SOLUTION: This is a prob/stat question. Hopefully someone will help since I am so lost right now. Trying to prepare for a test. In a study the carapace lengths (in mm) of Thenus lobster c

Question 763047: This is a prob/stat question. Hopefully someone will help since I am so lost right now. Trying to prepare for a test.
In a study the carapace lengths (in mm) of Thenus lobster caught near Singapore measured: 78, 66, 65, 63, 60, 60, 58, 56, 52, and 50 (the population data is approximately normal). Find (round to nearest hundredths):
a. The sample mean = __________ (s = 7.97)
b. The 90% confidence intervals for the population mean _________________.
c. The 70% confidence intervals for the population mean _________________.
d. The 80% confidence intervals for the population mean _________________.
e. How many would they need to sample to be 97% sure to be within 0.07 of the population mean? ______________.
f. How many more (than the original 10) would they need to sample to be 64% sure to be within 0.1 of the population mean? ______________.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
This is a prob/stat question. Hopefully someone will help since I am so lost right now. Trying to prepare for a test.
In a study the carapace lengths (in mm) of Thenus lobster caught near Singapore measured: 78, 66, 65, 63, 60, 60, 58, 56, 52, and 50 (the population data is approximately normal). Find (round to nearest hundredths):
a. The sample mean = 60.8____ (s = 7.97)
-------------------
b. The 90% confidence intervals for the population mean _________________.
ME = 1.645 = 4.416
CI::: 60.8-5.42 < u < 60.8+4.42
---------------------------
c. The 70% confidence intervals for the population mean _________________.
Has a different ME:
ME = 1.0364[7.97/sqrt(10)] = 2.61
CI: 60.8-2.61 < u < 60.8+2.61
---------------------------
d. The 80% confidence intervals for the population mean _________________.
Use a different ME
---------------------------------
e. How many would they need to sample to be 97% sure to be within 0.07 of the population mean? ______________.
n = [z*s/E]^2 = [2.17*7.97/0.07]^2 = 61,049 when rounded up
==========================
f. How many more (than the original 10) would they need to sample to be 64% sure to be within 0.1 of the population mean? ______________.
n = [0.9154*7.97/0.01]^2 = 532238 when rounded up
---
You would need 53228 more samples.
=========================
Cheers,
Stan H.
===============

RELATED QUESTIONS
Okay this is Very confusing to me since I just opened my Algebra book for my online class (answered by richard1234)

Hello, I'm working on the following for MATH 421A(Acedemic Grade 10). I'm on unit 5... (answered by richwmiller)

Hi, I am having problems with a chi-square related question. Here is the question: (answered by stanbon)

I am not sure which Algebra section this question fits under. Hopefully someone can help... (answered by jim_thompson5910)

HELP Please! I'm trying to help my 6th grader with his Algebra. What he is working on... (answered by rfer)

I wasn't sure what "type" of problem to put this under so hopefully you can help me.... (answered by galactus)

I am trying to figure this out and I am struggling so someone please help me. Why is... (answered by Edwin McCravy)

Write the slope for a graph with a line going from 8 to 0 to -8... I really need someone... (answered by fcabanski)

when I get the fraction 12/11...I understood all that pretty easily..but it's been a... (answered by ankor@dixie-net.com)