Question

Music and memory Refer to Exercise $87$.

a. Construct and interpret a $99\%$confidence interval for the true mean difference. If you already defined the parameter and checked conditions in Exercise$87$, you don’t need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise .

Short answer

a. $99\%$confidence interval for the true mean difference is $\left(0.2114,2.9286\right)$.

b. The confidence interval gives us the range of possible values for the true mean difference than the significance test.

Step-by-step solution

Part (a): Step 1: Given information

We need to Construct and interpret a $99\%$confidence interval for the true mean difference.

Part (a): Step 2: Explanation

Given:

$$\begin{gathered} n=Samplesize=30 \\ {\overline{x}}_d=Meandifferences=1.57 \\ {s}_d=S\tan darddeviationofdifferences=2.70 \\ c=confidencelevel=99\%=0.99 \end{gathered}$$

Now we will determine the confidence interval for the population mean difference.

Now at Degree of freedom

$$\begin{gathered} =29andc=99\% \\ {t}_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=2.756 \\ Themarginoferroris: \\ E=t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \frac{s_d}{\sqrt{n}}=2.756\times \frac{2.70}{\sqrt{30}} \\ \approx 1.3586 \\ Nowtheboundariesofconfidenceintervalwillbecome \\ {\overline{x}}_d-E=1.57-1.3586=0.2114 \\ {\overline{x}}_d+E=1.57+1.3586=2.9286 \\ Confidenceintervalis\left(0.2114,2.9286\right) \end{gathered}$$

Part (b): Step 1: Given information

We need to explain how the confidence interval provides more information than the test.

Part (b): Step 2: Explanation

The confidence interval gives us more information than the significance test because the confidence interval gives us a different range of possible values for true mean difference than the significance test because the significance test checks only one possible value for the true mean difference.