MATH 1401 · Section 8.2 · Confidence Intervals for a Population Mean (σ Unknown)

Confidence Intervals for a Population Mean (σ Unknown)

Section 8.2

The Structure of a Confidence Interval

Every confidence interval has the same basic structure:

Point Estimate ± Critical Value · Standard Error
p^
x¯
zα/2
zα/2
p^(1−p^)n
σn
PROPORTIONS
MEANS

Confidence Interval: p^±zα/2⋅p^(1−p^)n

Confidence Interval: x¯±zα/2⋅σn

What If We Don't Know σ?

If we knew σ:

Confidence Interval: x¯±zα/2⋅σn

Since we don't know σ:

Confidence Interval: x¯±?⋅sn

s estimates σ, but varies from sample to sample

Extra uncertainty means the standard normal curve no longer gives the right critical value.

It comes from something called the Student's t distribution.

What is a t Distribution?

t distributions are symmetric and unimodal, like the normal distribution, but more spread out.

Why the extra spread?

We are estimating σ with s, which introduces uncertainty. To maintain the same confidence level, we need a wider interval. The t distribution builds this in.

Why is there more than one?

Different sample sizes have different amounts of uncertainty. Each needs its own distribution.

We distinguish them by a quantity related to the sample size called degrees of freedom.

The number of degrees of freedom is n−1.

1 Degree of Freedom

4 Degrees of Freedom

10 Degrees of Freedom

Degrees of Freedom

Degrees of freedom =n−1

Fewer degrees of freedom → more uncertainty → wider distribution

More degrees of freedom → s is close to σ → nearly normal

The plan

What we'll do: drag the slider from 1 degree of freedom up to 30.

What to look for: how the dashed t curve compares with the standard normal curve.

As the degrees of freedom increase, the t curve gets closer to the standard normal curve.

  • By 30 degrees of freedom, the two curves nearly coincide.

Degrees of freedom: 1

Try it

Drag the degrees-of-freedom slider from 1 up to 30.

Confidence Interval When σ Is Unknown

Point Estimate ± Critical Value · Standard Error
x¯
tα/2
sn
from the t distributionwith n−1 degrees of freedom
Confidence interval for μ when σ is unknown

x¯±tα/2⋅sn

Critical values for 95% confidence
Sample size nDegrees of freedomtα/2
542.776
1092.262
30292.045
100991.984
Standard normal, zα/21.960

tα/2 is noticeably larger than zα/2 when n is small, but gets closer to it as n grows.

Why Not Just Use z?

What we'll do: draw samples of 5 from a normal population with μ=50 and σ=10, and build a 95% interval from each one two ways, with zα/2 on the left and tα/2 on the right.

What to look for: the percent of intervals in each column that contain μ.

In the long run, about 88% of the z intervals contain μ, but about 95% of the t intervals do.

  • So when σ is unknown, we use t.
x¯±zα/2⋅sn
x¯±tα/2⋅sn
Contain μPercent that contain μ
Contain μPercent that contain μ
Intervals drawn
contains μmisses μμ=50
Try it

Press Draw 50 twice.

Assumptions

When σ is known:

CLT → Sampling distribution of x¯ is normal

When σ is not known:

Same CLT conditions apply → Use t

Assumptions
The sample is a simple random sample.
Either n>30 or the population is assumed to be approximately normal.

Caveat:

When the sample size is small, we must check whether the population can be assumed to be normal. Draw a dotplot and look for skewness or outliers.

A hand lifts one folded paper slip from a glass jar filled with identical folded slips.
Example

Example 1: Grocery Delivery

Among 40 deliveries sampled by a grocery delivery service, the mean delivery time was 34.2 minutes with a standard deviation of 8.6 minutes. Construct a 95% confidence interval for the mean delivery time.

Solution

Check the assumptions:

Simple random sample ✓
Large sample (n>30) ✓

x¯=34.2

s=8.6

n=40

Confidence level = 0.95

Paper grocery bags filled with apples, oranges, leafy greens, a baguette and lettuce sit in the open back of a car.
Example

Example 1: Grocery Delivery

Recall

x¯=34.2

s=8.6

n=40

Example

Example 1: Grocery Delivery

TI-84 Plus
Recall

x¯=34.2

s=8.6

n=40

The TI-84 Plus STAT TESTS menu with 8: TInterval selected.The TI-84 Plus TInterval screen with Inpt set to Stats: x-bar 34.2, Sx 8.6, n 40, C-Level 0.95, and Calculate.The TI-84 Plus TInterval result: the interval 31.45 to 36.95, x-bar 34.2, Sx 8.6, n 40.

We are 95% confident that the mean delivery time is between 31.4 and 37.0 minutes.

Example

Example 2: Hours of Sleep

Seven college students were randomly selected and asked how many hours of sleep they got last night. The results were 7.0, 8.5, 6.5, 7.0, 5.0, 8.0, 6.0. Construct a 90% confidence interval for the mean hours of sleep.

Solution

Check the assumptions:

Simple random sample ✓
Small sample → Must check for normality
5.05.56.06.57.07.58.08.5Hours of Sleep

A dotplot reveals no outliers or strong skewness. It is reasonable to assume the population is normal. ✓

A young man sleeps on his side in a dorm bed with his head resting on his hand, a desk and a potted plant behind him.
Example

Example 2: Hours of Sleep

Recall

7.0, 8.5, 6.5, 7.0, 5.0, 8.0, 6.0

TI-84 Plus
The TI-84 Plus list editor with seven values in L1: 7, 8.5, 6.5, 7, 5, 8 and 6.The TI-84 Plus TInterval screen with Inpt set to Data: List L1, Freq 1, C-Level 0.90, and Calculate.The TI-84 Plus TInterval result: the interval 5.9903 to 7.7239, x-bar 6.857142857, Sx 1.180193689, n 7.

We are 90% confident that the mean sleep time is between 5.99 and 7.72 hours.

Example

Example 3: Electric Vehicle Range

An electric vehicle manufacturer claims that its newest model has a mean driving range of 310 miles on a full charge. A consumer advocacy group tested 70 randomly selected vehicles and found a mean range of 298.4 miles with a standard deviation of 24.1 miles. Construct a 95% confidence interval for the mean range. Does the interval contradict the manufacturer's claim?

Solution
Simple random sample ✓
Large sample (n>30) ✓
A woman smiles as she plugs a charging cable into the side of an electric car.
Example

Example 3: Electric Vehicle Range

Construct a 95% confidence interval for the mean range. Does the interval contradict the manufacturer's claim?

Recall

claim: 310 miles

x¯=298.4, s=24.1, n=70

claim: 310285290295300305310315292.7304.1x̄ = 298.4

We are 95% confident that the mean driving range is between 292.7 and 304.1 miles.

Because 310 is not in the interval, the confidence interval does contradict the manufacturer's claim.

Example

Example 4: Emergency Room Wait Times

A hospital claims that the mean wait time in its emergency room is less than an hour. A patient advocacy group surveyed 100 randomly selected patients and found a mean wait time of 54.2 minutes with a standard deviation of 17.6 minutes. Construct a 90% confidence interval. Does the interval contradict the hospital's claim?

Solution
Simple random sample ✓
Large sample (n>30) ✓
Two people sit quietly in a row of chairs in a hospital waiting area, one of them holding a paper cup.
Example

Example 4: Emergency Room Wait Times

Construct a 90% confidence interval. Does the interval contradict the hospital's claim?

Recall

claim: less than 60 minutes

x¯=54.2, s=17.6, n=100

claim: less than 60455055606551.357.1x̄ = 54.2

We are 90% confident that the mean waiting time is between 51.3 and 57.1 minutes.

Because all values in the confidence interval are less than one hour (60 minutes), the confidence interval does not contradict the hospital's claim.

Example

Example 5: Coffee Shop Tips

A coffee shop owner wants to estimate the mean tip amount for a given day. A random sample of 9 tips yielded the following amounts (in dollars):

0.750.251.000.820.801.5010.000.501.50

Construct a 95% confidence interval for the mean tip amount.

Solution

Check the assumptions:

Simple random sample ✓
Small sample → Must check for normality
012345678910Tip Amount (dollars)

A dotplot reveals an outlier and strong skewness.

It is not reasonable to assume the population is normal. ✗

A glass jar of folded paper bills sits on a wooden coffee shop counter, with cups on a shelf blurred behind it.
Check your understanding

Can We Use This Method?

For each situation, σ is unknown. Can we construct a confidence interval for μ using the methods of this section?

(a)A simple random sample of 12, from a population that is approximately normal

(a)Yes

(b)A simple random sample of 18, whose dotplot shows an outlier

(b)No

(c)A simple random sample of 75, from a population that is skewed

(c)Yes

(d)200 people who chose to answer an online poll

(d)No: a voluntary response sample

Case Study

A Real-World Case

The Question:

Among people hospitalized with COVID-19, do vaccinated patients have better outcomes than unvaccinated patients?

The Data:

GroupSample Size
Vaccinated7,847
Unvaccinated21,525

Source: Lee et al. (2023), 180+ hospitals, January to December 2021

A doctor and a nurse walk side by side down a bright hospital corridor, smiling as they talk.
Case Study

A Real-World Case

What Was Measured:

Average hospital length of stay
Percentage admitted to ICU

The Findings:

How long did patients stay hospitalized?
GroupMean95% CI
Vaccinated8.04 days(7.84, 8.24)
Unvaccinated9.49 days(9.34, 9.64)
7.58.08.59.09.510.0Mean length of stay (days)VaccinatedUnvaccinated
What proportion required ICU care?
GroupProportion95% CI
Vaccinated0.216(0.207, 0.225)
Unvaccinated0.269(0.263, 0.275)
0.200.220.240.260.28Proportion admitted to ICUVaccinatedUnvaccinated
Case Study

A Real-World Case

Conclusion

The confidence intervals do not overlap in either case.

Vaccinated Patients:

Spent about 1.5 fewer days in hospital
Were about 5 percentage points less likely to require ICU care

Keep in Mind:

This study only looked at patients already hospitalized. Vaccination also reduces the chance of infection, severe symptoms, and hospitalization in the first place.

Bottom Line:

By the numbers, there's very little doubt that vaccinated patients fared much better.

You Are Ready For:

8.2: Confidence Intervals for a Population Mean (σ Unknown)

Describe the Student's t distribution and its degrees of freedom

Check the assumptions, with a dotplot when the sample is small

Construct a confidence interval for μ with technology

Interpret it, and decide whether it contradicts a claim

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