Section 7-2 Hypothesis Testing for the Mean

(n 30)Objective: SWBAT Find a p value and use

them to test a mean ц. How to use P values for a z test. How to find critical values and rejection regions in a normal distribution. How to use rejection regions for a z test.

Using p values to make decisionsDecision Rule Based on P-value To use a P value to make a conclusion in a hypothesis test, compare the p value with α. 1. If P < α then reject Ho

2. If P > α then fail to reject Ho

Example 1Interpreting a P- valueThe P-value for a hypothesis test is P = 0.0237. What is your decision if the Level of significance is (1) α =0.05 and (2) α = 0.01?

Solution 1. Because 0.0237 < 0.05 you should reject the null hypothesis.2. Because 0.0237 > 0.01 you should fail to reject the null hypothesis.

Try it yourself The P – value for a hypothesis test is P = 0.0347. What is your decision if the Level of significance is (1) α = 0.01 and (2) α = 0.05

a. Compare the P-value with the level of significance.b. Make your decision.

Finding the P – value for a Hypothesis TestAfter determining the hypothesis test’s standardized test statistic and the test’s statistic’s area, Do one of the following to find the P-value.

a. For the left tailed test, P = (Area in left tail)b. For the right tailed test, P = (Area in right tail)c. For a two tailed test, P = 2(Area in tail of test statistic)

Example 2

Finding a P - Value for the left tailed testFind the P – value for a left tail hypothesis test with a test statistic of z = -2.23. Decide whether to reject Ho if the level of significance is α = 0.01

Solution:From table 4 in the front of your book the area corresponding to Z= -2.23 is 0.0129 which is the area in the left tail. So the P-value for a left tailed hypothesis test with a test statistic of z=-2.23 is P=0.0129. Because 0.0129 > 0.01. You should fail to reject Ho .

Try it Yourself:Find the P-value for a left tailed hypothesis test with a test statistic of z = -1.62. Decide whether to reject Ho if the level of significance is α = 0.05

Example 3

Finding a P - Value for a Two tailed testFind the P – value for a two tailed hypothesis test with a test statistic of z = 2.14. Decide whether to reject Ho if the level of significance is α = 0.05 Solution:From table 4 in the front of your book the area corresponding to Z= 2.14 is 0.9838 which is the area in the right tail. The area in the right tail is 1- .9838 = 0.0162. So the P-value for a two tailed hypothesis test with a test statistic of z=2.14 is P= 2(0.0162)= 0.0324 . Because 0.0324 < 0.05. You should reject Ho .

Try it Yourself:Find the P – value for a two tailed hypothesis test with a test statistic of z = 2.31. Decide whether to reject Ho if the level of significance is α = 0.01

The z-Test for a MeanThe z-test is a statistical test for a population mean. The z-test can be used: (1) if the population is normal and s is known or (2) when the sample size, n, is at least 30. The test statistic is the sample mean and the standardizedtest statistic is z.

When n 30, use s in place of .

A cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. You work for a national health service and are asked to test this claim. You find that a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg. At = 0.05, do you have enough evidence to reject the company’s claim?

1. Write the null and alternative hypothesis.

2. State the level of significance. = 0.05

3. Determine the sampling distribution.

Since the sample size is at least 30, the sampling distribution is normal.

The z-Test for a Mean (P-value)

4. Find the test statistic and standardize it.

5. Calculate the P-value for the test statistic.

Since this is a right-tail test, the P-value is the area found to the rightof z = 1.44 in the normal distribution. From the table P = 1 – 0.9251

n = 52

s = 10

Test statistic

z = 1.44

Area in right tail

P = 0.0749.

σ= 10

6. Make your decision.

7. Interpret your decision.

Compare the P-value to . Since 0.0749 > 0.05, fail to reject H0.

There is not enough evidence to reject the claim that the mean sodium content of one serving of its cereal is no more than 230 mg.

In other words we ACCEPT the Ho (Null hypothesis)

Example 4:

Hypothesis Testing Using P- ValuesIn an advertisement a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. I s there enough evidence to support the claim at α = 0.01? Use a P value.

Ho : ц > 30 minutes Ha: : ц < 30 min. the level of significance. α = 0.01

3.5 = .583336X n

28.5 30 2.57.5833

z

From Table 4 the P - value = .0051 Since the Alternative hypothesis indicates this Is a left tailed test. So P = .0051 and .0051 < .01 You should decide to reject the Ho (The null hypothesis).

Interpretation: At the 1% level of significance you have sufficient evidence to conclude that the mean delivery time is less than 30 min.

Rejection Area

-2.57

Sampling distribution for

The rejection region is the range of values for which the null hypothesis is not probable. It is always in the direction of the alternative hypothesis. Its area is equal to .

A critical value separates the rejection region from the non-rejection region.

Rejection Regions

Rejection Region

Critical Value z0z z0

The critical value z0 separates the rejection region from the non-rejection region. The area of the rejection region is .

Find z0 for a left-tail test with = .01.

Find z0 for a right-tail test with = .05.

Find –z0 and z0 for a two-tail test with = .01.

z0 = –2.33–z0 = –2.575 and z0 = 2.575

z0 = 1.645

Critical Values

z0 z0

Rejectionregion

Rejectionregion

z0z0

Rejectionregion

Rejectionregion

1. Write the null and alternative hypothesis.

2. State the level of significance.

3. Identify the sampling distribution.

Write H0 and Ha as mathematical statements. Remember H0

always contains the = symbol.

This is the maximum probability of rejecting the null hypothesis when it is actually true. (Making a type I error.)

The sampling distribution is the distribution for the test statistic assuming that the equality condition in H0 is true and that the experiment is repeated an infinite number of times.

Using the Critical Value to Make Test Decisions

6. Find the test statistic.

5. Find the rejection region.4. Find the critical value.

The critical value separates the rejection region of the sampling distribution from the non-rejection region. The area of the critical region is equal to the level of significance of the test.

Perform the calculations to standardize your sample statistic.

z0

Rejection Region

7. Make your decision.

8. Interpret your decision.

If the test statistic falls in the critical region, reject H0. Otherwise, fail to reject H0.

If the claim is the null hypothesis, you will either reject the claim or determine there is not enough evidence to reject the claim.

If the claim is the alternative hypothesis, you will either support the claim or determine there is not enough evidence to support the claim.

A cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. You work for a national health service and are asked to test this claim. You find that a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg. At = 0.05, do you have enough evidence to reject the company’s claim?

2. State the level of significance. = 0.05

3. Determine the sampling distribution.Since the sample size is at least 30, the sampling distribution is normal.

The z-Test for a Mean

1. Write the null and alternative hypothesis.

n = 52 = 232 s = 107. Make your decision.

6. Find the test statistic and standardize it.

8. Interpret your decision.

5. Find the rejection region.

Rejectionregion

Since Ha contains the > symbol, this is a right-tail test.

z = 1.44 does not fall in the rejection region, so fail to reject H0

There is not enough evidence to reject the company’s claim that there is at most 230 mg of sodium in one serving of its cereal.

1.645

4. Find the critical value.

z0

Using the P-value of a Test to Compare Areas

z0

Rejection area0.05

z0 = –1.645

z

Area to the left of z

0.1093

z = –1.23

For a critical value decision, decide if z is in the rejection region If z is in the rejection region, reject H0. If z is not in the rejection

region, fail to reject H0.

= 0.05

For a P-value decision, compare areas.

If reject H0. If fail to reject H0.

P = 0.1093

Homework 1-20 pgs. 357-358 Day2: 21-33 all pgs.358-359 35-45 all pg.360-361