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A headline that appeared in Woman's World stated "Black Currant Oil Curbs Hair Loss!" (Woman's World, April 4, 2016). This claim was based on an experiment described in the paper "Effect of a Nutritional Supplement on Hair Loss in Women" (Journal of Cosmetic Dermatology [2015]: 76-82). In this experiment, women with stage 1 hair loss were assigned at random to one of two groups. One group was a control group who did not receive a nutritional supplement. Of the 39 women in this group, 20 showed increased hair density at the end of the study period. Those in the second group received a nutritional supplement that included fish oil, black currant oil, vitamin E, vitamin C, and lycopene. Of the 80 women in the supplement group, 70 showed increased hair density at the end of the study period. a. Is there convincing evidence that the proportion with increased hair density is greater for the supplement treatment than for the control treatment? Test the appropriate hypotheses using a 0.01 significance level. b. Write a few sentences commenting on the headline that appeared in Woman's World. c. Based on the description of the actual experiment and the result from your hypothesis test in Part (a), suggest a more appropriate headline.

Step by step solution

01

State the null and alternative hypothesis

The null hypothesis (H鈧�) states that there is no difference in the proportion of increased hair density between the control group and the supplement group, and the alternative hypothesis (H鈧�) states that the supplement group has a higher proportion of increased hair density than the control group. H鈧�: p鈧� = p鈧� H鈧�: p鈧� < p鈧� where p鈧� is the proportion of increased hair density in the control group and p鈧� is the proportion of increased hair density in the supplement group.

02

Choose the significance level

The given significance level for this hypothesis test is 伪 = 0.01.

03

Calculate the sample proportions

From the provided data, we can calculate the sample proportions: - For the control group, there are 20 women with increased hair density out of a total of 39 women: - \(\hat{p}_{1} = \frac{20}{39}\) - For the supplement group, there are 70 women with increased hair density out of a total of 80 women: - \(\hat{p}_{2} = \frac{70}{80}\)

04

Calculate the pooled sample proportion and standard error

Calculate the pooled sample proportion by finding the overall proportion among the groups combined: \( \hat{p} = \frac{n_{1}\hat{p}_{1} + n_{2}\hat{p}_{2}}{n_{1} + n_{2}} \) Next, calculate the standard error (SE) using the pooled sample proportion: SE = \(\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_{1}} + \frac{1}{n_{2}})}\)

05

Compute the test statistic and p-value

The test statistic (z) is calculated as: \( z = \frac{(\hat{p}_{1} - \hat{p}_{2}) - 0}{SE} \) Using a Z-table or software, find the p-value corresponding to the test statistic.

06

Make a decision

If the p-value is less than the significance level (伪 = 0.01), we reject the null hypothesis and conclude that there is convincing evidence that the proportion with increased hair density is greater for the supplement treatment than for the control treatment. Otherwise, there is insufficient evidence to support that claim. #b. Comment on the Woman's World headline: # Based on the result from part (a), we can evaluate the appropriateness of the headline featured in Woman's World. #c. Suggest a more appropriate headline: # Finally, we will suggest a more appropriate headline based on the actual experiment and the result of our hypothesis test.

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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 is a statement that assumes no effect or no difference in the specific context of a hypothesis test. It acts as a starting point for statistical testing and is usually denoted as \( H_0 \). In this exercise, the null hypothesis states that there is no difference in the proportions of women experiencing increased hair density between those taking the supplement and those who are in the control group. It's mathematically expressed as: \( H_0: p_1 = p_2 \), where \( p_1 \) is the proportion in the control group and \( p_2 \) is the proportion in the supplement group. When conducting a hypothesis test, the null hypothesis serves as the benchmark that is tested against the evidence provided by the data. It is only rejected if there is strong enough evidence to indicate that the assumed condition (equality of proportions in this case) does not hold. Without this assumption, testing and comparisons would lack a reference point to measure the effect or difference.

Alternative Hypothesis

The alternative hypothesis is the statement that proposes a difference or effect that contrasts with the null hypothesis. Denoted as \( H_a \), it represents what a researcher seeks to prove through data analysis. In our context, the alternative hypothesis suggests that the proportion of women in the supplement group (\( p_2 \)) who show increased hair density is greater than the proportion in the control group (\( p_1 \)). It is mathematically written as: \( H_a: p_1 < p_2 \). This hypothesis is posited because it aligns with the claim made in the headline that black currant oil curbs hair loss. Therefore, the intention is to establish sufficient evidence against the null hypothesis in favor of this alternative perspective. The ultimate goal is to validate the experiment's indication that the supplement positively impacts hair density differently than the control.

Significance Level

The significance level, often represented by \( \alpha \), dictates the threshold for determining whether to reject the null hypothesis. In hypothesis testing, it is set in advance as a measure of the risk of concluding that a difference exists when there is none鈥攖he risk of a type I error. In this example, a 0.01 significance level is used, implying a 1% risk of rejecting the null hypothesis when it is in fact true. Choosing a lower significance level, like 0.01, indicates a stringent criterion for evidence in rejecting the null hypothesis. This strictness decreases the likelihood of making a false positive conclusion but also requires stronger evidence to claim that the treatment truly has a greater effect. The decision rule is straightforward: if the p-value obtained from the test is less than or equal \( \alpha \), the null hypothesis is rejected. Otherwise, it is accepted within the confines of the evidence available.

Proportions

Proportions represent the ratio of a part to the whole and are fundamental in assessing differences across groups in this study. Calculated as \( \hat{p}_1 = \frac{20}{39} \) for the control group, and \( \hat{p}_2 = \frac{70}{80} \) for the supplement group, they depict the fraction of women experiencing increased hair density in each condition. Understanding proportions helps in comparing group performances and identifying variations. It provides a statistic that encapsulates the essence of the dataset concerning the question posed鈥攚hether the supplement encourages more hair growth. The use of proportions makes complex data more manageable and interpretable, enhancing insight into potential differences and effects.

Test Statistic

The test statistic is a standardized value that results from a statistical test. It quantitatively measures the degree to which the observed data deviate from the null hypothesis. In this case, the test statistic is the z-value, calculated as: \[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{SE} \] where \( SE \) is the standard error, capturing the variability expected from the sample proportions. A key utility of the test statistic is its connection to the p-value, which helps determine the significance of results.This numerical expression provides clarity on whether the difference in proportions is likely due to chance or indicative of a genuine effect, as claimed. The larger the absolute value of the test statistic, the more the evidence veers away from the null hypothesis, paving the way for conclusions that either justify or nullify the headline's claim.