Chapter 11 Question 2 we\'re trying to fit a simple linear regression model for

Question

Chapter 11 Question 2 we're trying to fit a simple linear regression model for the the whole population:YsPo + ??? + c. (Assume ? are are independent and normally distributed with constant variance) We draw a random sample of n- 5 from the population and get a sample correlation r-0.5 Compute the 4 test statistics for testing the null Hoslope=0in the population. This can be written as Ho: ?1:0. It's also the same as testing whether rp pulation-0) (Round your final answers to 2 decimal places, but don't round during intermediate steps.) a. First compute: R2 0.25 1-R2 o.75 Computer's answer now shown above. You are correct. Previous Tries r receipt no. is 150-7457 b. Now compute the 4 statistics below: 1.29 = 1.00 1.00 puter's answer now shown Computer's answer now shown above. You are correct. Your receipt no. is 150-255 Computer's answer now shown above. You are correct. Your receipt no. is 150-9537 answer now shown Compute the values of the 4 test statistics. Previous Previous above. You are correct. Your receipt no. is 150-8934 above. You are correct. Your receipt no. is 150-7060 Previous Tries Tries Compute the p-values for each statistic. Assume the alternative for the Z and t test is 1-sided (Enter p-value as a decimal not a percent.): HA: slopepopulation>, and assume the alternative for the x2 and F is 2-sided HAi slopepopulation 0. ue = 0.18695 p-value0.10 value 0.20 many degrees of freedom? = 0.5 answer now shown Prevuse n is only -5 Computer's answer now shown above. You are correct. Your receipt no. is 150-5821 re the p-values for the Z and t. Why are they so different? Suamit Aremer Incorrect. Tries 1/3 Tries Previous above. You are correct. Because n is only -5 and Z and t give quite different results for very small samples. Because Z and t always yield very different p-values. Tries Tries r receipt no. is 150-7698 Submit Anewer Tries 2/4 Previous Tries

Explanation / Answer

For t statistic:

Degrees of freedom = n - 2 = 5 - 2 = 3

p - Value = 2*P(t)3) > 1) = 2*0.1955 = 0.3910

For F statistic:

Numerator Df = 1

Denominator Df = 4 - 1 = 3

Hence,

p - Value = P(F(1, 3) > 1) = 0.3910

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