Mathematics: Aprofessor 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.

Aprofessor records the difference between the marks

Ответ:

dearydn22

15.03.2022

Mathematics

$(10.82\ , 13.18)$

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 = $(10.82\ , 13.18)$

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Ответ:

jameslinimk

14.02.2023

Mathematics

"A large emerald with a mass of 812.04 grams was recently discovered in a mine. If the density of the emerald is 2.76grams over centimeters cubed, what is the volume? Round to the nearest hundredth when necessary and only enter numerical values, which can include a decimal point. "

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