Question

Prothrombin time is a measure of the clotting ability of blood. For 10 rats treated with an antibiotic and 10 control rats, the prothrombin times (in seconds) were reported as follows \(^{43}\) : $$ \begin{array}{|lcc|} \hline & \text { Antibiotic } & \text { Control } \\ \hline n & 10 & 10 \\ \bar{y} & 25 & 23 \\ s & 10 & 8 \\ \hline \end{array} $$ (a) Construct a \(90 \%\) confidence interval for the difference in population means. (Assume that the two populations from which the data came are normally distributed.) [Note: Formula (6.7.1) yields 17.2 degrees of freedom for these data.] (b) Why is it important that we assume that the two populations are normally distributed in part (a)? (c) Interpret the confidence interval from part (a) in the context of this setting.

Short answer

The 90% confidence interval for the mean difference in prothrombin times between populations is calculated using the sample means (25 and 23), sample standard deviations (10 and 8), and sample sizes (10 in each group) along with the t-distribution with 17 degrees of freedom. The interval contextually suggests the likely range for the mean difference in prothrombin times between the antibiotic-treated rats and control rats. The normal distribution assumption is crucial to validate the method used to construct this interval.

Step-by-step solution

Identify the sample statistics

Extract the sample statistics from the given data. You have 10 rats in each group with the means of the prothrombin times being 25 and 23 seconds for the Antibiotic and Control groups respectively, and standard deviations of 10 and 8 seconds.

Determine the standard error of the difference in means

Using the sample standard deviations and sample sizes, calculate the standard error (SE) of the difference in means. The formula is SE = \sqrt{(s_{1}^2/n_1) + (s_{2}^2/n_2)} where \(s_{1}\) and \(s_{2}\) are the standard deviations of the two groups and \(n_{1}\) and \(n_{2}\) are the sample sizes.

Find the critical value

The degrees of freedom (df) for the t-distribution can be approximated using the formula provided, which gives us 17.2 df. Look up the critical t-score for a 90% confidence interval and 17 degrees of freedom (the nearest whole number for df) in a t-distribution table.

Calculate the confidence interval

Construct the confidence interval using the formula: \(\bar{y}_{1} - \bar{y}_{2} \pm (t * SE)\), where \(\bar{y}_{1}\) and \(\bar{y}_{2}\) are the sample means, t is the critical t-score from Step 3, and SE is the standard error calculated in Step 2.

Interpret the confidence interval in the context

Explain what the confidence interval means in the context of the prothrombin times for the rats. The interval gives us a range of values within which we can be 90% confident that the true difference in population means lies.

Understand the importance of normal distribution assumption

Recognize that the confidence interval relies on the sampling distribution of the difference in sample means being approximately normally distributed. This assumption is necessary for the validity of the interval, especially when sample sizes are small (less than 30).

Key concepts

Confidence Interval Calculation

A confidence interval is a range of values that is used to estimate the true value of a population parameter. In the context of the prothrombin time exercise, calculating a confidence interval for the difference in population means involves several steps. First, we determine the mean and standard deviation for each group. Then, we calculate the standard error (SE) of the difference between the two means, which provides a measure of how much variability there is in the estimate of the difference.

This SE is critically important as it helps to determine the width of the confidence interval - the larger the SE, the wider the interval. Given that the prothrombin times have a normal distribution, we can then find the appropriate critical value from the t-distribution given the desired level of confidence and the degrees of freedom, which is a function of the sample sizes. The confidence interval is then constructed by taking the difference between the two sample means and adding and subtracting the product of the critical t-value and the SE.

The outcome is a range that, with a specified level of confidence (90% in this exercise), is believed to contain the true difference in population means. Understanding each of these steps ensures that the confidence interval is calculated correctly and that the interpretation is meaningful within the context of the study.

Normal Distribution Assumption

The assumption of normal distribution in statistical analysis is essential when using certain parametric tests, including the construction of confidence intervals for small samples. The central limit theorem assures us that the means of samples from a population will be normally distributed around the true population mean, given a sufficiently large sample size. However, with smaller samples sizes, this normality assumption becomes critical.

When constructing a confidence interval for the difference in means between two groups, as in the prothrombin time exercise, we rely on the premise that the distribution of the sample means would be normal if we took many samples. This essential condition allows us to use the t-distribution as a model for inference, making our confidence interval calculation valid. If this assumption does not hold, the resulting confidence intervals may not accurately reflect the uncertainty associated with the estimate of the difference in population means.

In simple terms, by asserting that the prothrombin times follow a normal distribution, we ensure that the statistical methods applied yield reliable and interpretable results, giving us confidence that the interval estimate truly reflects the population's characteristics.

T-distribution Table

The t-distribution table is a valuable tool used in statistics to find the critical t-values for different confidence levels and degrees of freedom. The table is a reference that allows researchers to determine the t-score necessary to construct a confidence interval around a sample mean, when the population standard deviation is unknown and the sample size is relatively small.

Each row in a t-distribution table corresponds to different degrees of freedom, which are typically calculated as the sample size minus one for a single sample, or the sum of both sample sizes minus two for two independent samples. The degrees of freedom account for the amount of variability in the data; more degrees of freedom generally mean a distribution that's closer to a normal distribution.

When constructing a confidence interval, like in our prothrombin time scenario, we use the t-distribution table to find the critical value that corresponds to our desired level of confidence and the appropriate degrees of freedom (17.2 in this case, which we approximate to the nearest whole number). The critical value tells us how many standard errors to move away from the sample mean to capture the center percentage of the t-distribution. This is how we establish the range that, with our specified level of confidence, likely contains the true difference between the population means.