Answer: [tex](10.82\ , 13.18)[/tex]
Step-by-step explanation:
The confidence interval for population mean is given by :-
[tex]\mu\pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
Given : Sample size : n= 35 , large sample (n>30)
Mean difference : [tex]\overline{x}=12[/tex]
Standard deviation : [tex]\sigma=3[/tex]
Significance level : [tex]\alpha=- 1-0.999=0.001[/tex]
Critical value : [tex]z_{\alpha/2}=z_{0.0005}=3.4807567[/tex]
Now, the 99.9% confidence interval for the mean difference between the marks scored last week and marks scored this week by all the students will be :-
[tex]12\pm\ (3.4807567)\dfrac{2}{\sqrt{35}}\\\\=12\pm1.18\\\\=(10.82\ , 13.18)[/tex]
Hence, the 99.9% confidence interval for the mean difference between the marks scored last week and marks scored this week by all the students = [tex](10.82\ , 13.18)[/tex]
