91Ó°ÊÓ

As mentioned in Exercise \(10.29\), a company claims that its medicine, Brand A, provides faster relief from pain than another company's medicine, Brand B. A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes. \begin{tabular}{cccc} \hline Brand & Sample Size & Mean of Relief Times & Standard Deviation of Relief Times \\ \hline A & 25 & 44 & 11 \\ B & 22 & 49 & 9 \\ \hline \end{tabular} Assume that the two populations are normally distributed with unknown and unequal standard deviations. a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at a \(1 \%\) significance level whether the mean relief time for Brand \(A\) is less than that for Brand B c. Suppose that the sample standard deviations were \(13.3\) and \(7.2\) minutes, respectively. Redo parts a and b. Discuss any changes in the results.

Step by step solution

01

Calculating Confidence Interval

To construct the 99% confidence interval for the difference between the mean relief times, apply the following formula: \[CI = (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2}(s_1^2/n_1 + s_2^2/n_2)^{0.5}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are sample means, \(n_1\) and \(n_2\) are sample sizes, \(s_1\) and \(s_2\) are sample standard deviations, and \(t_{\alpha/2}\) corresponds to the critical value for the desired level of confidence. In case of 99% confidence level, degrees of freedom will be \(n_1 + n_2 - 2 = 25 + 22 - 2 = 45\).

02

Hypothesis Testing

To test the claim at 1% significance level, perform a two-sample t-test. First, set up the null and alternative hypotheses. Let \(\mu_1\) and \(\mu_2\) represent the population means of Brand A and Brand B, respectively. The null hypothesis \(H_0\): \(\mu_1 - \mu_2 = 0\) (the mean relief times are equal), alternative hypothesis \(H_a\): \(\mu_1 - \mu_2 < 0\) (Brand A has less relief time than Brand B). Find the test statistic using: \[t = (\bar{x}_1 - \bar{x}_2) / (s_1^2/n_1 + s_2^2/n_2)^{0.5}\] and compare it to the critical t-value with 45 degrees of freedom and 1% significance level.

03

Additional Calculations

If the standard deviations were 13.3 for Brand A and 7.2 for Brand B, redo steps 1 and 2. The only things changing in the calculations are the values of \(s_1\) and \(s_2\) in the respective formulas. Compare these new results with previous ones and examine any changes and their potential implications.

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Key Concepts

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

Confidence Interval

A confidence interval gives us a range within which we expect the true difference between two population means to lie. In this exercise, we are dealing with the mean relief times of two brands of medicine. By using a 99% confidence interval, we are essentially saying that we are 99% confident that the true difference in mean relief times falls within this calculated range.
To calculate this, we use the formula:

• \(CI = (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2}(s_1^2/n_1 + s_2^2/n_2)^{0.5}\)

For calculating our confidence interval, we first need the sample means (\(\bar{x}_1, \bar{x}_2\)), the sample standard deviations (\(s_1, s_2\)), and the sample sizes (\(n_1, n_2\)). These values are substituted into the formula, and the result is our desired interval.
This interval helps us understand more about the variability in data and potential influences of sampling variation.

t-test

A t-test helps us test hypotheses concerning the difference between sample means. Here, it's used to determine if there's a significant difference in the mean relief times of two medicine brands. We begin by setting up our hypotheses.
The null hypothesis \(H_0\) suggests there's no difference in mean relief times: \(\mu_1 - \mu_2 = 0\). The alternative hypothesis \(H_a\) is \(\mu_1 - \mu_2 < 0\), indicating that Brand A might provide quicker relief than Brand B.
To compute the t-statistic, the formula is:

• \(t = (\bar{x}_1 - \bar{x}_2) / (s_1^2/n_1 + s_2^2/n_2)^{0.5}\)

The t-statistic is then compared against a critical t-value at the 1% significance level, considering the degrees of freedom. This comparison helps decide whether to reject \(H_0\) or not, affecting our conclusions about the claim.

Sample Standard Deviation

Sample standard deviation is a measure of the amount of variation or dispersion in a sample data set. In this study, different sample standard deviations were given for Brands A and B, which affects our calculations.
With different values for \(s_1\) and \(s_2\), the results of confidence interval and hypothesis testing will change as they are key components in our formulas. A higher standard deviation suggests more variability among data points, while a lower one indicates less variability.
This metric is crucial in understanding how much the individual data points differ from the mean, influencing the reliability of our comparison between the two brands.

Significance Level

The significance level, often denoted by \(\alpha\), tells us how confident we need to be in our testing. In this exercise, a 1% significance level is used, meaning there's a 1% risk of concluding that a difference in means exists when, in fact, there isn't one.
This low \(\alpha\) level shows that the test aims to minimize the probability of a Type I error, crucially when making important decisions based on this test. It affects the critical t-value and thus plays a key role in hypothesis testing, determining the threshold to which our sample evidence is compared.
A smaller \(\alpha\) results in larger critical regions, requiring stronger evidence against the null hypothesis to reach a significant conclusion.