Title: REVIEW
1- REVIEW
- Hypothesis Tests of Means
25 Steps for Hypothesis TestingTest Value Method
- Develop null and alternative hypotheses
- Specify the level of significance, ?
- Use the level of significance to determine the
critical values for the test statistic and state
the rejection rule for H0 - Collect sample data and compute the value of test
statistic - Use the value of test statistic and the rejection
rule to determine whether to reject H0
3When to use z and When to use t
- z and t distributions are used in hypothesis
testing. -
_
These are determined by the
distribution of X.
4General Form ofTest Statistics for Hypothesis
Tests
- A test statistic is nothing more than a
measurement of how far away the observed value
from your sample is from some hypothesized value,
v. - It is measured in terms of standard errors
- s known z-statistic with standard error
- s unknown t-statistic with standard error
- The general form of a test statistic is
5Example
- The average cost of all required texts for
introductory college English courses seems to
have gone up substantially as the professors are
assigning several texts. - A sample of 41 courses was taken
- The average cost of texts for these 41 courses is
86.15 - Can we conclude the average cost
- Exceeds 80?
- Is less than 90?
- Differs from last years average of 95?
- Differs from two years ago average of 78?
6CASE 1 z-tests for s Known
- Assume the standard deviation is 22.
- Because the sample size gt 30, it is not necessary
to assume that the costs follow a normal
distribution to determine the z-statistic. - In this case because it is assumed that s is
known (to be 22), these will be z-tests. -
7Example 1 Can we conclude µ gt 80?
1
- H0 µ 80
- HA µ gt 80
- Select a .05
- TEST Reject H0 (Accept HA) if z gt z.05
1.645 - z calculation
- Conclusion 1.790 gt 1.645
- There is enough evidence to conclude µ gt 80.
2
3
4
5
8Example 2 Can we conclude µ lt 90?
1
- H0 µ 90
- HA µ lt 90
- Select a .05
- TEST Reject H0 (Accept HA) if z lt-z.05 -1.645
- z calculation
- Conclusion -1.121 gt -1.645
- There is not enough evidence to conclude µ lt 90.
2
3
4
5
9Example 3 Can we conclude µ ? 95?
- H0 µ 95
- HA µ ? 95
- Select a .05
- TEST Reject H0 (Accept HA) if z lt-z.025 -1.96
or if z gt z.025 1.96 - z calculation
-
- Conclusion -2.578 lt -1.96
- There is enough evidence to conclude µ ? 95.
1
2
3
4
5
10Example 4 Can we conclude µ ? 78?
- H0 µ 78
- HA µ ? 78
- Select a .05
- TEST Reject H0 (Accept HA) if z lt-z.025 -1.96
or if z gt z.025 1.96 - z calculation
-
- Conclusion 2.372 gt 1.96
- There is enough evidence to conclude µ ? 78.
1
2
3
4
5
11P-values
- P-values are a very important concept in
hypothesis testing. - A p-value is a measure of how sure you are that
the alternate hypothesis HA, is true. - The lower the p-value, the more sure you are that
the alternate hypothesis, the thing you are
trying to show, is true. So - A p-value is compared to a.
- If the p-value lt a accept HA you proved your
conjecture - If the p-value gt a do not accept HA you failed
to prove your conjecture
Low p-values Are Good!
125 Steps for Hypothesis TestingP-value Method
- Develop null and alternative hypotheses
- Specify the level of significance, ?
- Collect sample data and compute the value of test
statistic - Calculate p-value Determine the probability for
the test statistic - Compare p-value and ?
- Reject H0 (Accept HA), if p-value lt ?
13Calculating p-values
- A p-value is the probability that, if H0 were
really true, you would have gotten a value - as least as great as the sample value for gt
tests - at most as great as the sample value for lt
tests - at least as far away from the sample value for
? tests - First calculate the z-value for the test.
- The p-value is calculated as follows
TEST P-value EXCEL
gt P(Zgtz) Area to the right of z 1-NORMSDIST(z)
lt P(Zltz) Area to the left of z NORMSDIST(z)
? For z lt 0 2(Area to the left of z) For z gt 0 2(Area to the right of z) 2NORMSDIST(z) 2(1-NORMSDIST(z))
14(No Transcript)
15Examples p-Values
- Example 1 Can we conclude µ gt 80?
- z 1.79
- P-value 1 - .9633 .0367 (lt a .05).
- Can conclude µ gt 80.
- Example 2 Can we conclude µ lt 90?
- z -1.12
- P-value .1314 (gt a .05).
- Cannot conclude µ lt 90.
- Example 3 Can we conclude µ ? 95?
- z -2.58
- P-value 2(.0049) .0098 (lt a .05).
- Can conclude µ ? 95.
- Example 4 Can we conclude µ ? 78?
- z 2.37
- P-value 2(1-.9911) .0178 (lt a .05).
- Can conclude µ ? 78.
16(No Transcript)
17(No Transcript)
18(No Transcript)
19(No Transcript)
20(No Transcript)
21(No Transcript)
22(No Transcript)
23CASE 2 t-tests for s Unknown
- Because the sample size gt 30, it is not necessary
to assume that the costs follow a normal
distribution to determine the t-statistic. - In this case because it is assumed that s is
unknown, these will be t-tests with 41-1 40
degrees of freedom. - Assume s 24.77.
-
24Example 1 Can we conclude µ gt 80?
1
- H0 µ 80
- HA µ gt 80
- Select a .05
- TEST Reject H0 (Accept HA) if t gtt.05,40
1.684 - t calculation
- Conclusion 1.590 lt 1.684
- Cannot conclude µ gt 80.
2
3
4
5
25Example 2 Can we conclude µ lt 90?
1
- H0 µ 90
- HA µ lt 90
- Select a .05
- TEST Reject H0(Accept HA) if tlt-t.05,40
-1.684 - t calculation
- Conclusion -0.995 gt -1.684
- Cannot conclude µ lt 90.
2
3
4
5
26Example 3 Can we conclude µ ? 95?
- H0 µ 95
- HA µ ? 95
- Select a .05
- TEST Reject H0 (Accept HA) if t lt-t.025,40
-2.021 or if t gt t.025,40 2.021 - t calculation
-
- Conclusion -2.288 lt -2.021
- Can conclude µ ? 95.
1
2
3
4
5
27Example 4 Can we conclude µ ? 78?
- H0 µ 78
- HA µ ? 78
- Select a .05
- TEST Reject H0 (Accept HA) if t lt-t.025,40
-2.021 or if t gt t.025,40 2.021 - t calculation
-
- Conclusion 2.107 gt 2.012
- Can conclude µ ? 78.
1
2
3
4
5
28The TDIST Function in Excel
- TDIST(t,degrees of freedom,1) gives the area to
the right of a positive value of t. - 1-TDIST(t,degrees of freedom,1) gives the area to
the left of a positive value of t. - Excel does not work for negative vales of t.
- But the t-distribution is symmetric. Thus,
- The area to the left of a negative value of t
area to the right of the corresponding positive
value of t. - TDIST(-t,degrees of freedom,1) gives the area to
the left of a negative value of t. - 1-TDIST(-t,degrees of freedom,1) gives the area
to the right of a negative value of t. - TDIST(t,degrees of freedom,2) gives twice the
area to the right of a positive value of t. - TDIST(-t,degrees of freedom,2) gives twice the
area to the right of a negative value of t.
29p-Values for t-Tests Using Excel
- P-values for t-tests are calculated as follows
HA TEST Sign of t EXCEL P-value
gt gt0 lt0 TDIST(t,degrees of freedom,1) Usual case 1-TDIST(-t,degrees of freedom,1)
lt lt0 gt0 TDIST(-t,degrees of freedom,1) Usual case 1-TDIST(t,degrees of freedom,1)
? lt0 gt0 TDIST(-t,degrees of freedom,2) TDIST(t,degrees of freedom,2)
30(No Transcript)
31(No Transcript)
32(No Transcript)
33(No Transcript)
34(No Transcript)
35(No Transcript)
36Test Value vs P-Value
- Example of one-tailed, positive test value
? is known ? is unknown
Test Value Compare ztest with zcritical Accept HA if ztest gt zcritical Compare ttest with tcritical Accept HA if ttest gt tcritical
P-value Compare p-value (based on normal distribution) with a Accept HA if p-value lt a Compare p-value (based on t distribution) with a Accept HA if p-value lt a
37Review
- When to use z and when to use t in hypothesis
testing - s known z
- s unknown t
- z and t statistics measure how many standard
errors the observed value is from the
hypothesized value - Form of the z or t statistic
- Meaning of a p-value
- z-tests and t-tests
- By hand
- Excel