Survey
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Degrees of freedom (statistics) wikipedia, lookup

Foundations of statistics wikipedia, lookup

Psychometrics wikipedia, lookup

Bootstrapping (statistics) wikipedia, lookup

Taylor's law wikipedia, lookup

Misuse of statistics wikipedia, lookup

Resampling (statistics) wikipedia, lookup

Student's t-test wikipedia, lookup

Transcript
```Notation
8-4 Testing a Claim About a Mean
This section presents methods for testing a claim about a
population mean.
n = sample size
x
= sample mean
μ x = population mean
When σ is not known, we use a “t test” that incorporates
the Student t distribution.
Test Statistic
Requirements
1) The sample is a simple random sample.
t=
2) Either or both of these conditions is satisfied:
The population is normally distributed or n > 30.
11
Running the Test
P-values: Use technology or use the Student t
distribution in Table A-3 with degrees of freedom
df = n – 1.
Critical values: Use the Student t distribution with
degrees of freedom df = n – 1.
x − μx
s
n
Important Properties of the
Student t Distribution
1. The Student t distribution is different for different sample
sizes (see Figure 7-5 in Section 7-3).
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 when s is used to
estimate σ.
3. The Student t distribution has a mean of t = 0.
4. The standard deviation of the Student t distribution varies
with the sample size and is greater than 1.
5. As the sample size n gets larger, the Student t distribution
gets closer to the standard normal distribution.
Example
Example - Continued
Listed below are the measured radiation emissions (in W/kg)
corresponding to a sample of cell phones.
Requirement Check:
1. We assume the sample is a simple random sample.
Use a 0.05 level of significance to test the claim that cell
phones have a mean radiation level that is less than 1.00
W/kg.
0.38
0.55
1.54
1.55
0.50
0.60
0.92
0.96
1.00
0.86
2. The sample size is n = 11, which is not greater than 30, so
we must check a normal quantile plot for normality.
1.46
The summary statistics are: x = 0.938 and s = 0.423 .
Example - Continued
Example - Continued
The points are reasonably close to a straight line and there is
no other pattern, so we conclude the data appear to be from a
normally distributed population.
Step 1: The claim that cell phones have a mean radiation
level less than 1.00 W/kg is expressed as μ < 1.00 W/kg.
22
Step 2: The alternative to the original claim is μ ≥ 1.00 W/kg.
Step 3: The hypotheses are written as:
H 0 : μ = 1.00 W/kg
H1 : μ < 1.00 W/kg
Example - Continued
Step 4: The stated level of significance is α = 0.05.
Step 5: Because the claim is about a population mean μ, the
statistic most relevant to this test is the sample mean: x .
Example - Continued
Step 6: Calculate the test statistic and then find the P-value or
the critical value from Table A-3:
t=
x − μ x 0.938 − 1.00
=
= −0.486
s
0.423
n
11
Example - Continued
Example - Continued
Step 7: Critical Value Method: Because the test statistic of
t = –0.486 does not fall in the critical region bounded by the
critical value of t = –1.812, fail to reject the null hypothesis.
Step 7: P-value method: Technology, such as a TI-83/84 Plus
calculator can output the P-value of 0.3191. Since the P-value
exceeds α = 0.05, we fail to reject the null hypothesis.
Example
Finding P-Values
Assuming that neither software nor a TI-83 Plus calculator is
available, use Table A-3 to find a range of values for the Pvalue corresponding to the given results.
Step 8: Because we fail to reject the null hypothesis, we
conclude that there is not sufficient evidence to support the
claim that cell phones have a mean radiation level that is less
than 1.00 W/kg.
33
Example – Confidence
Interval Method
We can use a confidence interval for testing a claim about μ.
For a two-tailed test with a 0.05 significance level, we
construct a 95% confidence interval.
For a one-tailed test with a 0.05 significance level, we
construct a 90% confidence interval.
a) In a left-tailed hypothesis test, the sample size is n = 12,
and the test statistic is t = –2.007.
b) In a right-tailed hypothesis test, the sample size is n = 12,
and the test statistic is t = 1.222.
c) In a two-tailed hypothesis test, the sample size is n = 12,
and the test statistic is t = –3.456.
Example – Confidence
Interval Method
Using the cell phone example, construct a confidence interval
that can be used to test the claim that μ < 1.00 W/kg,
assuming a 0.05 significance level.
Note that a left-tailed hypothesis test with α = 0.05
corresponds to a 90% confidence interval.
Using methods described in Section 7.3, we find:
0.707 W/kg < μ < 1.169 W/kg
Example – Confidence
Interval Method
Because the value of μ = 1.00 W/kg is contained in the
interval, we cannot reject the null hypothesis that μ = 1.00
W/kg .
Based on the sample of 11 values, we do not have sufficient
evidence to support the claim that the mean radiation level is
less than 1.00 W/kg.
44
```
Related documents