1.Suppose you fit the first-order multiple regression model y= 0+1x1+2x2+ n=25 d
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1.Suppose you fit the first-order multiple regression model y= 0+1x1+2x2+ n=25 data points and obtain the prediction equation ^y = 32.01+1.83x1+4.09x2. The estimated standard deviations of the sampling distributions of 1 and 2 are 0.26 and 0.53, respectively.
a. Test H0: 2=0 against Ha: 2>0. Use =0.01.
b. Find a 95% confidence interval for 1.
Interpret the interval.
a. The test statistic is _____ . (Round to three decimal places as needed.)
The p-value is ______ . (Round to three decimal places as needed.)
Either (1) Reject or (2) Do not reject the null hypothesis. There (3) is not or (4) is sufficient evidence to support the alternative hypothesis.
b. What is the confidence interval?
(___, ____) (Round to three decimal places as needed.)
Interpret this interval.
We are ___ % confident that the true value of 1 lies in this interval
2. The equation used to predict the total body weight (in pounds) of a female athlete at a certain school is ^y=126+3.57x1+1.57x2, where x1 is the female athlete's height (in inches) and x2 is the female athlete's percent body fat, measured as x2%. Use the multiple regression equation to predict the total body weight for a female athlete who is 62 inches tall and has 23% body fat.
The predicted total body weight for a female athlete who is 62 inches tall and has 23% body fat is ___ pounds. (Round to the nearest tenth as needed.)
3. Suppose a statistician built a multiple regression model for predicting the total number of runs scored by a baseball team during a season. Use the estimates to predict the number of runs scored by a team with 271 walks,760 singles,213 doubles,23 triples, and 115 home runs.
The model predicts ___ runs for the season.
Ind. Var. beta estimate Standard Error Intercept 3.63 19.08 Walks (x1) 0.35 0.03 Singles (x2) 0.42 0.06 Doubles (x3) 0.75 0.06 Triples (x4) 1.19 0.18 Home Runs (x5) 1.48 0.04Explanation / Answer
(1a) we use t-test and t=2/SE(2)=4.09/0.53=7.72 with n-3=25-3=22 df
the critcal t(0.01,22)=2.82 is less than calculated t=7.72 so we reject Ho:2=0
(1b)
(1-alpha)*100% confidence interval for =±t(alpha/2,n-3)*SE()
95% confidence interval for 1=1.83±t(0.05/2, 22)*0.26=1.83±2.82*0.26=1.83±0.7332=(1.0968,2.5632)
To interpret a confidence interval remember that the sample information is random - but there is a pattern to its behavior if we look at all possible samples. Each possible sample gives us a different sample proportion and a different interval. But, even though the results vary from sample-to-sample, we are "confident" because the margin-of-error would be satisfied for 95% of all samples
(2) ^y=126+3.57x1+1.57x2
for x1=62 and x2=23 , ^y=126+3.57*62+1.57*23=131.45
body weight for a female athlete who is 62 inches tall and has 23% body fat is given by 131.45
(3) let run=y
y=3.63+0.35x1+0.42x2+0.75x3+1.19x4+1.48x5
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