# 9.1: Two Population Means (Independent)

- Page ID
- 29617

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## Two Population Means (Independent)

## jkesler

- The two independent samples are simple random samples from two distinct populations.
- For the two distinct populations:
- if the sample sizes are small, the distributions are important (should be normal)
- if the sample sizes are large, the distributions are not important (need not be normal)

### NOTE

The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, $\bar X_1 – \bar X_2$, and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, of **the difference in sample means**, $\bar X_1 – \bar X_2$.

The standard error is:

$$\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}}$$

The test statistic (*t*-score) is calculated as follows:

$$t=\frac{(\bar x_1 – \bar x_2) – (\mu_1-\mu_2)}{\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}}}$$

where:

*s*_{1}and*s*_{2}, the sample standard deviations, are estimates of*σ*_{1}and*σ*_{2}, respectively.*σ*_{1}and*σ*_{2}are the unknown population standard deviations.- $\bar x_1$ and $\bar x_2$ are the sample means.
*μ*_{1}and*μ*_{2}are the population means, and the null hypothesis will always be*H*_{0}:*μ*=_{1}*μ*_{2 }

### Note

*μ*_{1} and *μ*_{2} are the population means, and the null hypothesis will always be *H*_{0}: *μ _{1}* =

*μ*This means that the calculation $(\mu_1-\mu_2)$ in our test statistic will always be zero, so we can simplify the test statistic calculation as

_{2 }.$$t=\frac{\bar x_1 – \bar x_2}{\sqrt{\frac{(s_1)^2}{n_1}+\frac{(s_2)^2}{n_2}}}$$

The number of degrees of freedom (*df*) requires a somewhat complicated calculation. However, for this introductory course, we will be using a simplified calculation. The test statistic calculated previously is approximated by the Student’s *t*-distribution with *df *that is 1 less than the smaller of $n_1$ and $n_2$ from your two samples. For example, if $n_1=14$ and $n_2=9$, we would use $df=8$.

Notice that the sample variances (*s*_{1})^{2} and (*s*_{2})^{2} are not pooled, meaning we don’t consider the standard deviation of the samples as if they were both just one big sample together. (If the question comes up, do not pool the variances for means.)

### Example 9.1

#### Independent groups

The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table 9.1. Each populations has a normal distribution.

Sample Size | Average Number of Hours Playing Sports Per Day | Sample Standard Deviation | |
---|---|---|---|

Girls | 9 | 2 | 0.866 |

Boys | 16 | 3.2 | 1.00 |

Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.

### Try It 9.1

Two samples are shown in Table 9.2. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.

Sample Size | Sample Mean | Sample Standard Deviation | |
---|---|---|---|

Population A | 25 | 5 | 1 |

Population B | 16 | 4.7 | 1.2 |

### Example 9.2

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is 4 math classes with a standard deviation of 1.5 math classes. College B samples 10 graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes that a student who graduates from college A **has taken more math classes,** on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.

- Is this a test of two means or two proportions?
- Are the populations standard deviations known or unknown?
- Which distribution do you use to perform the test?
- What is the random variable?
- What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.
- Is this test right-, left-, or two-tailed?
- What is the
*p*-value? - Do you reject or not reject the null hypothesis?
**Conclusion:**

### Try It 9.2

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.

- Are the population standard deviations known?
- Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?

### Example 9.3

A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table 9.3 and Table 9.4.

67.6 | 41.2 | 85.3 | 55.9 | 82.4 | 91.2 | 73.5 | 94.1 | 64.7 | 64.7 |

70.6 | 38.2 | 61.8 | 88.2 | 70.6 | 58.8 | 91.2 | 73.5 | 82.4 | 35.5 |

94.1 | 88.2 | 64.7 | 55.9 | 88.2 | 97.1 | 85.3 | 61.8 | 79.4 | 79.4 |

77.9 | 95.3 | 81.2 | 74.1 | 98.8 | 88.2 | 85.9 | 92.9 | 87.1 | 88.2 |

69.4 | 57.6 | 69.4 | 67.1 | 97.6 | 85.9 | 88.2 | 91.8 | 78.8 | 71.8 |

98.8 | 61.2 | 92.9 | 90.6 | 97.6 | 100 | 95.3 | 83.5 | 92.9 | 89.4 |

Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:

- Is this a test of two means or two proportions?
- Are the population standard deviations known or unknown?
- Which distribution do you use to perform the test?
- What is the random variable?
- What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
- Is this test right, left, or two tailed?
- What is the
*p*-value? - Do you reject or not reject the null hypothesis?
- At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.

(See the conclusion in Example 9.2, and write yours in a similar fashion)

### Note

Be careful not to mix up the information for Group 1 and Group 2!