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Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting cancer of any kind? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Short Answer

Expert verified
The Hormone Replacement Therapy (HRT) does not significantly influence the chances of a woman developing any form of cancer since the p-value (0.1208) is greater than the significance level (\(\alpha = 0.05\)) from which we fail to reject the null hypothesis.

Step by step solution

01

Setup the Hypotheses

Firstly, hypotheses for this analysis need to be set up. The null hypothesis will be \[ H_0: \] HRT does not affect the chance of getting cancer. The alternative hypothesis will be \[ H_a: \] HRT does affect the chance of getting cancer.
02

Calculate Observed Values

Next, the observed frequencies of cancer need to be noted down in a structured format: \[ Observations = \left[ \begin{array}{ccc} \text{HRT Group} & \text{Placebo Group} \ 502 & 458 \ 8004 & 7644 \end{array} \right] \] where the first row represents the frequencies of cancer and the second row represents the frequencies of non-cancer.
03

Compute Expected Values

Find the expected values for the Chi-square test. To compute these values, use the formula \[ E_{i,j} = \frac{(row_{i} Total * column_{j} Total)}{Total \space Sample \space Size} \] The Total Sample Size is \( 8506+8102 = 16608 \) and row and column totals are calculated by summing appropriate row and column values. These will give the expected values for this problem: \[ Expectations = \left[ \begin{array}{ccc} \text{HRT Group} & \text{Placebo Group} \ 505.16 & 454.84 \ 8000.84 & 7647.16 \end{array} \right] \]
04

Calculate the Chi-square Statistic

Compute the Chi-square statistic using the formula \[ X^2 = \sum \frac{(O_{i,j} - E_{i,j})^2}{E_{i,j}}\] where \(O_{i,j}\) are observed values and \(E_{i,j}\) are expected values. Carrying out these calculations gives the Chi-square statistic which equals to \(2.41\).
05

Determine the p-value

Determine the p-value for the chi-square statistic using a chi-square distribution table or calculator. For this problem, the degree of freedom is \( (R-1) \times (C-1) = (2-1) \times (2-1) = 1\). The p-value associated with a chi-square of 2.41 with 1 degree of freedom is approximately 0.1208.
06

Compare p-value to Significance Level

Now, the p-value, 0.1208, must be compared to the significance level, \( \alpha = 0.05 \). Since the p-value is greater than the significance level, it should not reject the null hypothesis that HRT does not influence the chance of a woman getting cancer.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used in statistics to determine whether there is enough evidence to reject a null hypothesis. The process begins by establishing two hypotheses: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)).
  • The null hypothesis generally states that there is no effect or no difference. In our study, the null hypothesis is that Hormone Replacement Therapy (HRT) does not affect the chance of getting cancer.
  • The alternative hypothesis states the opposite, suggesting that there is an effect or a difference. Here, it claims HRT does affect the chance of getting cancer.
Next, we collect data and compute statistics to test these hypotheses. The goal is to determine whether the data is consistent with the null hypothesis or if the alternative hypothesis is more plausible.
Hypothesis testing employs various methods, one of which is the Chi-square test, particularly useful for categorical data, like determining cancer occurrence in HRT and placebo groups.
p-value
The p-value is a crucial metric in hypothesis testing that helps you decide whether to reject the null hypothesis. It indicates the probability of observing the test results under the assumption that the null hypothesis is true.
  • If the p-value is small (typically less than 0.05), it suggests that the observed data is unlikely under the null hypothesis, leading to rejection of \( H_0 \).
  • Conversely, a large p-value implies that the data is consistent with the null hypothesis, so we lack strong evidence to reject it.
In our exercise, the p-value calculated from the Chi-square statistic was approximately 0.1208.
Since this p-value is greater than the alpha level (\( \alpha = 0.05 \)), we do not have sufficient evidence to reject the null hypothesis that HRT does not affect the chance of getting cancer. This suggests that any observed differences could be due to chance rather than a real effect of HRT.
HRT (Hormone Replacement Therapy)
Hormone Replacement Therapy (HRT) is a treatment used to relieve symptoms of menopause by replacing hormones that the body no longer produces after menopause.
  • HRT was widely used for years, but the 2002 large clinical trial mentioned in the exercise led to a massive reevaluation of its use. The trial suggested that HRT could increase the risk of certain conditions, such as breast cancer, which prompted many to stop using it.
  • The study within the exercise specifically looked at whether HRT was associated with a higher risk of cancer. It involved 8506 women in the HRT group and 8102 in the placebo group, aiming to capture any potential increased risk of cancer.
The controversies and studies around HRT highlight the importance of ongoing research and rigorous testing in medicine and healthcare. Understanding the precise risks and benefits associated with treatments like HRT is crucial to making informed health decisions.

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Most popular questions from this chapter

Drinking tea appears to offer a strong boost to the immune system. In a study introduced in Exercise 3.82 on page \(203,\) we see that production of interferon gamma, a molecule that fights bacteria, viruses, and tumors, appears to be enhanced in tea drinkers. In the study, eleven healthy non-tea- drinking individuals were asked to drink five or six cups of tea a day, while ten healthy nontea- and non-coffee-drinkers were asked to drink the same amount of coffee, which has caffeine but not the \(L\) -theanine that is in tea. The groups were randomly assigned. After two weeks, blood samples were exposed to an antigen and production of interferon gamma was measured. The results are shown in Table 6.23 and are available in ImmuneTea. The question of interest is whether the data provide evidence that production is enhanced in tea drinkers. (a) Is this an experiment or an observational study? (b) What are the null and alternative hypotheses? (c) Find a standardized test statistic and use the t-distribution to find the p-value and make a conclusion. (d) Always plot your data! Look at a graph of the data. Does it appear to satisfy a normality condition? (e) A randomization test might be a more appropriate test to use in this case. Construct a randomization distribution for this test and use it to find a p-value and make a conclusion. (f) What conclusion can we draw? $$ \begin{array}{lrrrrrr} \hline \text { Tea } & 5 & 11 & 13 & 18 & 20 & 47 \\ & 48 & 52 & 55 & 56 & 58 & \\ \hline \text { Coffee } & 0 & 0 & 3 & 11 & 15 & 16 \\ & 21 & 21 & 38 & 52 & & \\ \hline \end{array} $$

On page 1 1 in Section \(1.1,\) we describe studies to investigate whether there is evidence of pheromones (subconscious chemical signals) in female tears that affect sexual arousal in men. In one of the studies, \(^{71} 50\) men had a pad attached to the upper lip that contained either female tears or a salt solution dripped down the same female's face. Each subject participated twice, on consecutive days, once with tears and once with saline, randomized for order, and doubleblind. Testosterone levels were measured before sniffing and after sniffing on both days. While normal testosterone levels vary significantly between different men, average levels for the group were the same before sniffing on both days and after sniffing the salt solution (about \(155 \mathrm{pg} / \mathrm{mL}\) ) but were reduced after sniffing the tears (about \(133 \mathrm{pg} / \mathrm{mL}\) ). The mean difference in testosterone levels after sniffing the tears was 21.7 with standard deviation \(46.5 .\) (a) Why did the investigators choose a matchedpairs design for this experiment? (b) Test to see if testosterone levels are significantly reduced after sniffing tears? (c) Can we conclude that sniffing female tears reduces testosterone levels (which is a significant indicator of sexual arousal in men)?

Refer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting invasive breast cancer? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

(a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Table 6.10 gives flight arrival numbers from a random sample of flights for two airlines. Test whether there is a difference between the two airlines in the percent of flights that arrive late.

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 30 from a population with proportion 0.27

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