Question

Thank you, please!

Confidence Intervals for Means

4) The housing market recovered slowly from the economic crisis of 2008. Recently, in one large community, realtors randomly sampled 40 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss from the peak in 2008 was $8370 with a standard deviation of $2500.

a) Find a 98% confidence interval for the mean loss in value per home. (3 points)

b) Interpret the interval you found in part (a). (2 points)

c) Suppose the standard deviation of the losses had been $1500 instead of $2500. What would the smaller standard deviation do to the width of the confidence interval? (1 point)

d) If the confidence level were reduced to 95%, will the interval be wider or narrower? (1 point) e) If the sample size is increased to 80 bids, will the interval be wider or narrower? (

5) A medical researcher measured the body temperatures of a randomly selected group of adults. Here are the summaries of the data collected. The researcher wants to estimate the average (or “normal”) temperature among the adult population. Summary Statistics Temperature Count 52, Mean 98.285 , Median 98.2 ,St. Dev. 0.6824 Range 2.8, IQR 1.050

a) Find a 90% confidence interval for the mean body temperature. (3 points)

b) Interpret the interval you found in part (a). (2 points)

c) 98.6°F is commonly assumed to be “normal”. Do these data suggest otherwise? Explain.

Answer 1

Solution-4a:

xbar=8370

s=2500

n=40

alpha=1-0.98=0.02

alpha/2=0.02/2=0.01

df=n-1=40-1=39

t crit in excel

==T.INV(0.01,39)

=2.42584141

98% confidence interval for mean

xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)

8370-2.42584141*2500/sqrt(40),8370+2.42584141*2500/sqrt(40)

7411.102.,9328.898

98% confidence interval for mean loss in value per home

(7411.102.,9328.898)

Solution-4b:

we are 98% connfident that the true mean  loss in value per home lies in between 7411.102 and ,9328.898

Solution-4c:

8370-2.42584141*1500/sqrt(40),8370+2.42584141*1500/sqrt(40)

7794.661,8945.339

7794.7,8945.3

interval becoms narrower

smaller standard deviation leads to narrow  confidence interval

d) If the confidence level were reduced to 95%, will the interval be wider or narrower?

as you reduced from 98% to 95%

tcriical value reduces

hence interval becomes

narrower

e) If the sample size is increased to 80 bids, will the interval be wider or narrower? (

xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)

8370-2.42584141*2500/sqrt(80),8370+2.42584141*2500/sqrt(80)

7691.957,9048.04

interval becomes narrower