Question

8. [5 marks] You are studying two normally distributed populations with equal variances. A random sample of size 10 is taken from each population. The sample from the first population gives the following measurements: 16, 14, 19, 18, 19, 20, 15, 18, 17, 18. The sample from the second population gives the following measurements: 13, 19, 14, 17, 21, 14, 15, 10, 13, 15. Compute a 95% confidence interval for the difference between two population means.

Answer 1

Here we have given that

Xi: sample measurement form first population

Yi: sample measurement form second population

Xi	Yi
16	13
14	19
19	14
18	17
19	21
20	14
15	15
18	10
17	13
18	15

n1=1st sample size = 10

=1st sample mean =

   =

   =17.40   

S1=1st sample standard deviation =

                                     =

                                                      =1.90

n2 =2nd sample size=10

=2nd sample mean =

   =

                                   =15.10

S2=2nd sample standard deviation =

                                     =

                                                      =3.18

Here, The data follows the normal distribution and two population variance are equal. we are using the pooled standard deviation approach to find the 95% CI.

Now we want to find the 95% confidence interval for the difference in two population means .

The formula is as follows,

Where,

Sp=Pooled Sample standard deviation =

Now we find the S-pooled

Sp =

     =

     =2.619

Now, we can find the critical value

c=confidence level =0.95

=level of significance= 1-c =1-0.95=0.05

degrees of freedom = n1+n2-2=10+10-2=18

confidence interval is two tailed.

t-critical = 2.101 Using t table find the value corresponding to the D.F=18 and two tailed probablity 0.05

We get the 95% confidence interval is

The 95% confidence interval for the difference between two population means is (-0.16, 4.76).

Interpretation:

This confidence interval shows we are 95% confident that the difference in the two population mean will falls within that interval.