Question

A sample of 34 paired observations generates the following data: d− d − = 0.8 and s2D s D 2 = 4.4. Assume a normal distribution. (You may find it useful to reference the appropriate table: z table or t table) a. Construct the 99% confidence interval for the mean difference μD. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) b. Using the confidence interval, test whether the mean difference differs from zero. There is no evidence that the mean difference differs from zero. There is evidence that the mean difference differs from zero.

Answer 1

We observe, n=34,

The mean of the differences between two samples =0.8

The variance of the differences between two samples =4.4

a. The confidence interval is to be created for a paired t-test, since its based on paired observations and differences of means.

The confidence interval for the mean of the differences between the before and after scores is:

dbar±t (α/2,n-1) *(sd/√n)

where dbar- the mean of the differences between the two samples.

t (α/2,n-1) - the critical value of t at α/2 and degrees of freedom (n-1)

sd - the standard deviation of the differences between the two samples.

n- sample size.

From the information given in the problem,

n=34

dbar=0.8

sd= √4.4=2.0976

For 99% CI,α=1-0.99=0.01, α/2=0.01/2=0.005

degrees of freedom=(n-1)=34-1=33

t (α/2,n-1)=t(0.005,33)=2.7333

So, the 99% CI for the mean difference μD is

dbar±t (α/2,n-1) *(sd/√n)

= 0.8 ± 2.7333*2.0976/√34

=0.8 ± 0.9833

=(-0.1833, 1.7833)

Ans : The 99% CI is -0.1833< μD < 1.7833

b. From the above CI, we can see the interval contains 0, that implies that the mean difference doesn't differ from zero.

Hence, There is no evidence that the mean difference differs from zero.