The document discusses the t-test, which is used to test hypotheses about population means when the population standard deviation is unknown. It is similar to the z-test but uses the sample standard deviation instead of the population standard deviation. The t-test uses degrees of freedom to determine critical values from a t-distribution table. Examples are provided to demonstrate how to perform a t-test, including stating hypotheses, finding critical values, computing the test statistic, making a decision, and summarizing results.

1. T – TEST FOR A MEAN

2. When population standard
division is unknown, the z test
is not normally used for testing
hypothesis involving means.

3. Instead of the z-test being
used, we use the t-test.

4.  The t-test is a statistical test for a mean of a
population and is used when the population is
normally and approximately distributed and its
Population standard deviation is “unknown”.

5. THE FORMULA OF THE T-TEST IS SIMILAR TO Z-TEST.
Z-Test Formula: T-test Formula:
 (case 1)
(case 2)

6. The Formula for t-test is similar to
z – test.
But since the population standard
deviation σ is unknown, sample
standard deviation s is used.

7. THE CRITICAL VALUES FOR T-TEST

8.  Degrees of Freedom = n-1
 For a one tailed test, find the level by
looking at the top row of the table. The
degrees of freedom is looking down to
the left column.

9.  Ex.1)Find the critical t – value for Significance Level = 0.05
with d.f. = 16 for a left tailed test.
 Critical Value = +1.746.
 Ex.2)Find the critical values for a significance level = 0.10
with d.f.=18 for a two-tailed t-test.
 Critical values are +1.734,-1.734.

10.  Test hypotheses by using t-test (traditional
method). The procedure for t- test and z- test
are the same.
 Step 1: State the hypotheses and identify the
claim.
 Step 2: Find the critical value
 Step 3: Compute the test value.
 Step 4: Make the decision to reject or not
reject the null hypotheses.
 Step 5: Summarize the results.

11.  Ex. Hospital Infections
 A medical investigations claims that the average number of
infections per week at a hospital in southwestern
Pennsylvania is 16.3. A random sample of 10 weeks had a
mean number of 17.7 infections. The sample standard
deviation is 1.8. Is there enough evidence to reject the
investigators claim at a S level of = 0.05.
 Step 1 : H (Null):µ = 16.3(claim) and H(alternative):
µ≠16.3
 Step 2: The critical values are +2.262 and -2.262.
 Step 3: The test value is:

12.  Substiture Teachers Salaries.
 An educator claims that the average salary of
substitute teachers in school districts in
Allegheny County, Pennsylvania is less than
$60 per day. A random sample of eight school
districts is selected, and the daily salaries (in
dollars) are shown. Is there enough evidence
to support the educators claim at S level =
0.10?