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Section 9.2. Testing the Mean . Testing the Mean  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 .

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Section 9.2

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Section 9 2

Section 9.2

Testing the Mean 


Testing the mean when is known

Testing the Mean  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.

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

  • 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.


Example testing the mean when is known example

Example Testing the Mean When  is Known: Example

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.

Level of significance = α = 0.05


Sample test statistic

Sample Test Statistic


Sample results

Sample Results

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


P value and conclusion

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.


Testing the mean when is unknown

Testing the Mean  When  is Unknown

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.

  • 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 - 1

    3. 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.


Using table 4 to estimate p values

Using Table 4 to Estimate P-values

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


P value for one tailed tests

P-value for One-tailed Tests


Using table 4 to estimate p values1

Using Table 4 to Estimate P-values

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


P value for two tailed tests

P-value for Two-tailed Tests


Example testing the mean when is unknown

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%


Example cont

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:


Using table 4 with t 2 73 and d f 4

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

Sample t =  2.73


Section 9 2

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

Sample t =  2.73


Since we are performing a one tailed test we use the one tail area line of the chart

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

Sample t =  2.73


Since we are performing a one tailed test we use the one tail area line of the chart1

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

Sample t =  2.73


0 025 p value 0 050

0.025 < P-value < 0.050

Sample t = 2.73


0 025 p value 0 0501

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.


Critical region traditional method for hypothesis testing

Critical Region (Traditional) Method for Hypothesis Testing

  • An alternate technique to the P-value method

  • Logically equivalent to the P-value method


Critical region procedure for testing when is known

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.


Critical region method for testing the mean when is known

Critical Region Method for Testing the Mean  When  is Known

  • Use the test statistic:


Critical region method for testing the mean when is known1

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.


Most common levels of significance

Most Common Levels of Significance

  • α = 0.05 and

  • α = 0.01


Critical region s

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).


Concluding tests using the critical region method

Concluding Tests Using the Critical Region Method

  • 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.


Critical region for h 0 k left tailed test

Critical Region for H0:  = kLeft-tailed Test


Concluding tests using the critical region method1

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.


Critical region for h 0 k right tailed test

Critical Region for H0:  = k Right-tailed Test


Concluding tests using the critical region method2

Concluding Tests Usingthe Critical Region Method

  • 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.


Critical region for h 0 k two tailed test

Critical Region for H0:  = k Two-tailed Test


Critical values z 0 for 0 05 and 0 01 left tailed test

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


Critical values z 0 for 0 05 and 0 01 right tailed test

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


Critical values z 0 for 0 05 and 0 01 two tailed test

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


Testing the mean when is known example

Testing the Mean When  is Known: Example

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.


Hypothesis test example

Hypothesis Test Example

H0:m = 28

H1: m¹ 28

Perform a ________-tailed test.

two

Level of significance = α = 0.05


Sample test statistic1

Sample Test Statistic


Sample results1

Sample Results


Critical region for a two tailed test with 0 05

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


Our z 3 26 falls within the critical region

Our z =  3.26 falls within the critical region.

z =  3.26


Since the test statistic is in the critical region we

Since the test statistic is in the critical region we…

  • Reject the Null Hypothesis.


Conclusion

Conclusion

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


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