/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A 4-year experiment involving 45... [FREE SOLUTION] | 91Ó°ÊÓ

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A 4-year experiment involving 4532 women was conducted at 39 medical centers to study the benefits and risks of hormone replacement therapy (HRT). Half of the women took placebos and half used a widely prescribed type of hormone replacement therapy. There were 40 cases of dementia in the hormone group and 21 in the placebo group. \({ }^{19}\) Is there sufficient evidence to indicate that the risk of dementia is higher for patients using the HRT? Test at the \(1 \%\) level of significance.

Short Answer

Expert verified
Answer: Yes, there is sufficient evidence at the 1% level of significance to support that the risk of dementia is higher for patients using hormone replacement therapy compared to those on placebo.

Step by step solution

01

State the hypotheses

We will test the following hypotheses: Null hypothesis (H0): There is no difference in the risk of dementia for patients using HRT and those on placebo. This can be expressed as \(p_h - p_p = 0\) where \(p_h\) is the proportion of dementia cases in the hormone group and \(p_p\) is the proportion of dementia cases in the placebo group. Alternative hypothesis (H1): The risk of dementia is higher for patients using HRT. This can be expressed as \(p_h - p_p > 0\).
02

Find the sample proportions

Let's first find the sample proportions for both groups. For the hormone group: \(p_h = \frac{40}{2266}\) For the placebo group: \(p_p = \frac{21}{2266}\)
03

Calculate the test statistic

We use the following formula to calculate the test statistic (z-value): \(z = \frac{(p_h - p_p) - 0}{\sqrt{\frac{p(1 - p)}{n_h} + \frac{p(1 - p)}{n_p}}}\) where \(p\) is the pooled proportion defined as: \(p = \frac{x_h + x_p}{n_h + n_p} = \frac{40 + 21}{4532}\) Here, \(x_h\) and \(x_p\) are the number of dementia cases in the hormone and placebo groups respectively, and \(n_h\) and \(n_p\) are the sample sizes of the hormone and placebo groups respectively. Now, we can calculate the test statistic: \(z = \frac{(\frac{40}{2266} - \frac{21}{2266})}{\sqrt{\frac{(\frac{61}{4532})(1 - \frac{61}{4532})}{2266} + \frac{(\frac{61}{4532})(1 - \frac{61}{4532})}{2266}}}\) After evaluating the expression, we obtain: \(z \approx 2.648\)
04

Determine the critical value and make a decision

We are performing a one-tailed test at the 1% level of significance. The critical value (z-critical) for a one-tailed test at the 1% level of significance can be found using a Z-table or calculator, which is around 2.33. Since the calculated test statistic (\(z = 2.648\)) is greater than the critical value (\(z_{critical} = 2.33\)), we reject the null hypothesis.
05

Conclusion

There is sufficient evidence at the 1% level of significance to indicate that the risk of dementia is higher for patients using hormone replacement therapy compared to those on placebo.

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Key Concepts

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

Null and Alternative Hypothesis
Understanding the null and alternative hypothesis is a foundational block in hypothesis testing. In simple terms, the null hypothesis (\(H_0\)) is a statement that there is no effect or no difference - it represents the status quo or a point of skepticism. For instance, in the exercise focusing on hormone replacement therapy (HRT) and dementia, the null hypothesis asserts that HRT does not affect the risk of dementia. This is articulated as \(p_h - p_p = 0\) where \(p_h\) and \(p_p\) are the proportions of dementia in the hormone and placebo groups, respectively.

The alternative hypothesis (\(H_1\) or \(H_a\)), on the other hand, is the claim you hope to support with evidence. In our exercise, the alternative hypothesis posits that HRT does influence dementia risk and that this risk is higher compared to the placebo, stated as \(p_h - p_p > 0\). Crafting these hypotheses clearly is crucial because they guide the entire hypothesis testing process.
Test Statistic Calculation
Calculating the test statistic is a critical step that helps us to determine how far our sample result is from the null hypothesis. In hypothesis testing, the test statistic, often a Z-score in proportion comparison cases, is a way of quantifying the difference between groups. For the HRT study, the test statistic is computed using the formula:

\[\begin{equation} z = \frac{(p_h - p_p) - (\text{difference under }H_0)}{\sqrt{\frac{p(1 - p)}{n_h} + \frac{p(1 - p)}{n_p}}} \end{equation}\]\
with \(p\) as the pooled proportion of dementia cases over both groups, and \(n_h\) and \(n_p\) representing the sizes of the hormone and placebo groups. When the computed Z-score is extreme enough, which is determined by the context and the chosen level of significance, it suggests that the observed result is unlikely under the null hypothesis, leading us to consider the alternative hypothesis.
P-Value and Significance Levels
The p-value is a key concept in hypothesis testing, representing the probability of obtaining a result at least as extreme as the observed one, assuming the null hypothesis is true. A low p-value indicates that under the null hypothesis, our result is unusual. The significance level (\(\alpha\)), often set at 0.05 or 0.01, is a threshold to determine when a p-value is considered sufficiently low to reject the null hypothesis. In the HRT study, testing at the 1% level of significance (\(\alpha = 0.01\)) sets a high standard for evidence against the null hypothesis. The result is compared against a critical value from the Z-distribution, and if the test statistic exceeds this critical value, the p-value is considered small enough to reject the null hypothesis. This approach helps control the risk of incorrectly rejecting a true null hypothesis, known as Type I error.
Proportion Comparison in Statistics
Comparing proportions between groups is often essential in statistics to assess whether there is evidence for a difference. In the case of the HRT study, the objective is to compare the proportion of dementia cases between hormone users and non-users. Through a series of calculations—first identifying the sample proportions, then calculating the pooled proportion, and finally the test statistic—we can infer about the population proportions. By comparing this inference to the set significance level, we can judge whether the observed difference in sample proportions reflects a true difference in population proportions, or if it could be due to random chance in sample selection. When the groups' proportions differ significantly, it may indicate a real contrast in the rates of an outcome, such as the incidence of dementia in the medical study we examined.

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Most popular questions from this chapter

The weights of 3 -month-old baby girls are known to have a mean of 5.86 kilograms. \(^{2}\) Doctors at an inner city pediatric facility suspect that the average weight of 3 -month-old baby girls at their facility may be less than 5.86 kilograms. They select a random sample of 403 -month-old baby girls and find \(\bar{x}=5.56\) and \(s=0.70\) kilogram. a. Does the data indicate that the average weight of 3 -month-old baby girls at their facility is less than 5.86 kilograms? Test using \(\alpha=.05 .\) b. What is the \(p\) -value associated with the test in part a? Can you reject \(H_{0}\) at the \(5 \%\) level of significance using the \(p\) -value?

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given. State the null and alternative hypotheses to test for a difference in the two population proportions. Calculate the necessary test statistic, the p-value, and draw the appropriate conclusions with \(\alpha=.01 .\) $$ \begin{array}{lcc} \hline & {\text { Population }} \\ & 1 & 2 \\ \hline \text { Sample Size } & 800 & 640 \\ \text { Number of Successes } & 337 & 374 \\ \hline \end{array} $$

Analyses of drinking water samples for 100 homes in each of two different sections of a city gave the following information on lead levels (in parts per million): $$ \begin{array}{lcc} \hline & \text { Section 1 } & \text { Section 2 } \\ \hline \text { Sample Size } & 100 & 100 \\ \text { Mean } & 34.1 & 36.0 \\ \text { Standard Deviation } & 5.9 & 6.0 \end{array} $$ a. Calculate the test statistic and its \(p\) -value to test for a difference in the two population means. Use the \(p\) -value to evaluate the significance of the results at the \(5 \%\) level. b. Use a \(95 \%\) confidence interval to estimate the difference in the mean lead levels for the two sections of the city. c. Suppose that the city environmental engineers will be concerned only if they detect a difference of more than 5 parts per million in the two sections of the city. Based on your confidence interval in part b, is the statistical significance in part a of practical significance to the city engineers? Explain.

An experimenter has prepared a drug-dose level that he claims will induce sleep for at least \(80 \%\) of people suffering from insomnia. After examining the dosage we feel that his claims regarding the effectiveness of his dosage are too high. In an attempt to disprove his claim, we administer his prescribed dosage to 50 insomniacs and observe that 37 of them have had sleep induced by the drug dose. Is there enough evidence to refute his claim at the \(5 \%\) level of significance?

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation of \(s=.29 .\) Your research objective is to show that the population mean \(\mu\) exceeds 2.3. Use this information to answer the questions. Calculate \(\beta=P\left(\right.\) accept \(H_{0}\) when \(\left.\mu=2.4\right)\)

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