- √42.589/11= 1.968 r: r = −26.292/((12 − 1)(2.234)(1.968)) = -.544 c. moderate correlation between age and selling price, consider including: it not surprising that the correlation is negative. as the age of a car increases. the price of a car should decrease.

Given the following ANOVA table: Source DF SS MS F Regression 1 1,000.0 1,000.0 26.00 Error 13 500.0 38.46 Total 14 1,500.0 a. Determine the coefficient of determination. (Round your answer to 2 decimal places.) b. Assuming a direct relationship between the variables, what is the correlation coefficient?(Round your answer to 2 decimal places.) c. Determine the standard error of estimate.(Round your answer to 2 decimal places.)

a. The coefficient of determination is the percentage of the total variation explained by the regression equation. r2 = 1,000/1,500 = 0.667 b. The correlation coefficient is the square root of the coefficient of determination. The sign is positive when the relationship is direct. c. The standard error of estimate describes the dispersion about the regression line by the square root of the total squared differences between the observed and predicted values divided by the degrees of freedom. In this case it is 6.20, found by sy⋅x=√500/15-2

Following is information on the price per share and the dividend for a sample of 30 companies. Company Price per Share Dividend 1 $ 20.00 $ 3.14 2 22.01 3.36 29 77.91 17.65 30 80.00 17.36 a. Calculate the regression equation that predicts price per share based on the annual dividend.(Round your answers to 4 decimal places.) b-1. State the null and alternate hypotheses. b-2. State the decision rule. Use the 0.05 significance level.(Round your answers to 3 decimal places.) b-3. Compute the val

a. The regression equation is: Price = 26.8054 + 2.4082 dividend. For each additional dollar paid out in a dividend the per share price increases by $2.4082 on average. b.H0: β = 0; H1: β ≠ 0 At the 5% level, reject H0 if t is not between −2.048 and 2.048. t = 2.4082/0.3279 = 7.3446 Reject H0 and conclude slope is not zero. c.R2=Reg SS/Total SS =5,057.5543/7,682.7205=0.6583 65.83% of the variation in price is explained by the dividend. d. r=√0.6583=0.8114; 28 df; H0: ρ ≤ 0; H1: ρ > 0At the 5% level, reject H0 when t > 1.701. t=0.8114√30−2/√1−(0.8114)^2 =7.3457; using a p-value calculator, p-value is less than 0.00001.Thus H0 is rejected. The population correlation is positive. e. Price = 26.8054 + 2.4082 ($10) = $50.8874 f. $50.8874±2.048(9.6828)√1+1/30+(10−10.6777)^2/872.1023The interval is ($30.7241, $71.0507).

The owner of Maumee Ford-Volvo wants to study the relationship between the age of a car and its selling price. Listed below is a random sample of 12 used cars sold at the dealership during the last year. Car Age (years) Selling Price ($000) 1 9 8.1 2 7 6.0 3 11 3.6 4 12 4.0 5 8 5.0 6 7 10.0 7 8 7.6 8 11 8.0 9 10 8.0 10 12 6.0 11 6 8.6 12 6 8.0 a. Determine the standard error of estimate.(Round your answer to 3 decimal places.) b. Determine the coefficient of determination.(Round your answer to

a. The standard error of estimate describes the dispersion about the regression line as the total squared differences between the observed and predicted values divided by the degrees of freedom. For the first data point (9, 8.1) the predicted value for y is 6.87. It is a distance of 1.23 from the actual y value. The square of 1.23 is 1.52. Perform the same calculation for the other eleven points and add the squared differences to obtain 30. The standard error of estimate is 1.732, √.30.0018/12−2. b. The coefficient of determination is 0.296, found by (−0.544)^2. c. So, approximately 30 percent of the variation in the selling price is explained by the variation in the age of the car.

The following sample observations were randomly selected. (Do not round the intermediate values. Round your answers to 3 decimal places.) x: 4 5 3 6 10 y: 4 6 5 7 7 a. Determine the 95% confidence interval for the mean predicted when x = 7. b. Determine the 95% prediction interval for an individual predicted when x= 7.

a. When x = 7, the predicted y is 6.308, found by yˆ = 3.7671 + 0.3630(7). Since the sample size is 5, df = n − 2 = 3. For a 95% confidence interval t critical value is 3.182 and the standard error of estimate is 0.992.The sample mean is 5.6 and the sum of the squared deviations from the mean of x is 29.2. 6.308±3.182(0.993)√1/5+(7 − 5.6)^2/29.2 =6.308±1.633 The interval is from 4.675 up to 7.941 .b. The prediction interval formula has an extra 1 under the radical. 6.308 ± 3.182(0.993)√1+1/5+(7 − 5.6)^2/29.2 =6.308 ± 3.557 This interval is from 2.751 up to 9.865.

The following sample of observations were randomly selected. x: 5 3 6 3 4 4 6 8 y: 13 15 7 12 13 11 9 5 a. Determine the regression equation.(Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) x¯ = y¯ = sx = sy = r = b = a = b. Determine the value of yˆ when x is 7.(Round your answer to 3 decimal places.)

a. see image xbar: 39/8=4.875 ybar: 85/8=10.625 sx: √20.875/7=1.73 sy: √79.875/7=3.38 r: = −36.375/((8 − 1)(1.727)(3.378)) = -.089 b: -1.743 ( −0.891) (3.378)/1.727 =−1.743 a: 19.122 10.625-(−1.743)*4.875 =19.122 Y^ 19.122+-1.743 b. the value of y^ 6.921 19.122−1.743*7=6.921

Mr. James McWhinney, president of Daniel-James Financial Services, believes there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information. The x column indicates the number of client contacts last month and the y column shows the value of sales ($ thousands) last month for each client sampled. Number of Contacts X 14,12,20,16,46 Sales ($ thousands) Y 24,14,28,30,80 Number of Contac

a. see image x¯=334/10 =.334 y¯= 611/10 =61.1 sx =√(2,814.40/10-1) =17.68364 sy = √(14,248.90/10-1) = 39.78959 r = 6,176.6/((10 − 1)(17.68364)(39.789585)) =0.9754 b= (0.9753677)(39.789585)/17.68364 =2.19 a= 61.1-2.19456x33.4 = -12.198304 y^= -12.20+2.19 x 40 =75.4 a=-12.20 y^=-12.20 + 2.19x b. Y=-12.2+2.195*40 Y=75.6

The owner of Maumee Ford-Volvo wants to study the relationship between the age of a car and its selling price. Listed below is a random sample of 12 used cars sold at the dealership during the last year. Car Age (years) Selling Price ($000) 1 9 8.1 2 7 6.0 3 11 3.6 4 12 4.0 5 8 5.0 6 7 10.0 7 8 7.6 8 11 8.0 9 10 8.0 10 12 6.0 11 6 8.6 12 6 8.0 a. Determine the regression equation.(Round your answers to 3 decimal places. Negative amount should be indicated by a minus sign.) b. Estimate the selli

a=82.912−(−0.479)107/12 =11.177 b=−0.5441(.968/2.234) =−0.479 b. 6.390, found by 11.177 − 0.479(10) c. For each additional year the car decreases $479 in value.