Calculate an 80% confidence interval for the average stopping distance at 15 mph

Question

Calculate an 80% confidence interval for the average stopping distance at 15 mph

1) Last week we looked at data studying the (predictable) relationship between speed and stopping distance, and we found that stopping distances could be approximated with the least squares equation listed below. Also listed are some predictive standard errors needed to solve the following problems Vehicle Stopping Distances ft ý--17.8ft 3.946 mph n 50 when x 15mph SEji 2.181 SEy = 15.53 = 24mph when .1. 241111|1 SEi -4.183 SE,15.94 10 15 20 25 Speed (mph) 1a) Calculate an 80% confidence interval for the average stopping distance at 15 mph 1b) Calculate a 95% confidence interval for the average stopping distance at 15 mph 1.c) Why is your interval in 1.b wider than your interval in 1.a? d) Calculate a 95% prediction interval for the predicted stopping distance at 15 mph 1.e) Why is your interval in 1.d wider than your interval in 1.b?

Explanation / Answer

here n = 50

df = n-2 = 48

critical value =   1.29944

y^ = -17.8 + 3.946 * 15= 41.39

se(y^) =   when x = 24

hence confidence interval is

( 41.39- 1.29944* 15.53 ). (41.39 + 1.29944 * 15.53 )

=(21.2096, 61.570303)

Please rate my solution

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