confidence intervals for the mean (small samples) 1 larson/farber 4th ed
TRANSCRIPT
Interpret the t-distribution and use a t-distribution table
Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown
Larson/Farber 4th ed2
When the population standard deviation is unknown, the sample size is less than 30, and the random variable x is approximately normally distributed, it follows a t-distribution.
Critical values of t are denoted by tc.
Larson/Farber 4th ed3
-xtsn
1. The t-distribution is bell shaped and symmetric about the mean.
2. The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom. The degrees of freedom are the number of free choices left after a sample statistic such as is calculated. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size.
• d.f. = n – 1 Degrees of freedom
Larson/Farber 4th ed4
x
Larson/Farber 4th ed5
3. The total area under a t-curve is 1 or 100%.
4. The mean, median, and mode of the t-distribution are equal to zero.
5. As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t-distribution is very close to the standard normal z-distribution.
t0
Standard normal curve
The tails in the t-distribution are “thicker” than those in the standard normal distribution.d.f. = 5
d.f. = 2
Find the critical value tc for a 95% confidence when the sample size is 15.
Larson/Farber 4th ed6
Table 5: t-Distribution
tc = 2.145
Solution: d.f. = n – 1 = 15 – 1 = 14
95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = +2.145.
Larson/Farber 4th ed7
t
-tc = -2.145 tc = 2.145
c = 0.95
A c-confidence interval for the population mean μ
The probability that the confidence interval contains μ is c.
Larson/Farber 4th ed8
where c
sx E x E E t
n
Larson/Farber 4th ed9
1. Identify the sample statistics n, , and s.
2. Identify the degrees of freedom, the level of confidence c, and the critical value tc.
3. Find the margin of error E.
xxn
2( )1
x xsn
cE tn
s
d.f. = n – 1
x
In Words In Symbols
Larson/Farber 4th ed10
4. Find the left and right endpoints and form the confidence interval.
Left endpoint: Right endpoint: Interval:
x Ex E
x E x E
In Words In Symbols
You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 162.0ºF with a sample standard deviation of 10.0ºF. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed.
Larson/Farber 4th ed11
Solution:Use the t-distribution (n < 30, σ is unknown, temperatures are approximately distributed.)
n =16, x = 162.0 s = 10.0 c = 0.95 df = n – 1 = 16 – 1 = 15 Critical Value
Larson/Farber 4th ed12
Table 5: t-Distribution
tc = 2.131
Margin of error:
Confidence interval:
Larson/Farber 4th ed13
102.131 5.316cE t
n s
162 5.3
156.7
x E
162 5.3
167.3
x E
Left Endpoint:
Right Endpoint:
156.7 < μ < 167.3
156.7 < μ < 167.3
Larson/Farber 4th ed14
( )
• 162.0156.
7167.3
With 95% confidence, you can say that the mean temperature of coffee sold is between 156.7ºF and 167.3ºF.
Point estimate
xx E x E
No
Larson/Farber 4th ed15
Is n 30?
Is the population normally, or approximately normally, distributed? Cannot use the normal
distribution or the t-distribution. Yes
Is known?
No
Use the normal distribution with
If is unknown, use s instead.
cE zn
σYes
No
Use the normal distribution with
cE zn
σ
Yes
Use the t-distribution with
and n – 1 degrees of freedom.
cE tn
s
You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction cost?
Larson/Farber 4th ed16
Solution:Use the normal distribution (the population is normally distributed and the population standard deviation is known)