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Confidence Intervals: Teacher's Guide

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

Summary: This module is the complementary teacher's guide for the Confidence Intervals chapter of the Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.

Confidence intervals can be difficult for students. This chapter discusses confidence intervals for a single mean and for a single proportion. In this course, we do not deal with confidence intervals for two means or two proportions. For a single mean, confidence intervals are calculated when σσ is known and when σσ is not known (ss is used as an estimate for σσ).

Book notation:

  • CL = confidence level
  • EBM = error bound for a mean
  • EBP = error bound for a proportion

The student-t distribution in introduced in this chapter beginning with a little history:

history:

William Gossett derived the t-distribution in 1908. He needed a method for dealing with small samples (less than 30) in his research on temperature at the Guinness Brewery. Legend has it that the name Student-t comes from the fact that Gossett wrote a paper about the t-distribution and signed the paper Student because he was too modest to use his own name.

If you sample from a normal distribution in which σσ is not known, replace σσ with ss, the sample standard deviation, and use the Student-t distribution. The shape of the curve depends on the parameter degrees of freedom (dfdf). df=n−1df=n−1 where nn is the sample size.

Notation:

tdftdf designates the distribution. We use TT as the random variable. Value is an average.

The t-statistic (t-score)

t = value-μ (σ n) t = value-μ (σ n)

The relationship between the confidence interval for a single mean (when σσ and the confidence level can be shown in a picture as follows:

Figure 1: The α2α2 subscript indicates that the area to the right is α2α2.
Graph illustrating the distribution of a confidence interval when the mean value is known and the confidence level.

Formulas for the error bounds:

  • Single mean (known σσ): EBM=zα 2⋅( σn)EBM=zα 2⋅(σn)
  • Single mean (unknown σσ): EBM=tα 2⋅( sn)EBM=tα 2⋅(sn)
  • Binomial proportion: EBP=zα 2⋅p'q 'nEBP=zα 2⋅p'q 'n where q=1− p'q=1−p'

The confidence intervals have the form:

  • Single mean (unknown or known σσ ): (x- -EBM, x- +EBM)(x--EBM,x-+EBM)
  • Binomial proportion: (p' -EBP, p' +EBP)(p'-EBP,p'+EBP)

Example 1

Problem 1

The number of calories in fast food is always of interest. A survey was taken from 7 fast food restaurants concerning the number of calories in 4 ounces of french fries. The data is 296, 329, 306, 324, 292, 310, 350. Construct a 95% confidence interval for the true average number of calories in a 4 ounce serving of french fries.

Solution

You want a confidence interval for a single mean where σσ is not known. If you use the TI-83/84 series, enter the data into a list and then use the function TInterval, data option. C-level is 95. The confidence interval is (296.4, 334.2). This function also calculates the sample mean (315.3) and sample standard deviation (20.4). TInterval is found in STAT TESTS.

If you want the students to use the formulas for a normal or for the Student-t confidence interval, you will need to use a table for the z-score or the t-score. The book does not have the tables but the Internet has several. Do a search on "z-score table" and "Student-t table."

First, you need to calculate the sample mean and the sample standard deviation.

  • x- =315.29x- =315.29
  • s=20.40s=20.40

The confidence interval has the pattern : (x- -EBM, x- +EBM)(x--EBM,x-+EBM)

The error bound formula is : EBM = tα2 ⋅(sn )EBM =tα2⋅(sn)

CL = 0.95CL=0.95 so α=0.05α=0.05. Therefore, α2=0.025α2=0.025.

Using the Student-t table with df = 7-1=6df=7-1=6 , t.025 = 2.45t.025=2.45.

Figure 2:
Distribution curve with 90% of the data in the mean, and 10% in each adjacent deviation

EBM= tα 2⋅(s n)= t0.25⋅ 20.407= 2.45⋅20.407 =18.89EBM=tα 2⋅(s n)=t0.25⋅20.407=2.45⋅20.407=18.89

The confidence interval is (x- -EBM, x- +EBM)(x--EBM,x-+EBM) = (315.29 - 18.89, 315.29 +18.89) = (296.4, 334.2)

We are 95% confident that the true average number of calories in a 4 ounce serving of french fries is between 196.4 and 334.2 calories.

Example 2

Problem 1

At a local cabana club, 102 of the 450 families who are members have children who swam on the swim team in 1995. Construct an 80% confidence interval for the true proportion of families with children who swim on the swim team in any year.

Solution

You want a confidence interval for a single proportion. If you use the TI-83/84 series, use the function 1-PropZinterval. x=102x=102 , n=450n=450, C−level=80C−level=80. The confidence interval is (.2077, .2590)

If you want to use the formulas, first, you need to calculate the estimated proportion.

p'=xn =102450= 0.23p'=xn=102450=0.23

The confidence interval has the pattern (p' -EBP, p' +EBP)(p'-EBP,p'+EBP).

The error bound formula is EBP=zα 2⋅p'q 'nEBP=zα 2⋅p'q 'n where q=1− p'q=1−p'

CL=0.80CL=0.80 so α=0.20α=0.20. Therefore, α2= 0.10α2=0.10.

Using the normal table (find one on the Internet), z.10=1.28z.10=1.28 . (Remind students that 0.10 is the area to the right. The area to the left is 0.90.)

EBP=zα 2⋅p'q 'n =z.10⋅p'q 'n =z.10⋅ .23⋅.77n =1.28⋅ .23⋅.77450 =0.03EBP=zα 2⋅p'q 'n=z.10⋅p'q 'n=z.10⋅ .23⋅.77n=1.28⋅ .23⋅.77450=0.03

The confidence interval is : (p' -EBP, p' +EBP)(p'-EBP,p'+EBP) = (0.23 - 0.03, 0.23 + 0.03) = (0.20, 0.26)

We are 80% confident that the true proportion of families that have children on the swim team in any year is between 0.20 and 0.26.

Assign Practice

Assign the Practice 1, Practice 2, and Practice 3 in class to be done in groups.

Assign Homework

Assign Homework. Suggested homework: 1, 5, 9, 13, 15, 17, 21, 23, 24 - 31.

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