8.3: Testing a claim about a Mean when sigma is known

Introduction

Now that we have a collection of tools to use, we are going to apply them to our population parameters:Mean, Proportion and Standard Deviation/Variance. 

For the population parameter for the mean, we will review two cases much like we did with Confidence intervals. 

Recall that if we know the population standard deviation, \(\sigma\) (sigma), and the requirements for the hypothesis test are met, then we can use standard normal distribution, z-distribution, to calculate the critical values and the p-values for our test. 

Requirements for the z-test:

1. The sample observations are a simple random sample.
2. Either or both of these conditions are satisfied: The population the sample comes from is normally distributed or our sample size is 30 or more, \(n \geq 20\).

For these z-tests, we can use the Traditional method, which focuses on a comparison of test statistic to critical value or the p-value method, which focuses on a comparison of the significance level to the p-value. 

Outline of Steps of Hypothesis Testing

Traditional Method:	P-Value Method
Step 1: Setup the Hypotheses using Symbolic Notation.	Step 1: Setup the Hypotheses using Symbolic Notation.
Step 2: Calculate the test statistic: \(z = \dfrac{\bar{x} - \mu}{\dfrac{\sigma}{\sqrt{n}}}\)	Step 2: Calculate the test statistic: \(z = \dfrac{\bar{x} - \mu}{\dfrac{\sigma}{\sqrt{n}}}\)
Step 3: Identify the critical value using table or technology and the test type (left tail, right tail, two tail)	Step 3: Identify the p-value based on the table or technology and the test type (Left tail, right tail, two tail)
Step 4: Compare Test statistic to Critical Value to make a decision regarding the null hypothesis, \(H_0\).	Step 4: Compare the p-value to the significance level to make a decision regarding the null hypothesis, \(H_0\).
Step 5: Summarize your results, include the significance level.	Step 5: Summarize your results, include the significance level.

 For Step 3: Traditional Method, to find the critical value using a TI 83/84, the command is:

• Left tail test: \(z_{\alpha} = \text{invNorm}(\alpha)\) -- Expect a negative critical value since an area of \(\alpha\) is on the left side of the distribution. Creating a critical region to its left. 
• Right tail test: \(z_{\alpha} = \text{invNorm}(1 - \alpha)\) -- Expect a positive critical value since an area of \(1 - \alpha\) is on the right side of the distribution. Creating a critical region to its right. 
• Two tail test:\(z_{\frac{\alpha}{2}} = \text{invNorm}(\frac{\alpha}{2})\) -- The significance is split symmetrically on both sides of the distribution, so we expect a negative and positive critical value, creating two critical regions, one in each tail. 

For Step 3: p-value Method, to find the critical value using a TI 83/84, the command is: 

• Left tail test: p-value = normalcdf(\(-10^{99}\), Test Statistic) -- Left tail means we need the area to the left of the test statistic.
• Right tail test: p-value = normalcdf(Test Statistic, \(-10^{99}\)) -- Right tail means we need the area to the right of the test statistic. 
• Two tail test: P-vaule = 2 \(\cdot\) normalcdf((\(-10^{99}\), Test Statistic) if Test Statistic is negative and 2 \(\cdot\) normalcdf(Test Statistic, \(-10^{99}\)) if the Test Statistic is negative. We need to double the p-value since our significance level is split between the two tails. 

If you prefer to use a z-distribution table (z-table), then you would follow a similar technique when identifying the values on the table. 

For Step 4: Traditional Method - Compare a z-score to a z-score. 

• If the test statistic is within the critical region, then reject the null hypothesis. 
  • In the case of a left tail test, we reject \(H_0\) when the test statistic is to the left of the critical value in the critical region.
  • In the case of a right tail test, we reject \(H_0\) when the test statistic is to the right of the critical value in the critical region.
  • In the case of a two tail test, we reject \(H_0\) when the test statistic is in the critical region, which is both tails of the distribution.

• If the test statistic is outside the critical region, then do not reject the null hypothesis. 

For Step 4: P-value Method - Compare an area to an area

• If \(\text{p-value} \leq \alpha\), then reject the null hypothesis. 
• If \(\text{p-value} > \alpha\), then do not reject the null hypothesis. 

Full Hypothesis Test Examples

Example \(\PageIndex{1}\)

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal and Frank would like 5% significance.

Answer

Step 1: Set up the Hypothesis Test:

Since the problem is about a mean, our parameter is \(\mu\). The claim is "Jeffrey swims faster" so we can interpret this as his mean swim time decreases, which gives us a claim of \(\mu < 16.43\). This claim involves only an inequality, so we will assign it to the Alternate Hypothesis, \(H_{1}\). The null hypothesis is then its complement, but it gets simplified to just "=."

\(H_{0}: \mu = 16.43\)

\(H_{1}: \mu < 16.43\)

The "\(<\)" in our alternate hypothesis tells us this is a left-tailed test.

Determine the distribution needed:

Random variable: \(\bar{x} =\) the mean time to swim the 25-yard freestyle.

Distribution for the test: \(\bar{x}\) is normal and population standard deviation is known: \(\sigma = 0.8\)). Use the standard normal distribution.

Step 2: Find the Test Statistic:

\(\mu = 16.43\) comes from \(H_{0}\) and not the data. \(\sigma = 0.8\), and \(n = 15\).

\(z =\dfrac{\bar{X} - \mu}{\frac{\sigma_{x}}{\sqrt{n}}}\)

\(z =\dfrac{16 - 16.43}{\frac{0.8}{\sqrt{15}}}\)

\(z = -2.08172...\)

Step 3: Find the Critical value or the p-value

Side-by-Side of Traditional vs. p-value method.

Traditional Method	p-value method
\(z_{\alpha} = \text{invNorm}(\alpha)\) \(z_{0.05} = \text{invNorm}(0.05)\) = -1.64485... \(z_{0.05} \approx -1.64\)	p-value = normalcdf(\(-10^{99}\), Test Statistic) p-value = normalcdf(\(-10^{99}\), -2.0817) = 0.01868... p-value \(\approx\) 0.0187

Step 4: Make a decision

Side-by-Side of Traditional vs. p-value method.

Traditional Method

p-value method

Figure \(\PageIndex{1}\): Critical value v. Test Statistic Since we are just comparing z-scores, there is no need to draw the area of our standard normal curve. The number line with with the critical region shaded is enough to see that the Test Statistic (TS on the number line) is inside the critical region which is to the left of the Critical Value.  We reject the null hypothesis, \(H_{0}\).

Figure \(\PageIndex{2}\) We compare the area to the left of the test statistic to the area to the left of the significance level.  p-value vs. \(\alpha\) 0.0187 \(\leq\) 0.05 We reject the null hypothesis, \(H_{0}\). Note: What does the p-value mean? If \(H_{0}\) is true, there is a 0.0187 probability (1.87%) that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Step 5: Summarize the results. 

Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.

The Type I and Type II errors for this problem are as follows:

The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

Note

In this introductory course, you will see both method for hypothesis testing, however, the p-value method is the more generally accepted method of hypothesis testing due to the improvement in technology for calculating these number with improved accuracy. There are a few method to calculate the p-value, so consult your instructor for which techniques are allowed. 

Both techniques will be show in single population hypothesis tests for this chapter but the p-value will eventually be the only technique written out in examples. 

Exercise \(\PageIndex{1}\)

The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of three yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 42 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using \(\alpha = 0.05\). Assume the throw distances for footballs are normal.

Answer

Step 1: Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean. The claim is "throw farther than 40 yards" so the mean should be greater than 40. This doesn't have an equality, so we assign this claim to the alternate hypothesis. 

• \(H_{0}: \mu = 40\)
• \(H_{1}: \mu > 40\)

This is a right tail test. 

Step 2: Find the Test Statistic:

\(\mu = 40\) comes from \(H_{0}\) and not the data. \(\sigma = 3\), and \(n = 20\).

\(z =\dfrac{\bar{X} - \mu}{\frac{\sigma_{x}}{\sqrt{n}}}\)

\(z = 2.98142...\approx 2.98\)

Step 3: Find the Critical value or the p-value

Since the alternate hypothesis is a "Greater than" we want the critical region and p-value to be on the right side of the distribution.

Side-by-Side of Traditional vs. p-value method.

Traditional Method	p-value method
\(z_{\alpha} = \text{invNorm}(1 -\alpha)\) \(z_{0.05} = \text{invNorm}(0.95)\) = 1.64485... \(z_{0.05} \approx 1.64\)	p-value = normalcdf( Test Statistic,\(10^{99}\)) p-value = normalcdf(2.98142, \(10^{99}\)) = 0.0014346... p-value \(\approx\) 0.0014

Step 4: Make a decision

Side-by-Side of Traditional vs. p-value method.

Traditional Method	p-value method
Figure \(\PageIndex{3}\): Critical value v. Test Statistic The number line with with the critical region shaded is enough to see that the Test Statistic is inside the critical region which is to the right of the Critical Value.  We reject the null hypothesis, \(H_{0}\).	We compare the area to the right of the test statistic to the area to the right of the significance level.  p-value vs. \(\alpha\) 0.0014 \(\leq\) 0.05 We reject the null hypothesis, \(H_{0}\).

Step 5: Summarize the results. 

Using \(\alpha = 0.05\), there is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.

Example \(\PageIndex{2}\)

A college football coach thought that his players could bench press a mean weight of 275 pounds. It is known that the standard deviation is 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1).

Conduct a hypothesis test using a 5% level of significance to determine if the bench press mean is more than 275 pounds.

Answer

Step 1: Set up the Hypothesis Test:

Since the problem is about a mean weight, this is a test of a single population mean. The claim is "mean is more than 275." Since this doesn't involve any equality, we assign it to the alternate hypothesis, \(H_{1}\). 

• \(H_{0}: \mu = 275\)
• \(H_{1}: \mu > 275\)

This is a right-tailed test.

Step 2: Find the Test Statistic:

\(\mu = 275\) comes from \(H_{0}\) and not the data. \(\sigma = 55\), and \(n = 30\).

\(\bar{x}\) comes from the provided data by finding the mean for the 30 samples. \(\bar{x} = 286.1666...\approx 286.17\)

\(z =\dfrac{\bar{X} - \mu}{\frac{\sigma_{x}}{\sqrt{n}}}\)

\(z =\dfrac{286.17 - 275}{\frac{55}{\sqrt{30}}}\)

\(z = 1.112042...\approx 1.11\)

Step 3: Find the Critical value or the p-value

Since the alternate hypothesis is a "Greater than" we want the critical region and p-value to be on the right side of the distribution.

Side-by-Side of Traditional vs. p-value method.

Traditional Method	p-value method
\(z_{\alpha} = \text{invNorm}(1 -\alpha)\) \(z_{0.05} = \text{invNorm}(0.95)\) = 1.64485... \(z_{0.05} \approx 1.64\)	p-value = normalcdf( Test Statistic,\(10^{99}\)) p-value = normalcdf(1.11204, \(10^{99}\)) = 0.13305992... p-value \(\approx\) 0.1331

Step 4: Make a decision

Side-by-Side of Traditional vs. p-value method.

Traditional Method	p-value method
Figure \(\PageIndex{4}\): Critical value v. Test Statistic The number line with with the critical region shaded is enough to see that the Test Statistic is outside the critical region which is to the right of the Critical Value.  We do not reject the null hypothesis, \(H_{0}\).	We compare the area to the right of the test statistic to the area to the right of the significance level.  p-value vs. \(\alpha\) 0.1331 \(>\) 0.05 We do not reject the null hypothesis, \(H_{0}\).

Step 5: Summarize the results. 

At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.

References

1. Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
2. Data from Bloomberg Businessweek. Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
3. Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
4. Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
5. Data from Growing by Degrees by Allen and Seaman.
6. Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
7. Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
8. Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
9. Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm.
10. Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
11. Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
12. Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
13. Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1.
14. Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
15. Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
16. “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
17. Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
18. Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Glossary

Central Limit Theorem

Given a random variable (RV) with known mean \(\mu\) and known standard deviation \(\sigma\). We are sampling with size \(n\) and we are interested in two new RVs - the sample mean, \(\bar{X}\), and the sample sum, \(\sum X\). If the size \(n\) of the sample is sufficiently large, then \(\bar{X} - N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)\) and \(\sum X - N \left(n\mu, \sqrt{n}\sigma\right)\). If the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal \(n\) times the population mean. The standard deviation of the distribution of the sample means, \(\frac{\sigma}{\sqrt{n}}\), is called the standard error of the mean.

Contributors and Attributions

• Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/30189442-699...b91b9de@18.114.