We are studying mutual bond funds for the purpose of investing in several funds.

Question

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 five-year performance. The question is: Can the five-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 shown below. Assets Return Assets Return   Fund ($ millions) (%) Fund ($ millions) (%)   AARP High Quality Bond $622.2 10.8   MFS Bond A $494.5 11.6   Babson Bond L 160.4 11.3   Nichols Income 158.3 9.5   Compass Capital Fixed Income 275.7 11.4   T. Rowe Price Short-term 681.0 8.2   Galaxy Bond Retail 433.2 9.1   Thompson Income B 241.3 6.8   Keystone Custodian B-1 437.9 9.2 The regression equation is , the sample size is 9, and the standard error of the slope is 0.0032. Use the .05 significance level. Can we conclude that the slope of the regression line is less than zero?   (Click to select)Do not rejectReject H0 and conclude the slope may be (Click to select)less thanequal tomore than zero. Given the following ANOVA table:   Source DF SS MS F   Regression 1 1,800 1,800.00   24.00     Error 12 900 75.00      Total 13 2,700 a. Determine the coefficient of determination.(Round your answer to 2 decimal places.) Coefficient of determination b. Assuming a direct relationship between the variables, what is the correlation coefficient? (Round your answer to 2 decimal places.) Coefficient of correlation c. Determine the standard error of estimate. (Round your answer to 2 decimal places.)

Explanation / Answer

The regression output for the given data is

The regression line is y = -0.0004x + 9.9198

Can we conclude that the slope of the regression line is less than zero?

Here P-value is 0.9064, therefore dont reject null hypothesis.

hence conculde that the slope of the regression line is not less than zero.

coefficient of determination = r^2 = 0.002=0.00 (rounding of two decimal places)

Correlation coefficient = -0.046 = -0.05 (rounding off two decimal places

Standard error of estimate = 1.752 = 1.75

Regression Analysis r² 0.002 n   9 r   -0.046 k   1 Std. Error   1.752 Dep. Var. Return (%) (y) ANOVA table Source SS   df   MS F p-value Regression 0.0456 1   0.0456 0.01 .9064 Residual 21.4944 7   3.0706 Total 21.5400 8   Regression output confidence interval variables coefficients std. error    t (df=7) p-value 95% lower 95% upper Intercept 9.9198 1.3849 7.163 .0002 6.6449 13.1946 Assets ($mill) (x) -0.00039316 0.0032 -0.122 .9064 -0.00801896 0.00723265

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