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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.

User Tom Aarsen
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3 votes

Answer:
(10.82\ , 13.18)

Explanation:

The confidence interval for population mean is given by :-


\mu\pm\ z_(\alpha/2)(\sigma)/(√(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)(2)/(√(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)

User Rujul Gandhi
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