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# Fall 2012 (Total Marks15)

23:59 30th Jan   , 2013

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### Objective(s) of this Assignment:

• The assignment is being uploaded to build up the concepts about the Confidence interval and sampling distributions.

•  How to get interpretation about Confidence interval.

Assignment # 4 (Lessons 31- 36)

Question 1:                                                                                                                        (Marks: 2+5=7)

a. In a finite population with  and  , find the mean and variance for the sampling distribution of sample mean for n = 25.

b. The mean and standard deviation of the maximum loads supported by 50 cables are 13 tons and 0.80 tons respectively. Find the 99% confidence interval for the mean of the maximum loads of all cables produced by the company. Also interpret the findings.

Question 2:                                                                                                             (Marks: 5+3=8)

1. If  is normally distributed with  and .

Find the

1. A sample of people at the university cafeteria on meatless day showed 20% of them preferred vegetable dishes. How large a sample is needed if we want to be 95% confident that our estimate of population proportion is within 0.03?

-End-

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### Replies to This Discussion

koi tu complete ur accurate solution dy.

@ cute girl ye last Semester ki Assignment ha....

plz koi to proper accurate solution upload kare kuch samaj nai a ri uper se itne sare solutions ne dimagh khrab kar dya hai

koi batay ga k kon c file correct hy????

Check it out

Attachments:

sehar Q# 2 Part A wrong ha

Draaen to ni zalim bhai

Tou kisi ka bhi solution sahi nahi ? mein ne tou sab comments parh liye :\

# Z(alpha/2)

n=36
s=24
x-bar= 100
82 confidence level

find alpa and z alpha/2

Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval.

For large sample confidence intervals about the mean you have:

xBar ± z * sx / sqrt(n)

where xBar is the sample mean
z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α
sx is the sample standard deviation
n is the sample size

The sample mean xbar = 100
The sample standard deviation sx = 24
The sample size n = 36

The z score for a 0.82 confidence interval is the z score such that 0.09 is in each tail.
z = 1.340755
The confidence interval is:

( xbar - z * sx / sqrt( n ) , xbar + z * sx / sqrt( n ) )
( 94.63698 , 105.3630 )

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