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:
Normal distribution is: [latex]{\overline{X}}_{1}–{\overline{X}}_{2}\sim N\left[{\mu }_{1}–{\mu }_{2},\sqrt{\frac{{\left({\sigma }_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({\sigma }_{2}\right)}^{2}}{{n}_{2}}}\right][/latex]
The standard deviation is: [latex]\sqrt{\frac{{\left({\sigma }_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({\sigma }_{2}\right)}^{2}}{{n}_{2}}}[/latex]
The test statistic (z-score) is: [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]
Remember, use the probability table found in the Back Matter - Statistics Tables where needed.
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 the table below.
| 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} - {\overline{X}}_{2}[/latex] = difference in the mean number of months the competing floor waxes last.
[latex]H_0: \mu_1 \le \mu_2[/latex]
[latex]H_a: \mu_1 > \mu_2[/latex]
The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a “>” symbol and goes into [latex]H_a[/latex]. 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} \sim N\left(0,\sqrt{\frac{{0.33}^{2}}{20}+\frac{{0.36}^{2}}{20}}\right)[/latex]
Since [latex]\mu_1 \le \mu_2[/latex] then [latex]\mu_1 - \mu_2 \le 0[/latex] and the mean for the normal distribution is zero.
Calculate the p-value using the normal distribution: [latex]\text{p-value} = 0.1799[/latex]
Graph:
[latex]{\overline{X}}_{1} - {\overline{X}}_{2} = 3 - 2.9 = 0.1[/latex]
Compare [latex]\alpha[/latex] and the p-value: [latex]\alpha = 0.05[/latex] and [latex]\text{p-value} = 0.1799[/latex]. Therefore, [latex]\alpha \lt \text{p-value}[/latex].
Make a decision: Since [latex]\alpha \lt \text{p-value}[/latex], do not reject [latex]H_0[/latex].
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.
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. The table below 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 [latex]30 + 30 = 60[/latex], which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. The subscripts are 1: Democratic senators, 2: Republican senators
Random variable: [latex]{\overline{X}}_{1} - {\overline{X}}_{2} =[/latex] difference in the mean age of Democratic and Republican U.S. senators.
[latex]H_0: \mu_1 \le \mu_2[/latex], so [latex]H_0: \mu_1 - \mu_2 \le 0[/latex]
[latex]H_a: \mu_1 > \mu_2[/latex], so [latex]H_a: \mu_1 - \mu_2 > 0[/latex]
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 [latex]\mu_1 \le \mu_2[/latex], [latex]\mu_1 - \mu_2 \le 0[/latex] and the mean for the normal distribution is zero.
(Calculating the p-value using the normal distribution gives [latex]\text{p-value} = 0.4040[/latex])
Graph:
Compare [latex]\alpha[/latex] and the p-value: [latex]\alpha = 0.05[/latex] and [latex]\text{p-value} = 0.4040[/latex]. Therefore, [latex]\alpha \lt \text{p-value}[/latex].
Make a decision: Since [latex]\alpha \lt \text{p-value}[/latex], do not reject [latex]H_0[/latex].
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.
Videos
Below are helpful videos for the content covered in this Section. Videos are provided from YouTube.
Section 9.2 Review
A hypothesis test of two population means from independent samples where the population standard deviations are known will have these characteristics:
- Random variable: [latex]\overline{{X}_{1}}–\overline{{X}_{2}}[/latex] = the difference of the means
- Distribution: normal distribution
Formula Review
- Normal Distribution:
- [latex]{\overline{X}}_{1}–{\overline{X}}_{2}\sim N\left[{\mu }_{1}–{\mu }_{2},\sqrt{\frac{{\left({\sigma }_{1}\right)}^{2}}{{n}_{1}}+\frac{{\left({\sigma }_{2}\right)}^{2}}{{n}_{2}}}\right][/latex]
- Generally [latex]\mu_1 - \mu_2 = 0[/latex].
- Test Statistic (z-score):
- [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]
- Generally [latex]\mu_1 - \mu_2 = 0[/latex], where:
- [latex]\sigma_1[/latex] and [latex]\sigma_2[/latex] are the known population standard deviations.
- [latex]n_1[/latex] and [latex]n_2[/latex] are the sample sizes.
- [latex]{\overline{x}}_{1}[/latex] and [latex]{\overline{x}}_{2}[/latex] are the sample means.
- [latex]\mu_1[/latex] and [latex]\mu_2[/latex] are the population means.
Section 9.2 Practice
The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. The table below shows the result. Scouters believe that Rodriguez pitches a speedier fastball.
| Pitcher | Sample Mean Speed of Pitches (mph) | Population Standard Deviation |
|---|---|---|
| Wesley | 86 | 3 |
| Rodriguez | 91 | 7 |
- What is the random variable?
- State the null and alternative hypotheses.
- What is the test statistic?
- What is the p-value?
- At the 1% significance level, what is your conclusion?
A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller.
| Plant Group | Sample Mean Height of Plants (inches) | Population Standard Deviation |
|---|---|---|
| Food | 16 | 2.5 |
| No food | 14 | 1.5 |
- Is the population standard deviation known or unknown?
- State the null and alternative hypotheses.
- What is the p-value?
- Draw the graph of the p-value.
- At the 1% significance level, what is your conclusion?
Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point.
| Sample Mean Melting Temperatures (°F) | Population Standard Deviation | |
|---|---|---|
| Alloy Gamma | 800 | 95 |
| Alloy Zeta | 900 | 105 |
- State the null and alternative hypotheses.
- Is this a right-, left-, or two-tailed test?
- What is the p-value?
- Draw the graph of the p-value.
- At the 1% significance level, what is your conclusion?
Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1.
- Is this a test of means or proportions?
- What is the random variable?
- State the null and alternative hypotheses.
- What is the p-value?
- What can you conclude about the two operating systems?
A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3.
Use the Hypothesis Testing with Two Samples - Solution Sheet on the Introduction to Chapter 9 page. (Note: If you are using a Student's t-distribution, including for paired data, you may assume that the underlying population is normally distributed, but in a real situation, you must first prove that assumption, however.)
Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls.
Use the Hypothesis Testing with Two Samples - Solution Sheet on the Introduction to Chapter 9 page. (Note: If you are using a Student's t-distribution, including for paired data, you may assume that the underlying population is normally distributed, but in a real situation, you must first prove that assumption, however.)
Solution
Subscripts: 1 = boys, 2 = girls
- [latex]H_0: \mu_1 \le \mu_2[/latex]
- [latex]H_a: \mu_1 > \mu_2[/latex]
- The random variable is the difference in the mean auto insurance costs for boys and girls.
- normal
- test statistic: [latex]z = 2.50[/latex]
- p-value: 0.0062
- Check student’s solution.
- Correct decision, the reason for it, and an appropriate conclusion:
- [latex]\alpha: 0.05[/latex]
- Decision: Reject the null hypothesis.
- Reason for Decision: [latex]\text{p-value} \lt \alpha[/latex]
- Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls.
A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test.
Use the Hypothesis Testing with Two Samples - Solution Sheet on the Introduction to Chapter 9 page. (Note: If you are using a Student's t-distribution, including for paired data, you may assume that the underlying population is normally distributed, but in a real situation, you must first prove that assumption, however.)
One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). The table below contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife).
| Wife’s Score | 2 | 2 | 3 | 3 | 4 | 2 | 1 | 1 | 2 | 4 |
|---|---|---|---|---|---|---|---|---|---|---|
| Husband’s Score | 2 | 2 | 1 | 3 | 2 | 1 | 1 | 1 | 2 | 4 |
Use the Hypothesis Testing with Two Samples - Solution Sheet on the Introduction to Chapter 9 page. (Note: If you are using a Student's t-distribution, including for paired data, you may assume that the underlying population is normally distributed, but in a real situation, you must first prove that assumption, however.)
Solution
- [latex]H_0: \mu_d = 0[/latex]
- [latex]H_a: \mu_d \lt 0[/latex]
- The random variable [latex]X_d[/latex] is the average difference between husband’s and wife’s satisfaction level.
- [latex]t_9[/latex]
- test statistic: [latex]t = -1.86[/latex]
- p-value: 0.0479
- Check student’s solution
- Correct decision, the reason for it, and an appropriate conclusion:
- [latex]\alpha: 0.05[/latex]
- Decision: Reject the null hypothesis, but run another test.
- Reason for Decision: [latex]\text{p-value} \lt \alpha[/latex]
- Conclusion: This is a weak test because alpha and the p-value are close. However, there is insufficient evidence to conclude that the mean difference is negative.
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).
