91Ó°ÊÓ

Consider independent random samples from two populations that are normal or approximately normal, or the case in which both sample sizes are at least \(30 .\) Then, if \(\sigma_{1}\) and \(\sigma_{2}\) are unknown but we have reason to believe that \(\sigma_{1}=\sigma_{2}\), we can pool the standard deviations. Using sample sizes \(n_{1}\) and \(n_{2}\), the sample test statistic \(\bar{x}_{1}-\bar{x}_{2}\) has a Student's \(t\) distribution, where \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{s \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\) with degrees of freedom d.f. \(=n_{1}+n_{2}-2\) and where the pooled standard deviation \(s\) is $$s=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}$$ Note: With statistical software, select the pooled variance or equal variance options. (a) There are many situations in which we want to compare means from populations having standard deviations that are equal. This method applies even if the standard deviations are known to be only approximately equal (see Section \(11.4\) for methods to test that \(\sigma_{1}=\sigma_{2}\) ). Consider Problem 17 regarding average incidence of fox rabies in two regions. For region I, \(n_{1}=16\), \(\bar{x}_{1}=4.75\), and \(s_{1} \approx 2.82\) and for region II, \(n_{2}=15, \bar{x}_{2} \approx 3.93\), and \(s_{2} \approx\) 2.43. The two sample standard deviations are sufficiently close that we can assume \(\sigma_{1}=\sigma_{2}\). Use the method of pooled standard deviation to redo Problem 17 , where we tested if there was a difference in population mean average incidence of rabies at the \(5 \%\) level of significance. (b) Compare the \(t\) value calculated in part (a) using the pooled standard deviation with the \(t\) value calculated in Problem 17 using the unpooled standard deviation. Compare the degrees of freedom for the sample test statistic. Compare the conclusions.

Step by step solution

01

Calculate the Pooled Standard Deviation

The formula for the pooled standard deviation is given by:\[s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\]Plug in the given values: \( n_1 = 16, n_2 = 15, s_1 = 2.82, \) and \( s_2 = 2.43 \).\[ s = \sqrt{\frac{(16 - 1)(2.82)^2 + (15 - 1)(2.43)^2}{16 + 15 - 2}} = \sqrt{\frac{15 \cdot 7.9524 + 14 \cdot 5.9049}{29}} \approx \sqrt{\frac{119.286 + 82.6686}{29}} = \sqrt{\frac{201.9546}{29}} \approx \sqrt{6.964} \approx 2.64\]

02

Calculate the Test Statistic

The test statistic \( t \) is calculated using the formula:\[t = \frac{\bar{x}_1 - \bar{x}_2}{s \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]Substitute the values: \( \bar{x}_1 = 4.75, \bar{x}_2 = 3.93, n_1 = 16, n_2 = 15, \) and \( s \approx 2.64 \).\[t = \frac{4.75 - 3.93}{2.64 \sqrt{\frac{1}{16} + \frac{1}{15}}} = \frac{0.82}{2.64 \sqrt{0.0625 + 0.0667}} \approx \frac{0.82}{2.64 \times \sqrt{0.1292}} \approx \frac{0.82}{2.64 \times 0.3594} \approx \frac{0.82}{0.9489} \approx 0.86\]

03

Determine the Degrees of Freedom

The degrees of freedom for the test is calculated as:\[\text{d.f.} = n_1 + n_2 - 2 = 16 + 15 - 2 = 29\]

04

Compare the Test Statistic to Critical Values

Using a significance level of \( \alpha = 0.05 \) and \( 29 \) degrees of freedom, we look at a \( t \)-distribution table or calculator to find the critical value for a two-tailed test. The critical value is approximately \( \pm 2.045 \).Since the calculated \( t \approx 0.86 \), which is within the range of \(-2.045\) to \(2.045\), we fail to reject the null hypothesis. Thus, there is no evidence at the 5% significance level to conclude a difference between the population means.

05

Compare with Unpooled Results

The \( t \)-value and degrees of freedom from Problem 17 using unpooled standard deviations would generally differ due to calculating the standard error differently.For unpooled calculations: the degrees of freedom (d.f.) will be less due to the Welch's approximation, resulting in a slightly more conservative test.By contrast, using pooled standard deviations may result in a \( t \)-value closer to zero and a higher degree of freedom, potentially impacting the confidence intervals and significance results.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Random Samples

When dealing with statistical data analysis, independent random samples play a crucial role. These samples are drawn from two different populations, ensuring that each sample in one group is independent of the other. This is essential for achieving reliable comparisons between the two sets of data.

Independent samples imply that the scores or observations in one sample do not influence those in the other. This independence helps maintain clear and unbiased results. Thus, when applying statistical tests, it is assumed that the samples have no inherent connection affecting the data collected.

• Ensures that the findings from each group reflect only the differences that occur because of the population parameters and not the sampling method.
• Independence between samples is critical when comparing two population means, as it heavily impacts how conclusions are drawn regarding population differences.

Ensuring samples are independent can be achieved by appropriate randomization during sampling. In practical experiments, analysts place strong emphasis on maintaining this independence, ensuring the validity of results when comparing population means.

Student's t-distribution

The Student's t-distribution is a probability distribution used extensively in statistics when dealing with sample data. Particularly, it is applied in situations where the population standard deviation is unknown. This is common in real-world applications where you may not have access to full data about the population.

The t-distribution is similar to the normal distribution, but with heavier tails. This means it has more area in the tails compared to the normal distribution, making it useful for small sample sizes, where extreme values may occur.

• The shape of the t-distribution is determined by the degrees of freedom (df).
• A smaller df results in wider tails, while a larger df makes the t-distribution look more like the normal distribution.

When comparing means from two different populations using the pooled standard deviation method, the t-statistic follows a t-distribution. This accounts for additional variability introduced through sample measurements. The degrees of freedom in such contexts are calculated as the sum of the sample sizes minus two, which allows for proper scaling and comparison.

Population Means Comparison

Comparing population means is a fundamental aspect of statistical analysis. This is particularly important in hypothesis testing, where you wish to determine if differences between groups are statistically significant. To compare means from two populations effectively, several aspects must be considered.

Firstly, the basic tool used is the difference between sample means, which we assume is approximately normal in distribution when the conditions are met. However, several statistical assumptions underlie this process:

• Both samples should be approximately normally distributed, or at least large enough (typically at least 30 samples).
• The variances in the two populations are assumed to be either known to be equal or only approximately equal, allowing the use of a pooled variance estimate.
• The independence of the samples also plays a critical role in ensuring the integrity of comparisons.

By utilizing the pooled standard deviation, we combine the information from both samples to provide a more accurate estimate of the variability in the underlying populations. This enhances the reliability of the assessment of the difference in population means. Ultimately, this careful approach ensures that differences observed are concluded to be genuine, ruling out the effect of sample peculiarities or imbalances.