Confidence Intervals

The formula for a Confidence Interval

CI = Sample Mean ± Z ( standard deviation / sqrt (n))

Confidence Interval Calculator

http://www.socscistatistics.com/confidenceinterval/Default3.aspx

Formulas

• • μ = M ± Z(s_M)

    • where:

  • M = sample mean

  • Z = Z statistic determined by confidence level

  • s_M = standard error = √(s²/n)

Confidence Intervals where no Mean or Standard Deviation is available

CI = Sample Proportion ± Z [ sqrt ( ((p * (1-p)) / n)]

We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 74 had kids. Based on this, construct a 90% confidence interval for the proportion of adult residents who are parents in this county.

Provide the interval estimate in the form: Sample Proportion ± Margin of Error round to 3 places

There are Calculators for Proportions:

Confidence Interval for a Proportion is 0.668 +/- 0.812 http://www.sample-size.net/confidence-interval-proportion/

· But the question is asking for a SAMPLE Proportion for the Confidence Interval not Proportion

· Population formulas and Sample formulas are slightly different because a sample is not as accurate as a whole population and the slight change in the formulas balances that.

Here is how to calculate the correct answer to Sample Proportion± Margin of Error to 3 decimal places:

Our Data:

• n = sample size = 100

• x = the number found = 74

• 90% confidence interval = 1.645 z value for the proportion

Because you don’t have a mean or a standard deviation you must find the Critical Value of Z

And use the formula for The Sample Proportion = ^p = n/x = 74/100 = 0.74 (the first part of the answer)

Here is a link to the formula http://www.stat.wmich.edu/s216/book/node68.html

Here is a link to a Khan Academy video that explains the CONCEPT https://www.khanacademy.org/math/ap-statistics/estimating-confidence-ap/one-sample-z-interval-proportion/v/critical-value-for-a-given-confidence-level

TI-83/84 Calculator Videos to find Confidence Intervals

http://www.stat.wmich.edu/s216/cis/cis.html

You need the STANDARD ERROR because we do not have a Standard Deviation SE_p = sqrt [ p(1-p) / n ]

· SE = Standard Error = Z * SQRT ( p (1-p) / n )

· SE = SQRT (.74*.26) / 100 = .001924

Now we can find the MARGIN or ERROR

· ME = Z * SE

· ME = 1.645 * .001924

· ME = 0.072

Remember our Sample Proportion of ^p = n/x = 74/100 = 0.74

The correct answer for a Sample Proportion ± Margin of Error 0.74 ± 0.072

Confidence Interval for a SAMPLE Standard Deviation

- you need to use the T-Table instead of the Z=Table

CI =

Khan's video on using the T-Table CONCEPTS