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Statistics Tutorial: Hypothesis Test of the Mean
This lesson explains how to conduct a hypothesis test of a mean,
when the following conditions are met:
Generally, the sampling distribution will be approximately
normally distributed if any of the following conditions apply.
- The population distribution is normal.
- The sampling distribution is
symmetric,
unimodal, without
outliers,
and the sample size is 15 or less.
- The sampling distribution is moderately
skewed,
unimodal, without outliers,
and the sample size is between 16 and 40.
- The sample size is greater than 40, without outliers.
This approach consists of four steps: (1) state the hypotheses,
(2) formulate an analysis plan, (3) analyze sample data, and
(4) interpret results.
State the Hypotheses
Every hypothesis test requires the analyst
to state a
null hypothesis
and an
alternative hypothesis. The hypotheses are stated in such
a way that they are mutually exclusive. That is, if one is
true, the other must be false; and vice versa.
The table below shows three sets of hypotheses. Each makes a statement about how
the population mean μ is related to a specified
value M. (In the table, the symbol ≠ means " not equal to ".)
| Set
|
Null hypothesis |
Alternative hypothesis |
Number of tails |
| 1
|
μ = M |
μ ≠ M |
2 |
| 2
|
μ > M |
μ < M |
1 |
| 3
|
μ < M |
μ > M |
1 |
The first set of hypotheses (Set 1) is an example of a
two-tailed test, since an extreme value on either side of the
sampling distribution would cause a researcher to reject the null
hypothesis. The other two sets of hypotheses (Sets 2 and 3) are
one-tailed tests, since an extreme value on only one side of the
sampling distribution would cause a researcher to reject the null hypothesis.
Formulate an Analysis Plan
The analysis plan describes
how to use sample data to accept or reject the null
hypothesis. It should specify the following elements.
- Significance level. Often, researchers choose
significance levels
equal to
0.01, 0.05, or 0.10; but any value between 0 and
1 can be used.
- Test method. Use the
one-sample t-test
to determine whether the hypothesized mean differs
significantly from the observed sample mean.
Analyze Sample Data
Using sample data, conduct a one-sample t-test. This involves
finding the standard error, degrees of freedom,
test statistic, and the P-value associated with the test statistic.
Interpret Results
If the sample findings are unlikely, given
the null hypothesis, the researcher rejects the null hypothesis.
Typically, this involves comparing the P-value to the
significance level,
and rejecting the null hypothesis when the P-value is less than
the significance level.
Test Your Understanding of This Lesson
In this section, two sample problems illustrate how to conduct a
hypothesis test of a mean score. The first problem involves a
two-tailed test; the second problem, a one-tailed test.
Sample Planning Wizard
As you probably noticed, the process of testing a hypothesis about a mean
score can be complex and time-consuming. Stat Trek's Sample Planning Wizard
can do the same job quickly, easily, and error-free. In addition to
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Whenever you need to test a hypothesis, consider using the Sample Planning Wizard.
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Learn more
Problem 1: Two-Tailed Test
An inventor has developed a new, energy-efficient lawn mower engine.
He claims that the engine will run continuously for 5 hours
(300 minutes) on a single gallon of regular gasoline. Suppose a simple
random sample of 50 engines is tested. The engines run for an average
of 295 minutes, with a standard deviation of 20 minutes. Test the
null hypothesis that the mean run time is 300 minutes against
the alternative hypothesis that the mean run time is not 300
minutes. Use a 0.05 level of significance. (Assume that run times for
the population of engines are normally distributed.)
Solution: The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:
State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.
Null hypothesis: μ = 300
Alternative hypothesis: μ ≠ 300
Note that these hypotheses constitute a two-tailed test.
The null hypothesis will be rejected if the sample mean
is too big or if it is too small. Formulate an analysis plan. For this analysis,
the significance level is 0.05. The test method is a
one-sample t-test. Analyze sample data. Using sample data, we
compute the standard error (SE), degrees of freedom (DF),
and the t-score test statistic (t).
SE = s / sqrt(n)
= 20 / sqrt(50) = 20/7.07 = 2.83
DF = n - 1 = 50 - 1 = 49
t = (x - μ) / SE
= (295 - 300)/2.83 = 1.77
where s is the standard deviation of the sample,
x is the sample mean,
μ is the hypothesized population mean,
and n is the sample size.
Since we have a
two-tailed test, the P-value is the probability that the
t-score having 49 degrees of freedom is less than -1.77
or greater than 1.77.
We use the
t Distribution Calculator
to find P(t < -1.77) = 0.04, and
P(t > 1.75) = 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.
Interpret results. Since the P-value (0.08) is
greater than the significance level (0.05), we cannot reject the
null hypothesis.
Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling,
and the population was normally distributed.
Problem 2: One-Tailed Test
Bon Air Elementary School has 300 students. The principal of the
school thinks that the average IQ of students at Bon Air is at
least 110. To prove her point, she administers an IQ test to 20
randomly selected students. Among the sampled students, the average
IQ is 108 with a standard deviation of 10. Based on these
results, should the principal accept or reject her original
hypothesis? Assume a significance level of 0.01.
Solution: The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:
State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.
Null hypothesis: μ >= 110
Alternative hypothesis: μ < 110
Note that these hypotheses constitute a one-tailed test.
The null hypothesis will be rejected if the sample mean
is too small. Formulate an analysis plan. For this analysis,
the significance level is 0.01. The test method is a
one-sample t-test. Analyze sample data. Using sample data, we
compute the standard error (SE), degrees of freedom (DF),
and the t-score test statistic (t).
SE = s / sqrt(n)
= 10 / sqrt(20) = 10/4.472 = 2.236
DF = n - 1 = 20 - 1 = 19
t = (x - μ) / SE
= (108 - 110)/2.236 = -0.894
where s is the standard deviation of the sample,
x is the sample mean,
μ is the hypothesized population mean,
and n is the sample size.
Since we have a
one-tailed test, the P-value is the probability that the
t-score having 19 degrees of freedom is less than -0.894.
We use the
t Distribution Calculator
to find P(t < -0.894) = 0.19. Thus, the P-value is 0.19. Interpret results. Since the P-value (0.19) is
greater than the significance level (0.01), we cannot reject the
null hypothesis.
Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling,
and the population was normally distributed.
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