Question

Problem 5. Sampling from two populations produced the following results. ?̅ =385 Sx=30 nx =100

?? = 3 7 8 S y = 2 5 n y = 9 0

1. Find the standard error of (?̅ - ??). Show your calculations.

2. Find the 90% confidence interval estimate for the difference between the two population means.

   Show your calculations.

3. State clearly the assumption(s) that are necessary for your answer to (b) will be valid.

4. Interpret your result in (b).

5. It is claimed that the two population means are not different from each other. Do you reject this

   claim? Explain clearly.

Answer 1

Given

= 385   , sx = 30 , nx = n1 = 100

= 378 , sx = 25 , ny = n2 = 90

QTo find standard Error S.E

Formula : S.E =

where

sd2 =

       =

       = 144725 / 188 = 769.8138

sd2 = 769.8138

Hence

S.E =

      =

      =

S.E = 4.03133

Thus standard Error is S.E = 4.03133

QFind the 90% confidence interval estimate for the difference between the two population means .

It is given by

C.I = ( - - * S.E , - + * S.E )

Here is t-distributed with n1+n2-2 = 88 degree of freedom and =0.10

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

From R

> qt(1-0.1/2,df=188)
[1] 1.652999

90% confidence interval estimate for the difference between the two population means is

C.I = ( - - * S.E , - + * S.E )

   = ( 385 - 378 - 1.652999 * 4.03133 , 385 - 378   + 1.652999 * 4.03133 )

C.I = ( 0.3362155 , 13.66378 )

Thus 90% confidence interval estimate for the difference between the two population means is ( 0.3362155 , 13.66378 )

QState clearly the assumption(s) that are necessary for your answer to (b) will be valid.

The common assumptions made when doing a t-test or whicle estimate for the difference between the two population mean include those regarding the scale of measurement, random sampling, normality of data distribution, adequacy of sample size and equality of variance in standard deviation.

Here we have nx = 100 and ny = 90 so we have adequacate sample size

Since sample size are large (n>30) so by central limit theorm we can assume sample data follows normal distribution .

Hence assumption of random sampling and assuumption of normality are satisfied .

QIt is claimed that the two population means are not different from each other. Do you reject thisclaim? Explain clearly.

Now we have to test

H0 : u1 = u2    ( there is no difference in population means )

H1 : u1 u2 ( two population means are not different from each other )

Here we have obtained 90% confidence interval estimate for the difference between the two population means which was ( 0.3362155 , 13.66378 )

Now if confidence interval dose not include zero then we reject null hypothesis at =0.10 singificance

Now 90% confidence interval ( 0.3362155 , 13.66378 ) is greater than zero and dose not include number zero , hence we reject null hypothesis at 10% of level of significance .

So we reject the claim that two population means are not different from each other at given =0.10 singificance level .

Hence two population means may different from each other.

Note - But if we use =0.05 ( which is not given in above question hence can not be use ) , the result may be differenct , as we know that decreasing singificance level increases the confidence level , and for 0.10 singificance level lower bound of confidence interval was 0.336215 i.e ( 0.3362155 , 13.66378 ) which is very close to zero , so if we construct 95% confidence interval definatly it will be large than 90% confidence interval , and hence it may includse number zero , so at =0.05 significance level conclusion may be different , i.e we may not reject the claim .