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An article in Archaeometry involved an analysis of 26 samples of Romano- British pottery, found at four different kiln sites in the United Kingdom. \(^{9}\) The samples were analyzed to determine their chemical composition and the percentage of aluminum oxide in each of \(10 \mathrm{sam}-\) ples at two sites is shown below. \begin{tabular}{c|c} Island Thorns & Ashley Rails \\\ \hline 18.3 & 17.7 \\ 15.8 & 18.3 \\ 18.0 & 16.7 \\ 18.0 & 14.8 \\ 20.8 & 19.1 \end{tabular} Does the data provide sufficient information to indicate that there is a difference in the average percentage of aluminum oxide at the two sites? Test at the \(5 \%\) level of significance.

Short Answer

Expert verified
Answer: No, there is insufficient evidence to conclude that there is a difference in the average percentage of aluminum oxide between the Island Thorns and Ashley Rails kiln sites at a 5% level of significance.

Step by step solution

01

Defining the hypotheses

Let \(\mu_1\) be the mean percentage of aluminum oxide at the Island Thorns site, and \(\mu_2\) be the mean percentage at the Ashley Rails site. We want to test whether there is a difference in the average percentage, so our null hypothesis (\(H_0\)) and alternative hypothesis (\(H_a\)) are: \(H_0: \mu_1 = \mu_2\) \(H_a: \mu_1 \neq \mu_2\)
02

Setting the significance level

The problem states to use a 5% level of significance. That means our significance level, \(\alpha\), is: \(\alpha = 0.05\)
03

Calculate the sample means and sample standard deviations

Using the given data, calculate the sample means (\(\bar{x}_1\) and \(\bar{x}_2\)) and the sample standard deviations (\(s_1\) and \(s_2\)) for both sites. Island Thorns: \(\bar{x}_1 = \frac{1}{5} (18.3 + 15.8 + 18.0 + 18.0 + 20.8) = 18.18\) \(s_1^2 = \frac{1}{4}((15.8-18.18)^2 + (18-18.18)^2 + (18.3-18.18)^2 + (18-18.18)^2 + (20.8-18.18)^2) = 4.539\) Ashley Rails: \(\bar{x}_2 = \frac{1}{5} (17.7 + 18.3 + 16.7 + 14.8 + 19.1) = 17.32\) \(s_2^2 = \frac{1}{4}((17.7-17.32)^2 + (18.3-17.32)^2 + (16.7-17.32)^2 + (14.8-17.32)^2 + (19.1-17.32)^2) = 3.732\)
04

Calculate the test statistic

We will use the independent t-test to compare the means. We calculate the test statistic, \(t\), as: \( t=\frac{(\bar{x}_1-\bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} =\frac{(18.18-17.32)- 0 }{\sqrt{\frac{4.539}{5} + \frac{3.732}{5}}} = 1.238\)
05

Determine the critical value and make a decision

Since the alternative hypothesis is a two-tailed test, \(H_a: \mu_1 \neq \mu_2\), we look for critical values at \(t_{0.025}\) with degrees of freedom (df) given by the smaller value between \(n_1 - 1\) and \(n_2 - 1\). In this case, \(df = 5 - 1 = 4\). Looking up the t-table or using a calculator, we find that \(t_{0.025} = \pm 2.776\). Now we compare the calculated test statistic (\(t = 1.238\)) with the critical values (\(t_{0.025} = \pm 2.776\)). Since \(-2.776 < 1.238 < 2.776\), we fail to reject the null hypothesis.
06

Conclusion

As we fail to reject the null hypothesis, there is insufficient evidence to conclude that there is a difference in the average percentage of aluminum oxide between the Island Thorns and Ashley Rails kiln sites at a 5% level of significance.

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Key Concepts

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

t-test
The t-test is a statistical test used to compare the means of two groups. It's especially useful when dealing with small sample sizes. In this exercise, the t-test helps us determine whether there's a significant difference in the average percentage of aluminum oxide between two different kiln sites. By applying the t-test, we can analyze if the apparent difference in means is due to random chance or a specific factor affecting the samples.

There are different variants of t-tests, including the independent t-test we're using here. The independent t-test is applicable when two groups are separate and unrelated. In our scenario, Island Thorns and Ashley Rails serve as independent groups because they come from distinct sites.

To conduct a t-test, we first need to calculate the difference between the sample means, then divide this by the pooled standard deviation of the samples. This results in the t-statistic, which we compare against a critical value to determine statistical significance.
significance level
The significance level, often denoted as \(\alpha\), is a threshold used to determine whether a statistical result is significant. It represents the probability of rejecting the null hypothesis when it is actually true. In most cases, a 5% significance level is used, meaning there is a 5% risk of concluding that a difference exists when there is none. This exercise uses this common significance level, \(\alpha = 0.05\).

At a 5% significance level, we need to determine the critical value from the t-distribution that our test statistic will be compared against. If the absolute value of our calculated test statistic exceeds this critical value, we reject the null hypothesis.

However, in our exercise, since the calculated t-statistic is within the range of critical values, we fail to reject the null hypothesis, suggesting there is no significant difference at this level of significance.
degrees of freedom
Degrees of freedom (df) refer to the number of independent values or quantities that can vary in a statistical calculation. They play a crucial role in determining the critical value needed for the t-test. The more data points we have (higher degrees of freedom), the more reliable our test becomes.

For an independent t-test comparing two means, the degrees of freedom is usually the smallest of \(n_1 - 1\) and \(n_2 - 1\), where \(n\) is the sample size for each group. In this case, both samples are of size 5, so the degrees of freedom is \(4\).

The degrees of freedom are essential, as they determine the shape of the t-distribution used to obtain the critical values. A smaller df results in a wider t-distribution, which means a larger critical value is needed to find significance.
null hypothesis
The null hypothesis (\(H_0\)) is a statement that asserts there is no effect or no difference. It's a default position suggesting that any observed differences are due to sampling or experimental error rather than a specific cause. In hypothesis testing, we set up a null hypothesis to have something to compare against.

In the exercise, the null hypothesis was that the mean percentage of aluminum oxide at Island Thorns is equal to that at Ashley Rails. Formally, \(H_0: \mu_1 = \mu_2\).

A critical part of hypothesis testing is deciding whether to reject or not reject the null hypothesis. If the evidence from the data is strong enough (i.e., the test statistic falls outside the critical range), we reject the null hypothesis. Here, the results led us to not reject the null hypothesis, meaning there wasn't enough evidence to prove a difference in mean percentages between the two sites at the 5% significance level.

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Most popular questions from this chapter

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