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Verzani (2014) discusses a data set on healthy individuals, including their temperatures by gender. The data are in the file tempbygender. rda and the variables of interest are maletemp and femaletemp. Download this file from the site listed in the Preface. (a) Obtain comparison boxplots. Comment on the plots. Which, if any, gender seems to have lower temperatures? Based on the width of the boxplots, comment on the assumption of equal variances. (b) As discussed in Example 8.3.3, compute the two-sample, two-sided \(t\) -test that there is no difference in the true mean temperatures between genders. Obtain the \(p\) -value of the test and conclude in terms of the problem at the nominal \alpha-level of \(0.05 .\) (c) Obtain a \(95 \%\) confidence interval for the difference in means. What does it mean in terms of the problem?

Short Answer

Expert verified
By using the boxplot comparison, it could be indicated if one gender tends to have lower temperatures. The t-test would show if there's a significant difference in mean temperatures between genders. If the p-value is less than 0.05, we reject the null hypothesis and conclude that there's a significant difference. The 95% confidence interval provides a range of probable values for the mean difference. If the interval includes 0, it suggests there's no significant difference. Interpretations need to be made based on the actual data analysis.

Step by step solution

01

Obtain comparison boxplots

First, import the data file 'tempbygender.rda'. Then, use a tool such as the boxplot() function in the R software to create the comparison boxplots for the maletemp and femaletemp categories. Compare the two plots to identify which gender seems to have lower temperatures. Additionally, observe the width of the boxplots to assess the assumption of equal variances.
02

t-test

Perform a two-sample, two-sided t-test to verify if there is a difference in the true mean temperatures between genders. This can be done using the t.test() function in R. The null hypothesis assumes that there's no difference in the true mean temperatures between genders. A p-value less than 0.05 would imply that the null hypothesis can be rejected.
03

Obtain a 95% confidence interval for the difference in means

Use the data obtained from the t-test to determine the 95% confidence interval for the difference in mean temperatures between genders. In R, the confidence interval can be obtained from the output of the t.test() function. Analyzing this interval provides insights into the range of values within which the population mean difference could lie with 95% confidence.
04

Interpret Results

Interpret the results from the boxplots, t-test, and confidence interval in relation to the problem. The boxplots give an indication of which gender has generally lower temperatures. The t-test provides proof if there's a significant difference in true mean temperatures. The confidence interval offers the range where the true mean difference is likely to lie. If the confidence interval includes 0, it suggests there's no significant difference in mean temperatures between genders.

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Most popular questions from this chapter

Show that the likelihood ratio principle leads to the same test when testing a simple hypothesis \(H_{0}\) against an alternative simple hypothesis \(H_{1}\), as that given by the Neyman- Pearson theorem. Note that there are only two points in \(\Omega\).

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate normal distribution with \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}, \rho=\frac{1}{2}\), where \(\mu_{1}, \mu_{2}\), and \(\sigma^{2}>0\) are unknown real numbers. Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0, \sigma^{2}\) unknown against all alternatives. The likelihood ratio \(\Lambda\) is a function of what statistic that has a well- known distribution?

Let \(X\) be \(N(0, \theta)\) and, in the notation of this section, let \(\theta^{\prime}=4, \theta^{\prime \prime}=9\), \(\alpha_{a}=0.05\), and \(\beta_{a}=0.10 .\) Show that the sequential probability ratio test can be based upon the statistic \(\sum_{1}^{n} X_{i}^{2} .\) Determine \(c_{0}(n)\) and \(c_{1}(n)\).

Let \(X\) have the pdf \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), zero elsewhere. We test \(H_{0}: \theta=\frac{1}{2}\) against \(H_{1}: \theta<\frac{1}{2}\) by taking a random sample \(X_{1}, X_{2}, \ldots, X_{5}\) of size \(n=5\) and rejecting \(H_{0}\) if \(Y=\sum_{1}^{h} X_{i}\) is observed to be less than or equal to a constant \(c\). (a) Show that this is a uniformly most powerful test. (b) Find the significance level when \(c=1\). (c) Find the significance level when \(c=0\). (d) By using a randomized test, as discussed in Example \(4.6 .4\), modify the tests given in parts (b) and (c) to find a test with significance level \(\alpha=\frac{2}{32}\).

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a beta distribution with parameters \(\alpha=\beta=\theta>0\), find a best critical region for testing \(H_{0}: \theta=1\) against \(H_{1}: \theta=2\)

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