1. Statistical software was used to fit the model E y)-0 + 1x1 . 2x2 tons 20 dat
ID: 3293615 • Letter: 1
Question
1. Statistical software was used to fit the model E y)-0 + 1x1 . 2x2 tons 20 data ports. Complete parts a through h 1 Click the icon to see the software output a. What are the sample estimates of 0, 1, and 2? 2448.92 A1= -1547.1 297.82 Type integers or decimals.) b. What is the least squares prediction equation? y=( 244832 )+( -15471 Type integers or decimals.) )x1 + ( 1547.1+ -297.82 )x2 c. Find SSE, MSE, and s. Interpret the standard deviation in the context of the problem. SSE 1770090 MSE104123 s 322.681 (Type an integer or a decimal.) Interpret the value of s. The interval ± 2s will provide a rough approximation to the accuracy with which the model will predict future values of y for given values of x .Thus, it would be expected that the model will provide predictions of ±2s = ± 645.362 d. Test Ho: 1 :0 against Ha : 1 40, Use = 0.01 The test statistic is (Type an integer or a decimal.) The p-value is 001 y to within about -3.9 Type an integer or a decimal.) Draw a conclusion. Choose the answer below. A. Do not reject the null hypothesis. There is not sufficient evidence to support the alternative hypothesis Do not reject the null hypothesis. There is sufficient evidence to support the alternative hypothesis. Reject the null hypothesis. There is sufficient evidence to support the alternative hypothesis. B. C. D. Reject the null hypothesis. There is not sufficient evidence to support the alternative hypothesis. e. Use a 99% confidence interval to estimate 1. 2696.968 8· 37232 -397.232 ) (Round to three decimal places as needed.) (Round to three decimal places as needed.)Explanation / Answer
a) see the table in column coefficient
b) least square prediction estimate is already given in table .
It is b0 + b1*x1 +b2*x2 ....
c) E - represent error
hence SSE - see column SS and row E(residual error)
MSE - similarly column MS and row E
s = Sqrt(SSE/df_E) = 322.681 , also given in table
2s = 2 *322.681 = 645.362
d) TS = coef / se(coef)
already given in table {see column T}
p-value = 0.001
if p-value < level of significance
the variable is significant
here p-value < 0.05
hence option C) is correct
Post e) to h) part again .
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