Question

A developer collects a random sample of the ages of houses from two neighborhoods and finds that the summary statistics for each are as shown. Assume that the data come from a distribution that is Normally distributed. Complete parts a

Find a? 95% confidence interval for the mean? difference,?1minus??2?, in ages of houses in the two neighborhoods if

dfequals=62.3

Neighborhood 1 Neighborhood 2

n1= 30 n2= 35

y1= 53.8 y2= 42.4

s1=7.12 s2= 7.47

Answer 1

Given,

, , ,

, , DF = 62.3

Here population standard deviation's are unknown so we have to use t distribution.

T interval formula -

Where, is t critical value at given confidence interval.

Confidence level = 95% = 0.95

Significance level = = 1 - 0.95 = 0.05

= = 1.999            { Using Excel, =TINV( , DF ) = TINV(0.05 , 62.3 ) = 1.999 }

So, 95% confidence interval for ( ) is,

CI = ( 7.777, 15.022 )

So , 95% confidence interval for ( ) is ( 7.777, 15.022 )