91Ó°ÊÓ

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, \(P\) -value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. In a random sample of males, it was found that 23 write with their left hands and 217 do not. In a random sample of females, it was found that 65 write with their left hands and 455 do not (based on data from "The Left-Handed: Their Sinister History," by Elaine Fowler Costas, Education 91Ó°ÊÓ Information Center, Paper 399519). We want to use a \(0.01\) significance level to test the claim that the rate of left-handedness among males is less than that among females. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, is the rate of left-handedness among males less than the rate of lefthandedness among females?

Step by step solution

01

Define the Hypotheses

Identify the null and alternative hypotheses. Let \[ p_1 \] be the proportion of left-handed males, and \[ p_2 \] be the proportion of left-handed females.Null Hypothesis (\[ H_0 \]): \[ p_1 \, = \, p_2 \]Alternative Hypothesis (\[ H_1 \]): \[ p_1 \, < \, p_2 \]

02

Collect and Calculate Proportions

Calculate the sample proportions for each group:\[ \hat{p}_1 = \frac{23}{23 + 217} = \frac{23}{240} \approx 0.0958 \]\[ \hat{p}_2 = \frac{65}{65 + 455} = \frac{65}{520} \approx 0.125 \]

03

Calculate the Test Statistic

The test statistic for comparing two proportions can be calculated using the formula:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]where\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{23 + 65}{240 + 520} = \frac{88}{760} \approx 0.1158 \]The test statistic is:\[ z = \frac{0.0958 - 0.125}{\sqrt{0.1158(1 - 0.1158)(\frac{1}{240} + \frac{1}{520})}} \approx -1.05 \]

04

Find the P-value

Use the standard normal distribution (Z-distribution) to find the P-value. The P-value corresponding to \( z = -1.05 \) is about 0.1469 (using Z-tables or statistical software).

05

Compare the P-value to the Significance Level

The significance level is 0.01. Since the P-value (0.1469) is greater than the significance level (0.01), we fail to reject the null hypothesis.

06

State the Conclusion for Hypothesis Test

We fail to reject the null hypothesis. There is not enough evidence at the 0.01 significance level to support the claim that the rate of left-handedness among males is less than that among females.

07

Construct a Confidence Interval

Construct a 99% confidence interval for the difference in proportions. The formula for the confidence interval for the difference in proportions is:\[ (\hat{p}_1 - \hat{p}_2) \pm Z_{\alpha/2} \sqrt{\hat{p}_1(1-\hat{p}_1)/n_1 + \hat{p}_2(1-\hat{p}_2)/n_2} \]Using a Z value of 2.576 for a 99% confidence level:\[ CI = (0.0958 - 0.125) \pm 2.576 \sqrt{0.0958(1-0.0958)/240 + 0.125(1-0.125)/520} \approx -0.0292 \pm 0.0520 \]Which gives us:\[ CI \approx (-0.0812, 0.0228) \]

08

Conclusion for Confidence Interval

Since the confidence interval contains 0, we do not have sufficient evidence to conclude that the proportion of left-handed males is less than that of females.

09

Final Conclusion

Based on both the hypothesis test and the confidence interval, we conclude that there is not enough evidence to support the claim that the rate of left-handedness among males is less than that among females.

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Key Concepts

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

null hypothesis

The null hypothesis (often denoted as \(H_0\)) is a statement that there is no effect or no difference. It acts as a baseline or starting assumption for statistical testing. For example, in our problem, the null hypothesis is that the proportion of left-handed males (\( p_1 \)) is equal to the proportion of left-handed females (\( p_2 \)). In notation, this is written as: \[ H_0: p_1 = p_2 \] We use the null hypothesis to test whether any observed differences between groups (like males and females) are due to random chance.

alternative hypothesis

The alternative hypothesis (denoted as \( H_1 \)) is a statement that changes based on the claim we are testing. It represents a new effect or difference we're investigating. In our scenario, we want to test if the proportion of left-handed males is less than that of females. So, the alternative hypothesis states: \[ H_1: p_1 < p_2 \] If we find enough evidence against the null hypothesis, we'll support the alternative hypothesis, indicating a significant difference between the two proportions.

test statistic

The test statistic is a value determined from sample data, which we compare against a threshold to decide whether to reject the null hypothesis. For comparing two proportions, we use the Z-test formula: \[ z = \frac{\hat{p_1} - \hat{p_2}}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_1} + \frac{1}{n_2})}} \] Here, \( \hat{p_1} \) and \( \hat{p_2} \) are the sample proportions, and \( \hat{p} \) is the pooled proportion. For our exercise, the test statistic is calculated to be approximately -1.05. This negative value suggests that our sample mean of proportions aligns with our alternative hypothesis direction.

confidence interval

A confidence interval offers a range of values where we expect the true population parameter to lie, based on our sample data. For the difference in proportions, the confidence interval helps us determine if there's a significant difference. The formula for the confidence interval of the difference in proportions is: \[ (\hat{p_1} - \hat{p_2}) \pm Z_{\alpha /2} \sqrt{\hat{p_1}(1- \hat{p_1})/n_1 + \hat{p_2}(1- \hat{p_2})/n_2} \] For a 99% confidence level (Z-value of about 2.576), we found our interval to be approximately -0.0812 to 0.0228. Since this interval contains 0, we don't have enough evidence to claim a difference between male and female left-handedness proportions.

P-value

The P-value quantifies the evidence against the null hypothesis, representing the probability of observing the test result, or one more extreme, if the null hypothesis is true. For our test, the P-value is calculated from the Z-distribution. Given our Z-value of about -1.05, the corresponding P-value is approximately 0.1469. Because this P-value is greater than our significance level of 0.01, we do not reject the null hypothesis. This means there's not enough evidence to support the claim that the rate of left-handedness among males is less than that among females.