The plot below is a two way scatter plot of head circumference (cm) vs gestation

Question

The plot below is a two way scatter plot of head circumference (cm) vs gestational age (weeks) for a sample of 100 low birthweight infants, and the output from fitting a simple linear regression model to the data, regressing headcirc (Y) on gestage(X).

Analysis of Variance

Sum of Mean
Source DF Squares Square F Value Pr > F

Model 1 386.86737 386.86737 152.95 <.0001
Error 98 247.88263 2.52941
Corrected Total 99 634.75000


Root MSE 1.59041 R-Square 0.6095


Parameter Estimates

Parameter Standard
Variable Label DF Estimate Error t Value Pr > |t|

Intercept Intercept 1 3.91426 1.82915 2.14 0.0348
gestage gestage 1 0.78005 0.06307 12.37 <.0001

a) Estimate and provide a 95% confidence interval for the slope describing the relationship between head circumference and gestational age

b) Estimate and provide a 95% confidence interval for the expected (mean) head circumference corresponding to a gestational age of 29 weeks. You will need to know that the standard error of the estimated mean head circumference at 29 weeks is 0.159cm.

c) Suppose that a new newborn is selected from the underlying population of low birthweight infants with gestational age of 29 weeks. Give a 95% prediction interval for the new value of head circumference.

d) Explain the difference between a prediction interval and a confidence interval.

Explanation / Answer

Solution:

Solution:

Here, we have to find the 95% confidence interval for the slope describing the relationship between head circumference and gestational age. The confidence interval is given as below:

The estimate for the slope is given as 0.78005 and standard error is given as 0.06307. The critical value for 95% confidence interval is given as 1.96. The confidence interval is given as below:

Lower limit = 0.78005 – 1.96*0.06307 = 0.656433

Upper limit = 0.78005 + 1.96*0.06307 = 0.903667

Confidence interval = (0.656433, 0.903667)

Solution:

The critical value for 95% confidence interval is given as 1.96. The confidence interval is given as below:

Lower limit = 386.86 – 1.96*0.159 = 386.5484

Upper limit = 386.86 + 1.96*0.159 = 387.1716

Confidence interval = (386.5484, 387.1716)

Solution:

The critical value for 95% confidence interval is given as 1.96. The confidence interval is given as below:

Lower limit = 2.52 – 1.96*0.159 = 2.20836

Upper limit = 2.52 + 1.96*0.159 = 2.8316

Confidence interval = (2.20836, 2.8316)

Solution:

The confidence interval is the interval of the values associated with the parameter while the prediction interval is the interval of the values associated with random variable yet to be observed. In confidence interval we find out the limits for given confidence level and in the prediction interval we do the same for the purpose of prediction of the parameters.

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