calculate a 95 confidence interval for the average stopping distance at 24 mph 1

Question

calculate a 95 confidence interval for the average stopping distance at 24 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 y-17.8ft 3.946 n = 50 when x = 15mph SEi -2.181 SE, 15.53 when = 24mph SEj -4.183 SE, 15.94 10 15 20 25 Speed (mph) 1b) Calculate a 95% confidence interval for the average stopping distance at 15 mph. 1d) Calculate a 95% confidence interval for the average stopping distance at 24 mph. g) Calculate a 95% prediction interval for the predicted stopping distance at 24 mph. 1.h) Why is the interval in 1.f wider than the interval in 1.b? 1.i) Could we use this data to construct a prediction interval for the mean stopping distance at 30 mph? Why or why not? 1.j) Would our confidence intervals calculated here still be valid if we converted the data into km/hour and meters of stopping distance? How would we go about correcting our intervals for the new units?

Explanation / Answer

b)

when 95 % , t = 2.011

hence confidence interval is

( 41.39- 2.011 * 15.53 ). (41.39 + 2.011 * 15.53 )

=(10.15917,72.62083)

f)

here n = 50

df = n-2 = 48

critical value =   2.011

y^ = -17.8 + 3.946 * 24 = 76.904

se(y^) =   when x = 24

hence confidence interval is

(76.904 - 2.011 * 15.94 ). (76.904 + 2.011 * 15.94 )

=(44.84866,108.95934)

h) this is because 24 is more away from mean that 15 is.

more the deviation from mean is wider is the confidence interval .

i) no , because 30 is outside range of given data. { 30 > 25}

when the data is outside range , the result of regression can vary much,

hence it is not advisable to use it for x = 30 .

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