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

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

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

them with a t-test to test a mean

Critical values in a t- distribution

• In section 7.2 you learned how to perform a hypothesis test for a population mean when the sample size was 30 or more. In real life however it is often impractical to collect samples of size 30 or more. However if the population has a normal or nearly normal distribution you can still test the population mean ц. To do so you use the t-sampling distribution with n-1 degrees of freedom.

GuidelinesFinding critical values in a t- distribution1. 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-1Degrees 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.

Example1Finding critical values for tFind the critical value to for a left tailed test

given α= .05 and n=21

Solution:

The degrees of freedom are :

d. f. = n-1 = 21 – 1 =20 Find the critical value in the .05 column at the d.f = 20 the critical value 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. = 13 and α = .01 in the One tail α column.

b. Use a negative sign.

Example 2

Finding the critical values for tFind the critical value to for a right tailed test

with 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 it is 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 two

tailed 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

- =

/

Xt

s n

GuidelinesUsing 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 α3. Identify the degrees of freedom and d.f. = n - 1 sketch the sampling distribution4. 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.

- =

/

Xt

s n

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

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

Example 4A used car dealer says that the mean price of a

2002 Ford F-150 Super Cab is at least $18,800. You suspect that this claim is incorrect and find that a random sample of 14 similar has a mean price of $18000 and a standard deviation of $1250. Is there enough evidence to reject the dealer’s claim at α = .05

Assume the population is normally distributed.

Solution: The claim is that the mean price is at least $18,800. So the null and alternative hypothesis are

Ho: ц ≥ $18800 and H a < 18800

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

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

= ≈ -2.39

- =

/

Xt

s n

1800 18800

1250 / 14

Interpretation

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

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

Try it yourself

An industrial company claims that the pH level of the water in the nearby river is 6.8. You randomly select 19 water samples and measure the pH of each. The sample mean and the standard deviation are 6.7 and .24 respectively. Is there enough evidence to reject the company’s claim at α = .05?

Assume the population is normally distributed.

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

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 inleft tail

t0 t0

A university says the mean number of classroom hours per week for full-time faculty is 11.0. A random sample of the number of classroom 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’s claim?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 a t-distribution with 8 – 1 = 7 d.f.

Testing –Small Sample

t = –1.08 does not fall in the rejection region, so fail to reject H0 at = 0.01

n = 8 = 10.050 s = 2.485

7. Make your decision.

6. Find the test statistic and standardize it

8. Interpret your decision.

There is not enough evidence to reject the university’s claim that faculty spend a mean of 11 classroom hours.

5. Find the rejection region.

Since Ha contains the ≠ symbol, this is a two-tail test.

4. Find the critical values.

–3.499 3.499t0–t0

T-Test of the Mean

Test of = 11.000 vs not = 11.000

Variable N Mean StDev SE Mean T PHours 8 0.050 2.485 0.879 –1.08 0.32

Enter the data in C1, ‘Hours’. Choose t-test in the STAT menu.

Minitab reports the t-statistic and the P-value.

Since the P-value is greater than the level of significance (0.32 > 0.01), fail to reject the null hypothesis at the 0.01 level of significance.

Minitab Solution

Homework 1-25 odd. Pg. 370-371Day 2: 26-34 all pgs. 371-373