Home sales again in the previous exercise, you found a 95% confidence interval to estimate the average loss in home value. a) suppose the standard deviation of the losses had been $3000 instead of $1500. what would the larger standard deviation do to the width of the confidence interval (assuming the same level of confidence)? b) your classmate suggests that the margin of error in the interval could be reduced if the confidence level were changed to 90% instead of 95%. do you agree with this statement? why or why not? c) instead of changing the level of confidence, would it be more statistically

Step-by-step explanation:

Given that Home sales has 95% confidence interval to estimate the average loss in home value.

a) If std deviation of losses doubles as 3000 from 1500, we have margin of error also increases.  Because margin of error

= ±Critical value * Std error

= ±Critical value * Std dev/sqrt n

Hence we find that whenever std deviation increases the margin of error increases, for the same level of confidence.

b) Whenever confidence level increases, critical value increases and as a result margin of error increases.  Hence by reducing from 95% to 90% confidence interval would be reduced. True

c) Instead of changing conf level, increasing sample size would give more reliale and accurate results.

24 units

step-by-step explanation:

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)