Download 10.3

When the population standard deviation is unknown,
the sample taken is a simple random sample, and
either the sample size is at least 30 or the population
distribution is approximately normal, use the t
distribution (instead of z) for your hypothesis test.
Section 10.3
Hypothesis Testing for Population
Means (s Unknown)
And some added stuff
by D.R.S., University of Cordele.
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The “Test Statistic” in a t test
The presumed mean, according
to the null hypothesis.
The sample mean
x 
t
 s 
 
n
Since a “t” distribution
is being used, we’re
interested in degrees
of freedom, too:
d.f. = n - 1
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The sample’s standard deviation
The sample size
α = the level of significance
The Critical Value, tα,
which separates out an area of α
in the tail(s).
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A Left-Tailed Hypothesis Test: Ha < μ
tα
The critical t value is the negative z value which separates the left tail of area α.
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A Right-Tailed Hypothesis Test: Ha > μ
tα
The critical t value is the positive z value which separates the right tail of area α.
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A Two-Tailed Hypothesis Test: Ha ≠ μ
-tα/2
tα/2
The critical t values are the +/- t values which separate the two tails, area α/2 each.
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Or use the p-value method
1. Compute test statistic t.
2. What is p-value of that t?
3. Is p-value < the α (alpha)
level of significance?
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as opposed to the Critical value
method (below) in which
1. α implies a critical value tα or tα/2.
2. Compute your test statistic t.
3. Is your t more extreme than tα or
tα/2?
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Example 10.19: Finding the Critical t-Value for a
Right-Tailed Hypothesis Test
Find the critical t-score that corresponds with
14 degrees of freedom at the 0.025 level of significance
for a right-tailed hypothesis test.
Solution
Scanning through the table, we see that the row for
14 degrees of freedom and column for a one-tailed
area of a = 0.025 intersect at a critical t-score of 2.145.
Hence, ta  t0.025  2.145. TI-84: invT(area to left, degrees of freedom)
Here: invT( ______, ______) = ____________
and you have to fix up the __________.
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Example 10.19: Finding the Critical t-Value for a
Right-Tailed Hypothesis Test (cont.)
Area in One Tail
0.100
0.050
0.025
0.010
0.005
0.020
2.650
2.624
2.602
2.583
2.567
2.552
0.010
3.012
2.977
2.947
2.921
2.898
2.878
Area in Two Tails
df
13
14
15
16
17
18
0.200
1.350
1.345
1.341
1.337
1.333
1.330
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0.100
1.771
1.761
1.753
1.746
1.740
1.734
0.050
2.160
2.145
2.131
2.120
2.110
2.101
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Example 10.20: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Left-Tailed, s Unknown)
Nurses in a large teaching hospital have complained for
many years that they are overworked and understaffed.
The consensus among the nursing staff is that the
mean number of patients per nurse each shift is at least
8.0. The hospital administrators claim that the mean is
lower than 8.0. In order to prove their point to the
nursing staff, the administrators gather information
from a simple random sample of 19 nurses’ shifts. The
sample mean is 7.5 patients per nurse with a standard
deviation of 1.1 patients per nurse.
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Example 10.20: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Left-Tailed, s Unknown) (cont.)
Test the administrators’ claim using a = 0.025, and
assume that the number of patients per nurse has a
normal distribution.
Solution
Step 1: State the null and alternative hypotheses.
In order to determine the hypotheses, we
must first ask “What do the researchers want
to gather evidence for?”
H0: _______________
Ha: ________________
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Example 10.20: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Left-Tailed, s Unknown) (cont.)
Step 2: Determine which distribution to use for the
test statistic, and state the level of significance.
The ___-test statistic is appropriate to use in this case
because the claim is about a population
________, the population is normally
distributed, the population standard deviation
is ____known, and the sample is a simple
random sample.
In addition to determining which distribution to
use for the test statistic, we need to state the level of
significance. The problem states that a = ________
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Example 10.20: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Left-Tailed, s Unknown) (cont.)
Step 3: Gather data and calculate the necessary
sample statistics.
Substitute the information given in the
scenario into the formula to obtain the t-test
statistic.
Calculate:
𝑥−𝜇
𝑡=
𝑠
𝑛
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Example 10.20: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Left-Tailed, s Unknown) (cont.)
You should make a sketch of
a bell-shaped curve like this
and indicate the critical
value, shade the critical
region, and indicate where
the test statistic lands.
Label the t-axis with the critical
value and the test statistic value.
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Example 10.20: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Left-Tailed, s Unknown) (cont.)
CONCLUSION: { Reject / Fail to Reject } the null
hypothesis.
INTERPRETATION: We can interpret this conclusion to
mean that the evidence collected { is, is not } strong
enough at the 0.025 level of significance to reject the
null hypothesis in favor of the administrators’ claim
that the mean number of patients per nurse is less than
8.0.
IN OTHER WORDS – who wins? ____________
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p-Values
Conclusions Using p-Values
• If p-value ≤ a, then reject the null hypothesis.
• If p-value > a, then fail to reject the null
hypothesis.
Just like with z test, you have two ways of arriving at the same conclusion:
The Critical Value Method
The p-value method
α determines a rejection region and
a critical t value;
compare your t test statistic to the
critical value – “Is it more extreme?
Does it fall in the rejection region?”
Find the p-value of your test statistic:
the area that’s in the tail(s) beyond
your p-value. “Is the p-value less
than the level of significance, α?”
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Example 10.21: Performing a Hypothesis Test for a Population
Mean Using a TI-83/84 Plus Calculator (Right-Tailed, s Unknown)
A locally owned, independent department store has chosen
its marketing strategies for many years under the
assumption that the mean amount spent by each shopper
in the store is no more than $100.00. A newly hired store
manager claims that the current mean is higher and wants
to change the marketing scheme accordingly. A group of 27
shoppers is chosen at random and found to have spent a
mean of $104.93 with a standard deviation of $9.07.
Assume that the population distribution of amounts spent
is approximately normal, and test the store manager’s claim
at the 0.05 level of significance.
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Example 10.21: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Right-Tailed, s Unknown) (cont.)
Step 1: Hypotheses. H0: ________ Ha:___________
Step 2: Determine which distribution to use for the
test statistic, and state the level of significance.
A ______-distribution is appropriate to use in
this case because the claim is about a population
______, the population is normally distributed, the
population standard deviation is ____known, and the
sample is a simple random sample. The level of
significance is stated in the problem to be a = _______.
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Example 10.21: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Right-Tailed, s Unknown) (cont.)
Step 3: Gather data and calculate the necessary
sample statistics. STAT, TESTS, 2:T-Test
Data if raw data in list; Stats if summary here
the mean, according to the null hypothesis
the sample mean
the sample standard deviation
the sample
size
Highlight “Calculate”
and press ENTER.
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what is the
alternative
hypothesis
claiming?
Inputs:
μ0:
x:
Sx:
n:
μ: ≠μ0
<μ0
>μ0
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Example 10.21: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Right-Tailed, s Unknown) (cont.)
Step 3: Gather data and calculate the necessary
sample statistics. STAT, TESTS, 2:T-Test
Reminder of what the alternative
hypothesis says.
The t test statistic: 𝑡 = 𝑥 − 𝜇
𝑠
𝑛
The p-value of your
t test statistic.
Reminder of the sample
data you provided.
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Example 10.21: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Right-Tailed, s Unknown) (cont.)
Step 4: Draw a conclusion and interpret the decision.
Compare your p value, _______, to α, _______.
Is p < α ? { Yes, No }
So the decision is to { Reject , Fail to Reject } the null
hypothesis.
INTERPRETATION:
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Example 10.22: Performing a Hypothesis Test for a Population
Mean Using a TI-83/84 Plus Calculator (Two-Tailed, s Unknown)
The meat-packing department of a large grocery store
packs ground beef in two-pound portions. Supervisors are
concerned that their machine is no longer packaging the
beef to the required specifications. To test the claim that
the mean weight of ground beef portions is not 2.00
pounds, the supervisors calculate the mean weight for a
simple random sample of 20 packages of beef. The sample
mean is 2.10 pounds with a standard deviation of 0.33
pounds. Is there sufficient evidence at the 0.01 level of
significance to show that the machine is not working
properly? Assume that the weights are normally
distributed.
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Example 10.22: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Unknown) (cont.)
Step 1: hypotheses.
H0: _______________
Ha: ________________
Step 2: Why do we use t in this problem?
Because:
And for this problem, let’s choose the level of
significance ____ = 0.01.
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Example 10.22: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Unknown) (cont.)
Step 3: Gather data and compute.
STAT, TESTS, 2:T-Test
Inputs:
Outputs:
Inputs:
μ0:
x:
Sx:
n:
μ: ≠μ0
<μ0
>μ0
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Example 10.21: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Right-Tailed, s Unknown) (cont.)
Step 4: Draw a conclusion and interpret the decision.
Compare your p value, _______, to α, _______.
Is p < α ? { Yes, No }
So the decision is to { Reject , Fail to Reject } the null
hypothesis.
INTERPRETATION:
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Excel’s T.TEST is different
• It seems to deal with two samples. Maybe it’s a tool
for Chapter 9.
• Long story. Disregard it for purposes of this course.