91Ó°ÊÓ

A study of seat belt use involved children who were hospitalized after motor vehicle crashes. For a group of 123 children who were wearing seat belts, the number of days in intensive care units (ICU) has a mean of \(0.83\) and a standard deviation of \(1.77\). For a group of 290 children who were not wearing seat belts, the number of days spent in ICUs has a mean of \(1.39\) and a standard deviation of \(3.06\) (based on data from "Morbidity Among Pediatric Motor Vehicle Crash Victims: The Effectiveness of Seat Belts," by Osberg and Di Scala, American Journal of Public Health, Vol. 82, No. 3\() .\) a. Use a \(0.05\) significance level to test the claim that children wearing seat belts have a lower mean length of time in an ICU than the mean for children not wearing seat belts. b. Construct a confidence interval appropriate for the hypothesis test in part (a). c. What important conclusion do the results suggest?

Step by step solution

01

- Set Hypotheses

Define the null and alternative hypotheses. The null hypothesis (H0) is that the mean time in the ICU for children wearing seat belts (\( \mu_1 \)) is equal to the mean time for children not wearing seat belts (\( \mu_2 \)). The alternative hypothesis (H1) is that the mean time in the ICU for children wearing seat belts is less than that for children not wearing seat belts.\[H_0: \mu_1 = \mu_2\]\[H_1: \mu_1 < \mu_2\]

02

- Identify Test Statistic

Determine the test statistic to be used. For comparing two means, use the t-test for independent samples. The appropriate test statistic formula when population variances are unknown and assumed unequal is:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]

03

- Calculate Test Statistic

Substitute the given values into the test statistic formula:\(\bar{x}_1 = 0.83\), \(s_1 = 1.77\), \(n_1 = 123\)\(\bar{x}_2 = 1.39\), \(s_2 = 3.06\), \(n_2 = 290\)\[t = \frac{0.83 - 1.39}{\sqrt{\frac{1.77^2}{123} + \frac{3.06^2}{290}}} = \frac{-0.56}{\sqrt{\frac{3.1329}{123} + \frac{9.3636}{290}}} = \frac{-0.56}{\sqrt{0.0255 + 0.0323}} = \frac{-0.56}{0.239} \approx -2.34\]

04

- Determine Critical Value and P-value

The test is a one-tailed test with \( \alpha = 0.05 \). Using a t-distribution table or calculator with degrees of freedom approximated by:\[\]\[df \approx \min(n_1 - 1, n_2 - 1) = \min(122, 289) = 122\]the critical value for t at \( \alpha = 0.05 \) is approximately -1.658. Compare this to the test statistic calculated in Step 3 to get the p-value.

05

- Make Decision

Compare the calculated test statistic to the critical value. Since \(-2.34 < -1.658\), reject the null hypothesis. The p-value comparison also shows significance, as the p-value is less than 0.05.

06

- Construct Confidence Interval

Construct a 95% confidence interval for the difference in means, \( \mu_1 - \mu_2 \). Using the standard error from Step 3, the confidence interval is:\[(\bar{x}_1 - \bar{x}_2) \pm t_{critical} \times SE = -0.56 \pm 1.96 \times 0.239 = -0.56 \pm 0.468 \]\[CI = (-1.028, -0.092) \]

07

- Conclusion

Given the results from hypothesis tests and the confidence interval, the results suggest that children wearing seat belts spend significantly less time in the ICU compared to those not wearing seat belts.

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Key Concepts

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

t-test

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. This is a parametric test, which relies on the distribution of the data being approximately normal.
In the context of this exercise, we use the t-test to compare the mean number of days children spent in the ICU between those wearing seat belts and those not wearing seat belts. Specifically, we use the t-test for independent samples, since the groups being compared are exclusive of each other.
To perform the t-test, we need to calculate the test statistic using the formula:
\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]
Here, \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the standard deviations, and \(n_1\) and \(n_2\) are the sample sizes of the two groups.

significance level

The significance level, denoted by \(\alpha\), is the threshold at which we decide whether to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is, in fact, true.
In this exercise, the significance level is set to 0.05, meaning we are willing to accept a 5% chance of making a Type I error. When performing the t-test, if the calculated p-value is less than the significance level, we reject the null hypothesis.

confidence interval

A confidence interval provides a range of values that is likely to contain the population parameter (such as the mean difference) with a certain degree of confidence.
In this exercise, we construct a 95% confidence interval around the difference in means of ICU days for the two groups. The formula for the confidence interval in the context of this t-test is:
\[ (\bar{x}_1 - \bar{x}_2) \pm t_{critical} \times SE \]
where \(SE\) is the standard error calculated during the t-test process. This interval helps us understand the range within which the true mean difference lies with 95% confidence.

sample mean

The sample mean (\(\bar{x}\)) is a measure of central tendency that is used to estimate the population mean. It is calculated by summing all observations in a sample and dividing by the number of observations.
In this scenario, the sample mean for children wearing seat belts is 0.83 days in the ICU, while for children not wearing seat belts, it is 1.39 days. These means are pivotal in our t-test, as they are used to compare the two groups and help determine if seat belt use is associated with shorter ICU stays.

standard deviation

The standard deviation (\(s\)) measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates a wide range of values.
For this exercise, the standard deviation for ICU days is 1.77 for children wearing seat belts, and 3.06 for those not wearing seat belts. These values are used in the t-test formula to calculate the standard error and ultimately the test statistic, which in turn helps us determine if the difference in means is statistically significant.