Question

A professor records the difference between the marks scored by the students in the class test last week and those scored in the present week. The mean difference for 35 students was 12 with a standard deviation of 3. Construct a 99.9% confidence interval for the mean difference between the marks scored last week and marks scored this week by all the students.

Answer 1

Answer: $(10.82\ , 13.18)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\mu\pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 35  , large sample (n>30)

Mean difference : $\overline{x}=12$

Standard deviation : $\sigma=3$

Significance level : $\alpha=- 1-0.999=0.001$

Critical value : $z_{\alpha/2}=z_{0.0005}=3.4807567$

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 :-

$12\pm\ (3.4807567)\dfrac{2}{\sqrt{35}}\\\\=12\pm1.18\\\\=(10.82\ , 13.18)$

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 = $(10.82\ , 13.18)$