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A possible important environmental determinant of lung function in children is the amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups: Group 1 consists of 23 nonsmoking children 5-9 years of age, both of whose parents smoke, who have a mean forced expiratory volume (FEV) of 2.1 L and a standard deviation of \(0.7 \mathrm{L} ;\) group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have a mean FEV of \(2.3 \mathrm{L}\) and a standard deviation of \(0.4 \mathrm{L}\). What is the appropriate test procedure for the hypotheses in Problem 8.31?

Short Answer

Expert verified
Use a two-sample t-test for comparing the independent sample means.

Step by step solution

01

Understand the Hypotheses

The question involves comparing two independent means. Therefore, you need to determine the appropriate hypotheses. The null hypothesis ( H_0 ) states there is no difference in the mean FEV between the two groups, while the alternative hypothesis ( H_a ) states there is a difference in the means.
02

Select the Appropriate Test

Since we are comparing the means of two independent groups and the sample sizes are small (<30), the appropriate test is the two-sample t-test for independent samples. This test will help determine if the observed difference in means is statistically significant.
03

Check Test Assumptions

Ensure that the assumptions of the two-sample t-test are met: normally distributed populations or large sample sizes for the Central Limit Theorem to apply, independent samples, and approximately equal variances. Given the data, we should assume normality and check for variance equality if needed.
04

Information Needed for Test

Record the following information: - Group 1 Mean: 2.1 L, Standard Deviation: 0.7 L, Sample Size: 23 - Group 2 Mean: 2.3 L, Standard Deviation: 0.4 L, Sample Size: 20 You will use this data to perform the t-test.
05

Conduct the T-Test

Use the two-sample t-test formula:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where \(\bar{x}_1 = 2.1\), \(\bar{x}_2 = 2.3\), \(s_1 = 0.7\), \(s_2 = 0.4\), \(n_1=23\), \(n_2=20\). Calculate the t-value.
06

Decision Making

Compare the calculated t-value with the critical t-value from a t-distribution table at a chosen significance level (e.g., 0.05) and degrees of freedom. If the t-value exceeds the critical value, reject the null hypothesis.

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Key Concepts

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

Independent Samples
When dealing with independent samples, it's important to understand what this term means. Independent samples refer to groups that are not connected or related in any way. Each group is distinct, with no overlap in participants or units being studied.
In the context of the exercise given, we have two groups of children.
  • Group 1: Children whose parents smoke.
  • Group 2: Children whose parents do not smoke.
These two groups are independent because the membership of one child in a group does not influence or involve the membership of another child in the other group.
We can think of them as separate pools of data, and this is important because it affects how we analyze the data statistically. The independence of the samples allows us to compare their means using statistical tests such as the two-sample t-test.
Hypothesis Testing
Hypothesis testing is a fundamental part of statistical analysis that allows us to make inferences or judgments about data. In simple terms, it's used to evaluate the evidence provided by the sample data and determine if there is enough statistical evidence to support a certain belief or hypothesis about a population.
For the exercise at hand, the hypothesis testing process involves several steps: 1. **Formulate Hypotheses**: - **Null Hypothesis ( H_0 )**: This posits that there is no difference in mean FEV between the two groups of children. - **Alternative Hypothesis ( H_a )**: This suggests that there is a difference in mean FEV between the two groups. 2. **Significance Level**: Decide on the probability threshold (often 0.05) for rejecting the null hypothesis. 3. **Conduct Test**: Use statistical methods (like the t-test) to analyze the data and draw conclusions.
In essence, hypothesis testing helps us to determine whether any observed differences in data could have occurred by chance or if they are statistically significant.
Mean Comparison
Mean comparison is a straightforward yet essential concept in statistics used to identify differences between averages of two or more groups. By comparing means, we can ascertain if the average outcome for one group is statistically significantly different from another.
In the provided exercise, the mean forced expiratory volume (FEV) of children in households with smoking parents is compared to the mean FEV of children in non-smoking households.
  • Group 1 has a mean FEV of 2.1 L.
  • Group 2 has a mean FEV of 2.3 L.
To determine if this difference is meaningful, we apply the two-sample t-test, which evaluates whether these group means are different enough to rule out the possibility that they are just random sample variations.
By conducting this test, we gain insights into whether environmental factors in the home environment, like parental smoking, might affect children's respiratory health, concluding whether the observed mean difference is statistically significant or not.

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A possible important environmental determinant of lung function in children is the amount of cigarette smoking in the home. Suppose this question is studied by selecting two groups: Group 1 consists of 23 nonsmoking children 5-9 years of age, both of whose parents smoke, who have a mean forced expiratory volume (FEV) of 2.1 L and a standard deviation of \(0.7 \mathrm{L} ;\) group 2 consists of 20 nonsmoking children of comparable age, neither of whose parents smoke, who have a mean FEV of \(2.3 \mathrm{L}\) and a standard deviation of \(0.4 \mathrm{L}\). Assuming this is regarded as a pilot study, how many children are needed in each group (assuming equal numbers in each group) to have a \(95 \%\) chance of detecting a significant difference using a two-sided test with \(\alpha=.05\) ?

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