Chapter 9: Hypothesis Testing with Two Samples
9.2 Two Population Means with Known Standard Deviations
Learning Objectives
By the end of this section, the student should be able to:
- Conduct and interpret hypothesis tests for two population means, population standard deviations known.
Hypothesis Tests for Two Population Means, Population Standard Deviations Known
Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is [latex]\overline{{X}_{1}}–\overline{{X}_{2}}[/latex]. The normal distribution has the following format:
[latex]\sqrt{\frac{{\left({\sigma }_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({\sigma }_{2}\right)}^{2}}{{n}_{2}}}[/latex]
[latex]z=\frac{\left({\overline{x}}_{1}–{\overline{x}}_{2}\right)–\left({\mu }_{1}–{\mu }_{2}\right)}{\sqrt{\frac{{\left({\sigma }_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({\sigma }_{2}\right)}^{2}}{{n}_{2}}}}[/latex]
Example
Independent groups, population standard deviations known: The
mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have normal distributions. The data are recorded in [link].
| Wax | Sample Mean Number of Months Floor Wax Lasts | Population Standard Deviation |
|---|---|---|
| 1 | 3 | 0.33 |
| 2 | 2.9 | 0.36 |
Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of
significance.
Solution
This is a test of two independent groups, two population means, population standard deviations known.
Random Variable: [latex]{\overline{X}}_{1}\text{– }{\overline{X}}_{2}[/latex] = difference in the mean number of months the competing floor waxes last.
H0: μ1 ≤ μ2
Ha: μ1 > μ2
The words “is more effective” says that wax 1 lasts longer than wax 2, on average. “Longer” is a “>” symbol and goes into Ha. Therefore, this is a right-tailed test.
Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula, the distribution is:
[latex]{\overline{X}}_{1}–{\overline{X}}_{2}~N\left(0,\sqrt{\frac{{0.33}^{2}}{20}+\frac{{0.36}^{2}}{20}}\right)[/latex]
Since μ1 ≤ μ2 then μ1 – μ2 ≤ 0 and the mean for the normal distribution is zero.
Calculate the p-value using the normal distribution: p-value = 0.1799
Graph:
[latex]{\overline{X}}_{1}[/latex] – [latex]{\overline{X}}_{2}[/latex] = 3 – 2.9 = 0.1
Compare α and the p-value:α = 0.05 and p-value = 0.1799. Therefore, α < p-value.
Make a decision: Since α < p-value, do not reject H0.
Conclusion: At the 5% level of significance, from the sample data, there is not
sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.
Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ1: and arrow to > μ2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1799 and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw.
Your Turn!
The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. [link] shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance.
| Engine | Sample Mean Number of RPM | Population Standard Deviation |
|---|---|---|
| 1 | 1,500 | 50 |
| 2 | 1,600 | 60 |
Solution
The p-value is almost zero, so we reject the null hypothesis. There is sufficient evidence to conclude that Engine 2 runs at a higher RPM than Engine 1.
Example
An interested citizen wanted to know if Democratic U. S. senators are older than
Republican U.S. senators, on average. On May 26 2013, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years.
Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance.
Solution
This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators
Random variable:[latex]{\overline{X}}_{1}\text{ – }{\overline{X}}_{2}[/latex] = difference in the mean age of Democratic and Republican U.S. senators.
H0: µ1 ≤ µ2 H0: µ1 – µ2 ≤ 0
Ha: µ1 > µ2 Ha: µ1 – µ2 > 0
The words “older than” translates as a “>” symbol and goes into Ha. Therefore, this is a right-tailed test.
Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is: [latex]{\overline{X}}_{1}–{\overline{X}}_{2}\sim N\left[0,\sqrt{\frac{{\left(9.55\right)}^{2}}{30}+\frac{{\left(10.17\right)}^{2}}{30}}\right][/latex]
Since µ1 ≤ µ2, µ1 – µ2 ≤ 0 and the mean for the normal distribution is zero.
(Calculating the p-value using the normal distribution gives p-value = 0.4040)
Graph:
Compare α and the p-value:α = 0.05 and p-value = 0.4040. Therefore, α < p-value.
Make a decision: Since α < p-value, do not reject H0.
Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators.
References
Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf
Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbullying Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013).
“Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013).
Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013).
“State-Specific Prevalence of Obesity Among Adults—United States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013).
“Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/crime/txcrime.htm (accessed June 17, 2013).
