T – TEST FOR A MEAN

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

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

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”.

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

(case 2)

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.

THE CRITICAL VALUES FOR T-TEST

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.

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.

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.

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:

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?