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Chapter 6: Problem 53

Left-Handed Lawyers Approximately \(10 \%\) of Americans are left-handed (we will treat this as a known population parameter). A study on the relationship between handedness and profession found that in a random sample of 105 lawyers, 16 of them were left-handed. \({ }^{13}\) Test the hypothesis that the proportion of left-handed lawyers differs from the proportion of left-handed Americans. (a) Clearly state the null and alternative hypotheses. (b) Calculate the test statistic and p-value. (c) What do we conclude at the \(5 \%\) significance level? At the \(10 \%\) significance level?

Short Answer

Expert verified
Null hypothesis: \(P_l = 0.10\). Alternative hypothesis: \(P_l 鈮� 0.10\). Test statistic, \(Z \approx 2.04\). P-value \( \approx 0.0414\). At both \(5 \%\) and \(10 \%\) significance levels, we reject the null hypothesis. Thus, the proportion of left-handed lawyers is different from the proportion of left-handed Americans.

Step by step solution

01

State the Null and Alternative Hypotheses

Null Hypothesis (\(H_0\)): The proportion of left-handed lawyers is the same as the proportion of left-handed Americans, i.e., \(P_l = 0.10\). Alternative Hypothesis (\(H_1\)): The proportion of left-handed lawyers is different from the proportion of left-handed Americans, i.e., \(P_l 鈮� 0.10\).
02

Calculate the Test Statistic and p-value

To calculate the test statistic, we use the formula: \(Z = \frac{\(P_l - P_0\)}{\sqrt{P_0(1-P_0)/n}}\), where \(P_l\) is the sample proportion (16/105 = 0.152), \(P_0\) is the population proportion (0.10), and \(n\) is the sample size (105). Substitute the values into the formula to get: \(Z = \frac{0.152 - 0.10}{\sqrt{0.10(1-0.10)/105}} \approx 2.04\). The p-value can be found using the standard normal distribution table. Since we are looking for a two-tailed test (because our alternative hypothesis is \(P_l 鈮� 0.10\)), the p-value = 2(1 - 0.9793) 鈮� 0.0414 (0.9793 is the value in the Z table for 2.04).
03

Draw Conclusion Based on Significance Level

At the \(5 \%\) significance level, since the p-value (0.0414) is less than the significance level (0.05), we reject the null hypothesis. Hence, we can conclude that the proportion of left-handed lawyers is different from the proportion of left-handed Americans. At the \(10 \%\) significance level, since the p-value (0.0414) is less than the significance level (0.10), we again reject the null hypothesis, leading to the same conclusion.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that there is no effect or no difference, and it serves as the starting point for statistical analysis. It represents the idea that any observed effect is due to chance.
This hypothesis is typically denoted as \(H_0\). In our exercise, the null hypothesis states that the proportion of left-handed lawyers is the same as the proportion of left-handed Americans, \(P_l = 0.10\).
  • The null hypothesis provides a baseline for comparison.
  • If the data shows a significant difference, the null hypothesis may be rejected.
Rejecting the null hypothesis suggests that there is enough evidence to support the alternative hypothesis.
Significance Level
The significance level, often denoted by \(\alpha\), is a threshold used to determine whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error.
Common significance levels are 0.05 (5%) and 0.10 (10%).
  • At a 5% significance level, there is a 5% risk of concluding that a difference exists when there isn't one.
  • At a 10% significance level, this risk increases to 10%.
In our exercise, we compare the p-value to the significance levels to decide whether to reject the null hypothesis. If the p-value is less than \(\alpha\), we reject \(H_0\). The study concludes that the proportion of left-handed lawyers is different from the general population at both significance levels.
Test Statistic
The test statistic helps determine how far the observed data is from the null hypothesis. It is calculated using a formula that depends on the type of test being conducted.
For proportion tests, the Z-test statistic is used, calculated as follows: \[Z = \frac{P_l - P_0}{\sqrt{P_0(1-P_0)/n}}\]Here, \(P_l\) is the sample proportion, \(P_0\) is the population proportion, and \(n\) is the sample size. In the exercise, we find:- \(P_l = \frac{16}{105} \approx 0.152\)- \(P_0 = 0.10\)- \(n = 105\)Substituting these into the formula gives \(Z \approx 2.04\).
This Z-value is then used to find the p-value, which shows how extreme the observed data is under the null hypothesis. The smaller the p-value, the greater the evidence against \(H_0\). The exercise's p-value of 0.0414 leads to the rejection of \(H_0\) at both significance levels.

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