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Lecture 17
Section 8.2
Objectives:
•Tests concerning hypotheses about means
− One sample tests
− Two sided tests and Confidence Intervals
− Two sample tests (independent samples)
− Paired t-tests
Tests for a population mean
To test the hypothesis H0 : µ = µ0 based on an SRS of size n from a
Normal population with unknown mean µ and known standard deviation
σ, we rely on the properties of the sampling distribution N(µ, σ/√n).
The P-value is the area under the sampling distribution for values at
least as extreme, in the direction of Ha, as that of our random sample.
Sampling
distribution
Again, we first calculate a z-value
and then use Table A.
x 
z
 n
σ/√n
x
µ
defined by H0
P-value in one-sided and two-sided tests
One-sided
(one-tailed) test
Two-sided
(two-tailed) test
To calculate the P-value for a two-sided test, use the symmetry of the
normal curve. Find the P-value for a one-sided test and double it.
One sample t-test
Suppose that an SRS of size n is drawn from an N(µ, σ) population.

When  is known, the sampling distribution of x bar is N(,/√n).

When  is estimated from the sample standard deviation s, the
sampling distribution follows a t distribution t(, s/√n) with degrees
of freedom n − 1.
x
t
s n
is the one-sample t statistic.
The P-value is the probability, if H0 is true, of randomly drawing a
sample like the one obtained or more extreme, in the direction of Ha.
The P-value is calculated as the corresponding area under the curve,
one-tailed or two-tailed depending on Ha:
One-sided
(one-tailed)
Two-sided
(two-tailed)
x  0
t
s n
Example
The medical director of one large company is concerned about the
effects of stress on the company’s younger executives. The mean
systolic blood pressure for males 35 to 44 years of age (national
average) is 128 and standard deviation is 15. They examine the records
of 72 executives in this age group and finds that their mean systolic
blood pressure is 129.93. Assume that the population distribution is
normal. Is this evidence that the mean blood pressure for all the
company’s young male executives is higher than the national average?
Test this at α=0.05 significance level.
a. State hypotheses.
b. Compute test statistic and P-value.
c. Make a decision and state a conclusion in terms of the problem.
Example
The recommended daily dietary (RDA) allowance for zinc among males
older than 50 years is 15mg/day. The following data on zinc intake for a
sample of males age 65-74 years:
n=115, sample mean=11.3, sample sd=6.43
Does this suggest that true average daily zinc intake for the entire
population of males age 65-74 is less than the recommended allowance?
Use α=0.05.
Example
Consider the following data on the fill amounts in "16oz." ketchup bottles:
15.39, 15.62, 16.05, 15.90, 15.47, 15.83, 15.80, 15.65.
Assume that the population distribution of the fill amount is normal with
unknown σ .
Test H0: μ = 15.5 vs. Ha: μ > 15.5 at the α = 0.01 significance level.
Confidence intervals to test hypotheses
Because a two-sided test is symmetrical, you can also use a
confidence interval to test a two-sided hypothesis.
In a two-sided test,
C = 1 – α.
C confidence level
α significance level
α /2
α /2
Packs of cherry tomatoes (σ = 5 g): H0 : µ = 227 g versus Ha : µ ≠ 227 g
Sample average 222 g. 95% CI for µ = 222 ± 1.96*5/√4 = 222 g ± 4.9 g
227 g does not belong to the 95% CI (217.1 to 226.9 g). Thus, we reject H0.
Logic of confidence interval test
Ex: Your sample gives a 99% confidence interval of x  m  0.84  0.0101 .
With 99% confidence, could samples be from populations with µ = 0.86? µ = 0.85?
Cannot
 reject
H0:  = 0.85
Reject H0 :  = 0.86
99% C.I.
x

A confidence interval gives a black and white answer: Reject or don't reject H0.
But it also estimates a range of likely values for the true population mean µ.
A P-value quantifies how strong the evidence is against the H0. But if you reject
H0, it doesn’t provide any information about the true population mean µ.
Example
Consider the following weights of some runners are expressed in
kilograms.
67.8 61.9 63.0 53.1 62.3 59.7 55.4 58.9 60.9 69.2 63.7 68.3
64.7 65.6 56.0 57.8 66.0 62.9 53.6 65.0 55.8 60.4 69.3 61.7
Assume that the population (weights of all runners) has normal distribution
with mean μ (unknown) and the population standard deviation σ = 4.5 kg.
a. Give a 95% confidence interval for the mean weight of the population of
all such runners.
b. Based on this confidence interval, does a test of
H0: μ = 61.3 kg
Ha: μ ≠ 61.3 kg
reject H0 at the 5% significance level?
Comparing two samples
(A)
Population 1
Population 2
Sample 2
Sample 1
Which
is it?
(B)
Population
We often compare two
treatments used on
independent samples.
Sample 2
Sample 1
Is the difference between both
treatments due only to variations
from the random sampling (B),
Independent samples: Subjects in one samples are
completely unrelated to subjects in the other sample.
or does it reflect a true
difference in population means
(A)?
Two-sample z statistic
We have two independent SRSs (simple random samples) possibly
coming from two distinct populations with (1,1) and (2,2). We use x
and
x 2 to estimate the unknown 1 and 2.
When both populations are normal, the sampling distribution
 of (x1- x2)
 12
is also normal, with standard deviation :
n1
Then the two-sample z statistic
has the standard normal N(0, 1)
sampling distribution.
z

 22
n2
 
( x1  x2 )  ( 1   2 )
 12
n1

 22
n2
1
Two independent samples t distribution
We have two independent SRSs (simple random samples) possibly
coming from two distinct populations with (1,1) and (2,2) unknown.
We use ( x1,s1) and ( x2,s2) to estimate (1,1) and (2,2), respectively.


To compare the means, both populations should be normally
distributed. However, in practice, it is enough that the two distributions
have similar shapes and that the sample data contain no strong outliers.
Two-sample t significance test
The null hypothesis is that both population means 1 and 2 are equal,
thus their difference is equal to zero.
H0: 1 = 2 <>1 − 2 0
with either a one-sided or a two-sided alternative hypothesis.
We find how many standard errors (SE) away
from (1 − 2) is ( x1− x 2) by standardizing with t:
Because in a two-sample test H0
poses (
1 −2) 0, we simply use
s12 s22 
  
n1 n 2 
df 
2 2
2 2


1 s1
1 s2
  
 
n1 1 n1  n 2 1 n 2 
2

(x1  x 2 )  (1  2 )
t
SE
t
x1  x 2
2
1
2
2
s
s

n1 n 2
Example
Consider the lifetimes of two kinds of light bulbs:
Take a random sample of size n1=40 from the population with
mean μ1 and standard deviation σ1=26,
Take independently another random sample of size n2=50 from
the population with mean μ2 and standard deviation σ2=22.
Test : H0: μ1 −μ2 =0 vs. Ha: μ1 −μ2 ≠ 0 at α = .05 .
Example
In the comparison of two kinds of paint, a consumer testing service finds that
34 1-gallon cans of Benjamin Moore paint cover on the average 5480 square
feet with a standard deviation of 62 feet, whereas 41 1-gallon cans of
Pittsburgh paint cover on the average 5452 square feet with a standard
deviation of 51 feet. Test to see whether or not the Pittsburgh paints cover a
larger area on average, at the 1% significance level.
Example
The following data on tensile strength (psi) of linear specimens both when a
certain fusion process was used and when this process was not used:
1. No fusion: 2784 2700 2655 2822 2511 3149 3257 3213 3220 2753
2. Fused: 3027 3356 3359 3297 3125 2910 2889 2902
Test if the fusion process increased the average tensile strength at the 5%
significance level.
Paired t-test
In these cases, we use the paired data to test the difference in the two
population means. The variable studied becomes Xdifference = (X1 − X2),
and
H0: µdifference= 0 ; Ha: µdifference>0 (or <0, or ≠0)
Conceptually, this is not different from tests on one population.
Example
The following data was obtained from a sample of n=16 subjects. Each
observation is the amount of time, expressed as a proportion of total time
observed, during which arm elevation was below 30o. The two
measurements from each subject were obtained 18 months apart. During
this period, work conditions were changed, and subjects were allowed to
engage in a wider variety of work tasks
a. Does the data suggest that true average time during which elevation is
below 30o differs after the change from what it was before the change?
b. Does it appear the change in work conditions decreases true average time
by more than 5?