MARIO F. TRIOLA - PowerPoint PPT Presentation

1 / 17
About This Presentation
Title:

MARIO F. TRIOLA

Description:

Fail to reject the null hypothesis if the P-value is greater than the significance level ... we make an assumption (null hypothesis) that contains equality. ... – PowerPoint PPT presentation

Number of Views:204
Avg rating:3.0/5.0
Slides: 18
Provided by: Addi58
Category:

less

Transcript and Presenter's Notes

Title: MARIO F. TRIOLA


1

STATISTICS
ELEMENTARY
Section 7-3 Testing a Claim about a Mean
Large Samples
MARIO F. TRIOLA
EIGHTH
EDITION
2
Two Methods Discussed
  • 1) P-value method
  • 2) Confidence intervals

3
Assumptions
  • for testing claims about population means
  • The sample is a simple random sample.
  • The sample is large (n gt 30).
  • a) Central limit theorem applies
  • b) Can use normal distribution
  • If ? is unknown, we can use sample standard
    deviation s as estimate for ?.

4
P-Value Methodof Testing Hypotheses
  • Goal is to determine whether a sample result is
    significantly different from the claimed value
  • Finds the probability (P-value) of getting a
    result and rejects the null hypothesis if that
    probability is very low

5
Rare Event Rule for Inferential Statistics
  • If, under a given assumption,
  • the probability of an observed event
  • is exceptionally small,
  • we conclude that
  • the assumption is probably not correct.

6
P-Value Methodof Testing Hypotheses
  • Definition
  • P-Value (or probability value)
  • The probability of getting a value of the sample
    test statistic that is at least as extreme as the
    one found from the sample data, assuming that the
    null hypothesis is true

7
TRADITIONAL METHOD
critical region
  • Use significance level to determine critical
    region.

Z0
  • Calculate test statistic by converting sample
    mean to a z-score.

critical value
  • Reject H0 if the test statistic falls in the
    critical region.

P-value area of this region
P-VALUE METHOD
  • Calculate p-value of sample mean, assuming that
    H0 is true.

_ x
µ (according to H0)
  • Reject H0 if the p-value is less than the
    significance level ?.

8
P-value Method of Testing Hypotheses
  1. Write the CLAIM in symbolic form.
  1. Write H0 and H1. If the claim contains equality,
    it becomes H0. Otherwise it becomes H1. The other
    hypothesis is the symbolic form that must be true
    when the original claim is false.
  1. Select the significance level ? based on the
    seriousness of a type I error. Make ? small if
    the consequences of rejecting a true H0 are
    severe. Values of 0.05 and 0.01 are very common.
    Often the significance level will be given.
  1. Identify the appropriate distribution (normal
    distribution or t distribution).
  1. Find the P-value and draw a graph.
  1. Make a DECISION
  • Reject the null hypothesis if the P-value is less
    than or equal to the significance level ?
  • Fail to reject the null hypothesis if the P-value
    is greater than the significance level ?
  1. Write the CONCLUSION in simple non-technical
    terms.

9
Wording of Final Conclusion
Start
Only case in which original claim is rejected
Claim contains equality?
There is sufficient evidence to reject the
claim that. . . (original claim).
Yes
Yes Claim becomes H0
Reject H0?
No
There is not sufficient evidence to reject
the claim that (original claim).
No Claim becomes H1
Only case in which original claim is
supported
There is sufficient evidence to support the
claim that . . . (original claim).
Yes
Reject H0?
No
There is not sufficient evidence to support
the claim that (original claim).
10
Example The body temperatures of 106 healthy
adults were recorded. The mean was 98.2o and s
was 0.62o. At the 0.05 significance level, test
the claim that the mean body temperature of ALL
healthy adults is equal to 98.6o.
  1. CLAIM µ 98.6
  1. Graph and P-value
  1. H0 µ 98.6

H1 µ ? 98.6
? x 98.2
µ 98.6
  1. ? .05 (given in problem)
  1. Normal distribution (n gt 30)

11
To Determine the P-value
STAT, TESTS, Z-Test Inpt Stats µ0 98.6
(population mean according to H0) s 0.62 (can
use s because n gt 30) x 98.2 (sample mean) n
106 (sample size) µ ? µ0 (according to
H1) CALCULATE (Can use DRAW to see the graph)
Calculator returns P 3.1039E-11, which is
equal to 0.000000000031039. Round to 0.
12
Example The body temperatures of 106 healthy
adults were recorded. The mean was 98.2o and s
was 0.62o. At the 0.05 significance level, test
the claim that the mean body temperature of ALL
healthy adults is equal to 98.6o.
  1. CLAIM µ 98.6
  1. Graph and P-value
  1. H0 µ 98.6

H1 µ ? 98.6
? x 98.2
µ 98.6
  1. ? .05 (given in problem)

P 0
  • Decision
  • Reject H0 because P lt .05
  1. Normal distribution (n gt 30)

13
Wording of Final Conclusion
Start
Only case in which original claim is rejected
Claim contains equality?
There is sufficient evidence to reject the
claim that. . . (original claim).
Yes
Yes Claim becomes H0
Reject H0?
No
There is not sufficient evidence to reject
the claim that (original claim).
No Claim becomes H1
Only case in which original claim is
supported
There is sufficient evidence to support the
claim that . . . (original claim).
Yes
Reject H0?
No
There is not sufficient evidence to support
the claim that (original claim).
14
Example The body temperatures of 106 healthy
adults were recorded. The mean was 98.2o and s
was 0.62o. At the 0.05 significance level, test
the claim that the mean body temperature of ALL
healthy adults is equal to 98.6o.
  1. CLAIM µ 98.6
  1. Graph and P-value
  1. H0 µ 98.6

H1 µ ? 98.6
? x 98.2
µ 98.6
  1. ? .05 (given in problem)

P 0
  • Decision
  • Reject H0 because P lt .05
  1. Normal distribution (n gt 30)
  1. Conclusion There is sufficient evidence to
    reject the claim that the mean body temperature
    for healthy adults is 98.6.

15
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.

16
Testing Claims with Confidence Intervals
Claim mean body temperature 98.6º, where n
106, x 98.2º and s 0.62º
  • 95 confidence interval of body temperature data
  • 98.08º lt µ lt 98.32º
  • 98.6º is not in this interval
  • Therefore it is very unlikely that µ 98.6º
  • Thus we reject claim µ 98.6º

17
Rationale of Hypotheses Testing
  • Based on RARE EVENT PRINCIPLE If, under a given
    assumption, the probability of getting an
    observed result is very small, we conclude that
    the assumption is probably not correct.
  • When testing a claim, we make an assumption
    (null hypothesis) that contains equality. We
    compare the assumption and the sample results and
    form one of the following conclusions
  • If the sample results can easily occur when the
    assumption (null hypothesis) 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 (null hypothesis) is true, we
    explain the relatively large discrepancy between
    the assumption and the sample by concluding that
    the assumption is not true.
Write a Comment
User Comments (0)
About PowerShow.com