1. Assumptions For testing a claim about the mean of a single population
1) Sample is large (n > 30)
a) Central limit theorem applies
b) Can use normal distribution
2) Can use sample standard deviation s as estimate for s if s is unknown

2. Three Methods Discussed
1) Traditional method
2) P-value method
3) Confidence intervals
Note: These three methods are equivalent, I.e., they
will provide the same conclusions.

3. Procedure
Figure 7-4
1. Identify the specific claim or hypothesis to be tested, and put it in symbolic form.
2. Give the symbolic form that must be true when the original claim is false.
3. Of the two symbolic expressions obtained so far, put the one you plan to reject in the null hypothesis H0 (make the formula with equality). H1 is the other statement.
Or, One simplified rule suggested in the textbook: let null hypothesis H0 be the one that contains the condition of equality. H1 is the other statement.

4. 4. Select the significant level a based on the seriousness of a type I error. Make a small if the consequences of rejecting a true H0 are severe. The values of 0.05 and 0.01 are very common.
5. Identify the statistic that is relevant to this test and its sampling distribution.
6. Determine the test statistic, the critical values, and the critical region. Draw a graph and include the test statistic, critical value(s), and critical region.
7. Reject H0 if the test statistic is in the critical region. Fail to reject H0 if the test statistic is not in the critical region.
8. Restate this previous decision in simple non-technical terms. (See Figure 7-2)

5. In a P-value method, procedure is the same except for steps 6 and 7
Step 6: Find the P-value
Step 7: Report the P-value
Reject the null hypothesis if the P-value is less than or equal to the significance level a
Fail to reject the null hypothesis if the P-value is greater than the significance level a

6. Testing Claims with Confidence Intervals
A confidence interval estimate of a population parameter contains the likely values of that parameter. We should therefore reject a claim that the population parameter has a value that is not included in the confidence interval.

7. Testing Claims with Confidence Intervals
Claim: mean body temperature = 98.6°,
where n = 106, x = 98.2° and s = 0.62°
95% confidence interval of 106 body temperature data (that is, 95% of samples would contain true value µ )
98.08º < µ < 98.32º
98.6º is not in this interval
Therefore it is very unlikely that µ = 98.6º
Thus we reject claim µ = 98.6

8. Underlying Rationale of Hypotheses Testing
When testing a claim, we make an assumption (null hypothesis) that contains equality. We then compare the assumption and the sample results and we form one of the following conclusions:
If the sample results can easily occur when the assumption is true, we attribute the relatively small discrepancy between the assumption and the sample results to chance.
If the sample results cannot easily occur when the assumption is true, we explain the relatively large discrepancy between the assumption and the sample by concluding that the assumption is not true.

9. Testing a Claim about a Mean: Small Samples
Section 7-4
M A R I O F. T R I O L A
Copyright © 1998, Triola, Elementary Statistics, Addison Wesley Longman

10. the population essentially
Figure Choosing between the Normal and Student t-Distributions when Testing a Claim about a Population Mean µ
Start
Use normal distribution with
x – µx Is
n > 30 ?
Yes
Z =
s/ n
(If s is unknown use s instead.) No
Is the
distribution of
the population essentially
normal ? (Use a
histogram.) No
Use non-parametric methods, which don’t require a normal distribution.
Yes
Use normal distribution with
Is s
known ?
x – µx
Z =
s/ n No
(This case is rare.)
Use the Student t-distribution
with
x – µx
t =
s/ n

11. The distribution of sample means will be NORMAL and you can use Table A-2 :
a) For any population distribution (shape) where, n > 30 (use s for a if a is unknown)
b) For a normal population distribution where
1) s is known (n any size), or
2) s is unknown and n > 30 (use s for s)

12. The Student t-distribution and Table A-3 should be used when:
Population distribution is essentially normal
s is unknown
n £ 30

13. Test Statistic (TS) for a Student t-distribution
x – µx
t = s n
Critical Values (CV)
Found in Table A-3
Formula card, back book cover, or Appendix
Degrees of freedom (df) = n – 1
Critical t values to the left of the mean are negative

14. Important Properties of the Student t Distribution
1. The Student t distribution is different for different sample sizes (see Figure 6-5 in Section 6-3).
2. The Student t distribution has the same general bell shape as the normal distribution; its wider shape reflects the greater variability that is expected with small samples.
3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean of z = 0).
4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1 (unlike the standard normal distribution, which has a s = 1).
5. As the sample size n gets larger, the Student t distribution get closer to the normal distribution. For values of n > 30, the differences are so small that we can use the z values instead of developing a much larger table of critical t values. (The values in the bottom row of Table A-3 are equal to the corresponding critical z values from the standard normal distribution.)

15. All three methods. 1) Traditional method. 2) P-value method
All three methods ) Traditional method ) P-value method ) Confidence intervals and the testing procedure Step 1 to Step are still valid, except that the test statistic (therefore corresponding Table) is different.

16. The larger Student t-distribution values shows that with small samples the sample evidence must be more extreme before the difference is significant.

17. P-Value Method Table A-3 includes only selected values of a
Specific P-values usually cannot be found
Use Table to identify limits that contain the P-value
Some calculators and computer programs will find exact P-values

18. Example Conjecture: “the average starting salary for a
computer science gradate is $30,000 per year”.
For a randomly picked group of 25 computer
science graduates, their average starting salary
is $36,100 and the sample standard deviation
is $8,000.

19. Example Solution Step 1: µ = 30k
Step 2: µ > 30k (if believe to be no less than 30k)
Step 3: H0: µ = 30k versus H1: µ > 30k
Step 4: Select a = 0.05 (significance level)
Step 5: The sample mean is relevant to this test and
its sampling distribution is t-distribution with
( ) = 24 degrees of freedom.

20. t = x – µx n t-distribution (DF = 24) Assume the conjecture is true!
(Step 6)
Assume the conjecture is true!
x – µx
t =
Test Statistic:
S / n
Critical value =
1.71 * 8000/ = 32736
Fail to reject H0
Reject H0
30 K
( t = 0)
32.7 k
( t = 1.71 )

21. t = x – µx s / n t-distribution (DF = 24)
(Step 7)
Assume the conjecture is true!
x – µx
t =
Test Statistic:
s / n
Critical value =
1.71 * 8000/ = 32736
Sample data: t =
x = 36.1k
Fail to reject H0
Reject H0 or
30 K
( t = 0)
32.7 k
( t = 1.71 )

22. Example
Step 8:
Conclusion: Based on the sample set, there is sufficient evidence to warrant rejection of the claim that “the average starting salary for a computer science gradate is $30,000 per year”.

23. t = t-distribution (DF = 24) x – µx S / n
(Step 6)
Assume the conjecture is true!
x – µx
t =
Test Statistic:
S / n
P-value = area
to the right of the
test statistic
30 K
36.1 k
P-value =
(by a computer program)
t = =
8 / 5

24. Highly statistically significant
t-distribution (DF = 24)
(Step 7)
P-value = area
to the right of the
test statistic
30 K
36.1 k
P-value =
t = =
8 / 5
P-value < 0.01
Highly statistically significant
(Very strong evidence against the null hypothesis)