Section 7-3 Hypothesis Testing for the Mean (Small Samples)

Section 7-3 Hypothesis Testingfor the Mean (Small Samples)

Objective: SWBAT How to find critical values ina t- distribution.

How to use the t-test to test a mean

How to use technology to find P- values and usethem with a t-test to test a mean

Critical values in a t- distribution

•In section 7.2 you learned how to perform ahypothesis test for a population mean whenthe sample size was 30 or more. In real lifehowever it is often impractical to collectsamples of size 30 or more. However if thepopulation has a normal or nearly normaldistribution you can still test the populationmean ц. To do so you use the t-samplingdistribution with n-1 degrees of freedom.

Guidelines

Finding critical values in a t- distribution

1.Identify the level of significance α.

2.Identify the degrees of freedom d. f. = n-1

3.Find the critical values Using the t- table with n-1

Degrees of freedom. If the hypothesis test is

a. Left tailed use One tail α column with a negative sign

b. right tailed use One tail α column with a positive sign

c. Two tailed use Two tails α column with a negative sign

and a positive sign.

Example1

Finding critical values for t

Find the critical value to for a left tailed testgiven α= .05 and n=21

Solution:

The degrees of freedom are :

d. f. = n-1 = 21 – 1 =20 Find the critical value inthe .05 column at the d.f = 20 the criticalvalue is negative.

So to = - 1.725

Try it yourself

Find a critical value for a left tailed test with

α = .01 and n = 14

a.Find the t- value in the t- table use d.f. = 13and α = .01 in the One tail α column.

b. Use a negative sign.

Example 2

Finding the critical values for t

Find the critical value to for a right tailed testwith α = .01 and n = 17

Solution : the degrees of freedom are:

d. f. = n -1 = 17 – 1 = 16

Use the t- table we can use α = .01 and d. f. =16

So to = 2.583 ( the value is positive because itis for a right tailed test)

Try it yourself

Find the critical value for a right tailed test with

α = .05 and n =9

a.Find the t value in the t- table using d.f = 8,

And in the One tailed column

Example 3

Finding critical values for t

Find the critical values for to and – to fro a twotailed test with α = .05 and n = 26

Solution :the degrees of freedom are:

n-1 = 26 – 1 = 25

because this is a two tailed test one value is

positive and one is negative.

So to = 2..060 and - to = -2.060

Try it yourself

Find the critical values ± to for a two tailed test

With α = .01 and n = 16

a.Find the t value in the t- table using d .f. =15

and α = .01 in the two tail α column.

The t-test for a mean ц(n < 30,σ unknown)

t = (sample mean – Hypothesized mean)

Standard error

The degrees of freedom are

d.f. = n - 1

Guidelines

Using the t test for a mean ц (Small Sample)

1.Stat the claim mathematically identify Ho and Ha

The null and alternative hypothesis

2.Idetify the level of significance Identify α

3.Identify the degrees of freedom and d.f. = n - 1

sketch the sampling distribution

4.Determine any critical values Use the t – table

5. Determine any rejection regions

6. Find the standardized test statistic.

7. Make a deciasion to reject or fail to If t is in the rejection region

reject the null hypothesis. Reject Ho Otherwise fail to

reject Ho.

8. Interpret the decision in the context of the

Original claim.

Remember that when you make a decision thepossibility of a type I or a type II error exists.

We will later discuss how to use a P value for at- test for a mean ц (small sample).

Example 4

A used car dealer says that the mean price of a2002 Ford F-150 Super Cab is at least$18,800. You suspect that this claim isincorrect and find that a random sample of 14similar has a mean price of $18000 and astandard deviation of $1250. Is there enoughevidence to reject the dealer’s claim at α = .05

Assume the population is normally distributed.

Solution: The claim is that the mean price is atleast $18,800. So the null and alternativehypothesis are

Ho: ц ≥ $18800 and H a < 18800

The test is a left tailed test the level ofsignificance is α = .05 and there are

d.f. = 14-1=13 degrees of freedom. So thecritical value is to =-1.771 the rejection regionis t < - 1.771

= ≈ -2.39

Interpretation

Because t is in the rejection region you decide toreject the null hypothesis.

There is enough evidence at the 5% level ofsignificance to reject the claim that the meanprice of a 2002 For F-150 Super Cab is at least$18800.

Try it yourself

An industrial company claims that the pH levelof the water in the nearby river is 6.8. Yourandomly select 19 water samples andmeasure the pH of each. The sample meanand the standard deviation are 6.7 and .24respectively. Is there enough evidence toreject the company’s claim at α = .05?

Assume the population is normally distributed.

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Find the critical value t0 for a left-tailed test given  = 0.01 and n =18.

Find the critical values –t0 and t0 for a two-tailed test given

d.f. = 18 – 1 = 17

t0 = –2.567

d.f. = 11 – 1 = 10

–t0 = –2.228 and t0 = 2.228

The t Sampling Distribution

= 0.05 and n = 11.

Area in

left tail

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A university says the mean number of classroom hours per week forfull-time faculty is 11.0. A random sample of the number ofclassroom hours for full-time faculty for one week is listed below.You work for a student organization and are asked to test this claim.At  = 0.01, do you have enough evidence to reject the university’sclaim?

11.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1

1. Write the null and alternative hypothesis

2. State the level of significance

= 0.01

3. Determine the sampling distribution

Since the sample size is 8, the sampling distribution is at-distribution with 8 – 1 = 7 d.f.

Testing –Small Sample