Question

Given two independent random samples with the following results:

n1=18x‾1=180s1=21n1=18x‾1=180s1=21   n2=12x‾2=163s2=30n2=12x‾2=163s2=30

Use this data to find the 99%99% confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed.

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Step 3 of 3 :  

Construct the 99%99% confidence interval. Round your answers to the nearest whole number.

Answer 1

Given two independent random samples with the following results:

n1=18   ;    1 = 180 ;   s1=21   

n2=12   ;    2 = 163 ;   s2=30

Assume that the population variances are not equal and that the two populations are normally distributed.

To find the 99% confidence interval for the true difference between the population .

99% confidence interval for the true difference between the population is given by

CI = { ( 1 - 2 ) - * SE , ( 1 - 2 ) + * SE }

Calculation :-

sample difference = 1 - 2 = 180 - 163 = 17

Standard Error SE :-

S.E =

where =

But since variances are not equal , then standard error are given by

Thus   S.E =    = = 9.974969

Hence

SE = 9.974969

Since variances are not equal degree of freedom will be given by

df =

     =

df = 18.11002

Now is t-distributed with df = 18.11 degree of freedom and =0.01 {for 99% confidence}

It can be computed from statistical book or more accurately from any software like R,Excel

From R

> qt(1-0.01/2,df=18.11)
[1] 2.876411

Thus = 2.876411

99% confidence interval for the true difference between the population is given by

CI = { ( 1 - 2 ) - * SE , ( 1 - 2 ) + * SE }

     = { ( 180 - 163 ) - 2.876411 * 9.974969 , ( 180 - 163 ) + 2.876411 * 9.974969 }

   = { 17 - 2.876411 * 9.974969 , 17 + 2.876411 * 9.974969 }

   = { -11.69211 , 45.69211 }

Thus 99% confidence interval for the true difference between the population is { -11.69 , 45.69 } or

{ -12 , 46 }