Question

Birth weight and gestational age.  The Child Health and Development Studies considered pregnancies among women in the San Francisco East Bay area. Researchers took a random sample of 50 pregnancies and used statistical software to construct a linear regression model to predict a baby's birth weight in ounces using the gestation age (the number of days the mother was pregnant). A portion of the computer output and the scatter plot is shown below. Round all calculated results to four decimal places.

Coefficients	Estimate	Std. Error	t value	Pr(>|t|)
Intercept	51.7814	48.867	1.0596	0.2946
gestation	0.2428	0.1745	1.3911	0.1706

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Residual standard error: 16.5836 on 48 degrees of freedom

Multiple R-squared: 0.0388, Adjusted R-squared: 0.0187

1. Use the computer output to write the estimated regression equation for predicting birth weight from length of gestation.

Birth weight =  +  * gestation

2. Using the estimated regression equation, what is the predicted birth weight for a baby with a length of gestation of 282 days?

3. The recorded birth weight for a baby with a gestation of 282 days was 128 ounces. Complete the following sentence:

The residual for this baby is _______. This means the birth weight for this baby is higher than  the birth weight predicted by the regression model.

4. Complete the following sentence:

3.88 % of the variation in Birth weight can be explained by the linear relationship to Gestation age.

Do the data provide evidence that gestational age is associated with birth weight? Conduct a t-test using the information given in the R output and the hypotheses

?0:?1=0H0:β1=0 vs. ??:?1≠0HA:β1≠0

3. Test statistic =

4. Degrees of freedom =

5. P-value =

6. Based on the results of this hypothesis test, there is_____evidence of a linear relationship between the explanatory and response variables.

7. Calculate a 95% confidence interval for the slope, ?1β1. (  ,   )

***need help with #7 please***

Answer 1

1) birth weight=51.7814+0.2428*gestation  
2) birth weight=51.7814+0.2428*282=   120.251
3) residual = 128-120.251=   7.749
4)  
test stat=0.2428/0.1745=   1.3911
df = 48  
p value=0.1706  
6) not enough evidence  

7)

confidence interval for slope                      
                      
n =   50                  
alpha,α =    0.05                  
estimated slope=   0.2428                  
std error =    0.1745                  
                      
Df = n-2 =   48                  
t critical value =    2.0106   [excel function: =t.inv.2t(α,df) ]              
                      
margin of error ,E = t*std error =    2.0106   *   0.1745   =   0.35086  
                      
95%   confidence interval is ß1 ± E                   
lower bound = estimated slope - margin of error =    0.2428   -   0.3509   =   -0.1081  
upper bound = estimated slope + margin of error =    0.2428   +   0.3509   =   0.5937