99.3k views
3 votes
A company produces precision 1 meter (1000 mm) rulers. the actual distribution of lengths of the rulers produced by this company is normal, with mean μ and standard deviation σ = 0.02 mm. suppose i select a simple random sample of four of the rulers produced by the company and i measure their lengths in mm. the sample yields = 1000. a 90% confidence interval for μ is:

User Ccny
by
8.0k points

1 Answer

4 votes

Let X be the length of rulers. X follows Normal distribution with mean μ and standard deviation σ = 0.02. A sample of size n=4 is drawn from population and sample mean m=1000. We have to find 90% confidence interval for population mean μ.

Here population standard deviation is given to us so we will use z confidence interval for mean. It is given as

(Sample mean - Margin of error , Sample mean + Margin of error)

Where margin of error for z confidence interval is

ME = σ
(z_(\alpha/2))/(√(n))

Where σ = population standard deviation = 0.02

n= sample size =4


z_(\alpha/2) = Critical z score value for given confidence level

We have to find here 90% confidence interval so confidence level c= 0.9

α = 1 -c =1-0.9 = 0.1


z_(\alpha/2) =
z_(0.1/2) =
z_(0.05)

Here we have to find z score value such that area below it is -z is 0.05 and above z is 0.05

Using excel function to find z score value

=NORM.S.INV(0.05) = -1.645

For calculating confidence interval we consider positive z score value which is 1.645

So the margin of error is

ME =
(0.02 * 1.645)/(√(4))

ME = 0.01645

90% confidence interval for mean is

(Sample mean - ME, Sample mean+ME)

1000- 0.01645, 1000+0.01645

(1000.01645, 999.9836)

Rounding interval upto 3 decimal places

(1000.016, 999.984)

90% confidence interval for population mean is (1000.016, 999.984)

User Tavakoli
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories