A company sampled 40 of the springs that it manufactures for use in small engines and discovered that 20% of the springs

Question

did not meet specifications.  Construct a 99% confidence interval for the population proportion of springs that do not meet specifications.

Answer 1

Hello, thank you for taking the time to post your question!

The underlying formula that you want to use here is

Confidence Internal = p-hat +/- [z* x SE]

For this question we have

p-hat = sample proportion = 0.20

z* = 2.576 for a 99% level

SE = sqrt(p-hat(1 – p-hat)/n)) = sqrt(0.20(1 – 0.20)/40) = 0.0632

Putting it all together then gives us

CI = 0.20 +/- [2.576 *0.0632]

=0.20 +/- 0.1629

=0.0371 to 0.3629

So the 99% confidence interval for this scenario is (0.0371 to 0.3629)

I hope that helps get you moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

x= 20% of 40 = .20(40)=8

Confidence Interval/Level= .99

Using TI-83 Plus Graphing Calculator:

Press STAT,TESTS, A: 1-PropZInt, x:8, C-Level: .99, Calculate

answer: (.03709, .36291)

 

To calculate without graphing calculator, use formula Confidence Interval for proportions(one population):

p ±( z_α/2)(√pq/n)

 

Since p=.20

Then  q=.80 *Note:(q= 1- p)

n=40

99%CI ==> α = .01

z_.01/2 = z_{.005 =}± 2.575

 

Now, substitute those value into formula:

(.20)±2.575(√(.20)(.80)/40 =.20±2.575√.16/40 =.20±2.575√.004 = .20±0.1628573 = 0.0371, 0.3629.