- 64 Views
- Uploaded on
- Presentation posted in: General

Section 9.2

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Testing the Mean

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.

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

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

- 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.

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.

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

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

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%

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:

Sample t = 2.73

Sample t = 2.73

Sample t = 2.73

Sample t = 2.73

…

Sample t = 2.73

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.

- An alternate technique to the P-value method
- Logically equivalent to the P-value method

- 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.

- Use the test statistic:

- 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.

- α = 0.05 and
- α = 0.01

- The values of x for which we will reject the null hypothesis.
- The critical values are the boundaries of the critical region(s).

- 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.

- 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.

- 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.

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:m = 28

H1: m¹ 28

Perform a ________-tailed test.

two

Level of significance = α = 0.05

z = 3.26

- Reject the Null Hypothesis.

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