Hypothesis Test for a Mean
This lesson explains how to conduct a hypothesis test of a mean,
when the following conditions are met:
 The sampling distribution is normal or nearly normal.
Generally, the sampling distribution will be approximately
normally distributed if any of the following conditions apply.
 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
twotailed 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
onetailed 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
onesample ttest
to determine whether the hypothesized mean differs
significantly from the observed sample mean.
Analyze Sample Data
Using sample data, conduct a onesample ttest. This involves
finding the standard error, degrees of freedom,
test statistic, and the Pvalue 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 Pvalue to the
significance level,
and rejecting the null hypothesis when the Pvalue is less than
the significance level.
Test Your Understanding
In this section, two sample problems illustrate how to conduct a
hypothesis test of a mean score. The first problem involves a
twotailed test; the second problem, a onetailed test.
Sample Planning Wizard
As you probably noticed, the process of hypothesis testing
can be complex. When you need to test a hypothesis about a mean score, consider using the
Sample Planning Wizard. The Wizard is fairly easy to use, and it
is free. You can find the Sample Planning Wizard in Stat Trek's
main menu under the Stat Tools tab. Or you can tap the button below.
Sample Planning Wizard
Problem 1: TwoTailed Test
An inventor has developed a new, energyefficient lawn mower engine.
He claims that the engine will run continuously for 5 hours
(300 minutes) on a single gallon of regular gasoline. From his stock of 2000 engines,
the inventor selects a simple
random sample of 50 engines for testing. 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 twotailed 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
onesample ttest.
 Analyze sample data. Using sample data, we
compute the standard error (SE), degrees of freedom (DF),
and the t statistic 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
twotailed test, the Pvalue is the probability that the
t statistic 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 > 1.77) = 0.04. Thus, the Pvalue = 0.04 + 0.04 = 0.08.
 Interpret results. Since the Pvalue (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,
the population was normally distributed, and the sample size was small relative to the
population size (less than 5%).
Problem 2: OneTailed Test
Bon Air Elementary School has 1000 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. (Assume that test scores in
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 >= 110
Alternative hypothesis: μ < 110
Note that these hypotheses constitute a onetailed 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
onesample ttest.
 Analyze sample data. Using sample data, we
compute the standard error (SE), degrees of freedom (DF),
and the t statistic 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.
Here is the logic of the analysis: Given the alternative hypothesis (μ < 110),
we want to know whether the observed sample mean is small enough to cause us to
reject the null hypothesis.
The observed sample mean produced a t statistic test statistic of 0.894. We use the
t Distribution Calculator to find P(t < 0.894) = 0.19.
This means we would expect to find a sample mean of 108 or smaller in 19 percent
of our samples, if the true population IQ were 110. Thus the Pvalue in this analysis is 0.19.
 Interpret results. Since the Pvalue (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,
the population was normally distributed, and the sample size was small relative to the
population size (less than 5%).