Problem 2 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

The data in Table 28.4 represent the duration of pregnancy for 1669 women who gave birth in a maternity hospital in Newcastle-upon-Tyne, England, in 1954 . The durations are measured in complete weeks from the beginning of the last menstrual period until delivery. The pregnancies are divided into those where an admission was booked for medical reasons, those booked for social reasons (such as poor housing), and unbooked emergency admissions. For the three groups the sample means and sample variances are Medical: \(\quad 775\) observations with \(\bar{x}=39.08\) and \(s^{2}=7.77\), Emergency: 261 observations with \(\bar{x}=37.59\) and \(s^{2}=25.33\), Social: \(\quad 633\) observations with \(\bar{x}=39.60\) and \(s^{2}=4.95\). Suppose we view the datasets as realizations of random samples from normal distributions with expectations \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\sigma_{3}^{2}\), where \(\mu_{i}\) represents the duration of pregnancy for the women from the \(i\) th group. We want to investigate whether the duration differs for the different groups. For each combination of two groups test the null hypothesis of equality of \(\mu_{i}\). Compute the values of the test statistic and report your conclusions.

Step-by-Step Solution

Answer

The groups differ significantly in pregnancy duration.

Step 1: Set Up Hypotheses

For each pair of groups, the null hypothesis is that the mean duration of pregnancy is the same, i.e., \(H_0: \mu_i = \mu_j\). The alternative hypothesis is that the means are different, i.e., \(H_a: \mu_i eq \mu_j\).

Step 2: Calculate the Test Statistics for Medical vs Emergency

For the Medical (\(n_1 = 775\), \(\bar{x}_1 = 39.08\), \(s_1^2 = 7.77\)) and Emergency (\(n_2 = 261\), \(\bar{x}_2 = 37.59\), \(s_2^2 = 25.33\)) groups, the test statistic for testing equality of means is calculated as follows:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}\]\[t = \frac{39.08 - 37.59}{\sqrt{7.77/775 + 25.33/261}} = 3.618\]This value is compared against the critical value from a t-distribution.

Step 3: Calculate the Test Statistics for Medical vs Social

For the Medical (\(n_1 = 775\)) and Social (\(n_3 = 633\), \(\bar{x}_3 = 39.60\), \(s_3^2 = 4.95\)) groups, the test statistic is:\[t = \frac{\bar{x}_1 - \bar{x}_3}{\sqrt{s_1^2/n_1 + s_3^2/n_3}}\]\[t = \frac{39.08 - 39.60}{\sqrt{7.77/775 + 4.95/633}} = -2.746\]We compare this to the t-distribution critical value for a specified significance level.

Step 4: Calculate the Test Statistics for Emergency vs Social

For the Emergency (\(n_2 = 261\)) and Social (\(n_3 = 633\)) groups, calculate the statistic:\[t = \frac{\bar{x}_2 - \bar{x}_3}{\sqrt{s_2^2/n_2 + s_3^2/n_3}}\]\[t = \frac{37.59 - 39.60}{\sqrt{25.33/261 + 4.95/633}} = -3.140\]This result is assessed using the t-distribution critical value at the desired confidence level.

Step 5: Determine Critical Values and Conclusion

Assuming a significance level \(\alpha = 0.05\) for a two-tailed test, we look up critical t-values for the corresponding degrees of freedom, which can be approximated using statistical tables or software.- From the calculated t-values: - Medical vs Emergency: \(t = 3.618\) (significant) - Medical vs Social: \(t = -2.746\) (significant) - Emergency vs Social: \(t = -3.140\) (significant)Since all the calculated \(|t|\) values exceed the critical \(t\) value for the respective degrees of freedom, we reject the null hypothesis for all group comparisons.

Key Concepts

Pregnancy Duration

Pregnancy duration refers to the total time, measured in weeks, from the first day of the woman's last menstrual period until the time of delivery. A full-term pregnancy typically lasts about 40 weeks, but this can vary between individuals. Understanding the duration is crucial for prenatal care and predicting delivery dates.
In the study at Newcastle-upon-Tyne, the pregnancy durations were gathered for different groups: Medical, Emergency, and Social. Each group had varying average durations:

• The Medical group had a mean duration of 39.08 weeks.
• The Emergency group's mean was 37.59 weeks.
• The Social group had the longest duration, with a mean of 39.60 weeks.

These differences in means hint at possible variations in factors such as health care access or underlying health conditions affecting each group. The aim is to statistically determine whether these differences in pregnancy duration are significant or merely due to chance.

T-Test

The t-test is a statistical method used to compare two groups to determine if there is a significant difference between their means. It is particularly useful when dealing with small sample sizes and unknown population variances.
In our exercise, the t-test helps in comparing the mean pregnancy durations between different groups:

• Medical vs Emergency: Here, the calculated t-value was 3.618, indicating a significant difference in mean durations.
• Medical vs Social: The t-value obtained was -2.746, pointing to a significant difference as well.
• Emergency vs Social: With a t-value of -3.140, there is evidence of a significant mean difference.

The formula for the t-test statistic is \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}\]where \( \bar{x}_i \) are the sample means, \( s_i^2 \) are the sample variances, and \( n_i \) are the sample sizes.

Significance Level

Significance level is a critical concept in hypothesis testing that helps in making decisions about the null hypothesis. It is usually denoted by \( \alpha \) and represents the probability of rejecting the null hypothesis when it is actually true.
In the context of our problem, the significance level was set at \( \alpha = 0.05 \). A two-tailed test approach was adopted, implying that we are interested in deviations in both directions from the null value.

• If the calculated t-value is greater than the critical t-value derived from statistical tables (depending on degrees of freedom), the null hypothesis is rejected.
• The significance level determines the critical t-values. For \( \alpha = 0.05 \), smaller critical values mean higher statistical evidence against the null hypothesis.

In our comparisons:

• All t-values (3.618, -2.746, -3.140) exceeded the critical values, leading to the rejection of null hypotheses across all groups.

This demonstrates statistically significant differences in the mean pregnancy durations between each pair of groups.