Section 9.2

Testing the Mean

9.2 / 1

Testing the Mean When is KnownLet x be the appropriate random variable. Obtain a simple random

sample (of size n) of x values and compute the sample mean x. 1. State the null and alternate hypotheses and set the level of

significance α.2. If x has a normal distribution, any sample size will work. If we

cannot assume a normal distribution, use n > 30.Use the test statistic:

3 Use the standard normal distribution and the type of test (one-tailed or two-tailed) to find the P-value corresponding to the test statistic.4. If the P-value < α, then reject H0. If the P-value > α, then do not reject

H0.

5. State your conclusion.

/

xts n

3

Your college claims that the mean age of its students is 28 years. You wish to check the validity of this statistic with a level of significance of a = 0.05. Assume the standard deviation is 4.3 years.A random sample of 49 students has a mean age of 26 years.

H0: = 28 H1: ¹ 28 Perform a two-tailed test.

Example Testing the Mean When is Known: Example

Level of significance = α = 0.05

4

Sample Test Statistic

n

xz

statistic test samplex

used being size samplen

ondistributix the of deviation standardσ

H in specified mean

where

0

5

For a two-tailed test: P-value = 2P(z < 3.26) = 2(0.0006) = 0.0012

26.373.4

2826

n

xz

Sample Results

:z statistictest the Calculate

.3.2,26 sx

P-value and Conclusion

• P-value = 0.0012

• α = 0.05. Since the P-value < α , we reject the null hypothesis.

• We conclude that the true average age of students is not 28.

9.2 / 6

Let x be the appropriate random variable. Obtain a simple random sample (of size n) of x values and compute the sample mean x.

1. State the null and alternate hypotheses and set the level of significance α.

2. If x has a mound shaped symmetric distribution, any sample size will work. If we cannot assume this, use n>30

Use the test statistic: with d.f. = n - 13. Use the Student’s t distribution and the type of test (one-tailed

or two-tailed) to find (or estimate) the P-value corresponding to the test statistic.

4. If the P-value < α, then reject H0. If the P-value > α, then do not reject H0.

5. State your conclusion.

Testing the Mean When is Unknown

/

xts n

Using Table 4 to Estimate P-values

Use one-tailed areas as endpoints of the interval containing the P-value for one-tailed tests.

9.2 / 8

P-value for One-tailed Tests

9.2 / 9

10

Use two-tailed areas as endpoints of the interval containing the P-value for one-tailed tests.

Using Table 4 to Estimate P-values

P-value for Two-tailed Tests

9.2 / 11

Example Testing the Mean When is Unknown

The Parks Department claims that the mean weight of fish in a lake is 2.1 kg. We believe that the true average weight is lower than 2.1 kg. Assume that the weights are mound-shaped and symmetric and a sample of five fish caught in the lake weighed an average of 1.99 kg with a standard deviation of 0.09 kg.

Determine the P-value when testing the claim that the mean weight of fish caught in a lake is 2.1 kg (against the alternate that the weight is lower).

A sample of 5 fish weighed an average of 1.99 kg with a standard deviation of 0.09 kg.

Test the Claim Using α = 10%9.2 / 12

Example (cont.) Null Hypothesis: H0: = 2.1 kg

Alternate Hypothesis: H1: < 2.1 kg

Level of significance: α = 0.10

We will complete a left-tailed test with:

The Test Statistic t:

1.99, 0.09, 5, . . 4x s n d f

1.99 2.12.73

/ 0.9/ 5

xt

s n

9.2 / 13

Using Table 4 with t = 2.73 and d.f. = 4

Sample t = 2.739.2 / 14

The t value is between two values in the chart.Therefore the P-value will be in a corresponding interval.

Sample t = 2.739.2 / 15

Sample t = 2.73

Since we are performing a one-tailed test, we use the “one-tail area” line of the chart.

9.2 / 16

Sample t = 2.73

Since we are performing a one-tailed test, we use the “one-tail area” line of the chart.

9.2 / 17

Sample t = 2.73

…

0.025 < P-value < 0.050

9.2 / 18

0.025 < P-value < 0.050

Since the range of P-values is less than a (10%), we reject the null hypothesis.

Interpret the results:At level of significance 10% we rejected the null

hypothesis that the mean weight of fish in the lake was 2.1 kg.

Based on our sample data, we conclude that the true mean weight is actually lower than 2.1 kg.

9.2 / 19

Critical Region (Traditional) Method for Hypothesis Testing

• An alternate technique to the P-value method

• Logically equivalent to the P-value method

9.2 / 20

Critical Region Procedure for Testing When is Known

• Let x be the appropriate random variable. Obtain a simple random sample (of size n) of x values and compute the sample mean x.

• State the null and alternate hypotheses and set the level of confidence α.

• If x has a normal distribution, any sample size will work. If we cannot assume a normal distribution, use n > 30.

9.2 / 21

Critical Region Method for Testing the Mean When is Known

• Use the test statistic:

n

xz

9.2 / 22

Critical Region Method for Testing the Mean When is Known

• Using the level of significance α and the alternate hypothesis, show the critical region and critical values on a graph of the sampling distribution.

• Conclude the test. If the test statistic is in the critical region, then reject H0. If not, do not reject H0.

• State your conclusion.

9.2 / 23

Most Common Levels of Significance

• α = 0.05 and• α = 0.01

9.2 / 24

Critical Region(s)

• The values of x for which we will reject the null hypothesis.

• The critical values are the boundaries of the critical region(s).

26

• Compare the sample test statistics to the critical value(s)

• For a left-tailed test:• If the sample test statistic is < critical value,

reject H0.• If the sample test statistic is > critical value,

fail to reject H0.

Concluding Tests Using the Critical Region Method

Critical Region for H0: = kLeft-tailed Test

9.2 / 27

Concluding Tests Using the Critical Region Method

• Compare the sample test statistics to the critical value(s)

• For a right-tailed test:• If the sample test statistic is > critical value,

reject H0.• If the sample test statistic is < critical value,

fail to reject H0.

9.2 / 28

Critical Region for H0: = k Right-tailed Test

9.2 / 29

30

• Compare the sample test statistics to the critical value(s)

• For a two-tailed test:• If the sample test statistic lies beyond the

critical values, reject H0.• If the sample test statistic lies between the

critical values, fail to reject H0.

Concluding Tests Usingthe Critical Region Method

Critical Region for H0: = k Two-tailed Test

9.2 / 31

32

Critical Values z0 for α = 0.05 and α = 0.01: Left-tailed Test

33

Critical Values z0 for α = 0.05and α = 0.01: Right-tailed Test

34

Critical Values z0 for α = 0.05 and α = 0.01: Two-tailed Test

35

Your college claims that the mean age of its students is 28 years. You wish to check the validity of this statistic with a level of significance of a = 0.05. Assume the standard deviation is 4.3 years.

A random sample of 49 students has a mean age of 26 years.

Testing the Mean When is Known: Example

two

H0: m = 28H1: m ¹ 28

Perform a ________-tailed test.

Level of significance = α = 0.05

Hypothesis Test Example

9.2 / 36

37

Sample Test Statistic

n

xz

statistic test samplex

used being size samplen

ondistributix the of deviation standardσ

H in specified mean

where

0

38

26.373.4

2826

n

xz

Sample Results

:z statistictest the Calculate

.3.2,26 sx

Critical Region for a Two-tailed Test with α = 0.05

9.2 / 39

Our z = 3.26 falls within the critical region.

z = 3.269.2 / 40

Since the test statistic is in the critical region we…

• Reject the Null Hypothesis.

9.2 / 41

Conclusion

• We conclude that the true average age of students is not 28.

9.2 / 42