Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)

Seat Belts 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?

Short answer

a.There is enough evidence tosupport the claim that thechildren wearing seat belts have a lower mean length of time in an ICU than the mean for children not wearing seat belts.

b.The 90% confidence interval is equal to\( - 0.9583 < \left( {{\mu _1} - {\mu _2}} \right) < - 0.1617\).

c.It can be concluded that the mean time length of children in the ICU who were wearing seat belts is less than the mean time length of children in the ICU who were not wearing seat belts.

Step-by-step solution

Given information

In a sample of 123 children wearing seat belts, the mean number of days in ICU is equal to 0.83, and the standard deviation is equal to 1.77. In another sample of 290 children not wearing seat belts, the mean number of days spent in ICU is equal to 1.39, and the standard deviation is equal to 3.06.

It is claimed that the mean number of days spent in an ICU by children who were wearing a seatbelt is less than the mean number of days in the ICU by children who were not wearing a seatbelt.

Hypotheses

Null Hypothesis: The population mean time length of children in the ICU who were wearing seat belts is equal to the population mean time length of children in the ICU who were not wearing seat belts.

Symbolically,

\({H_0}:{\mu _1} = {\mu _2}\)

Since the original claim does not include equality, the alternate hypothesis becomes:

Alternative Hypothesis: The population mean time length of children in the ICU who were wearing seat belts is less than the population mean time length of children in the ICU who were not wearing seat belts.

Symbolically,

\({H_1}:{\mu _1} < {\mu _2}\)

Compute the test statistic

Apply the t-test to compute the test statistic using the formula,\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\)

Here,

\({n_1}\)denotes the sample size of children who were wearing seat belts and got hospitalized after an accident. Its value is equal to 123.

\({n_2}\)denotes the sample size of children who were not wearing seat belts and got hospitalized after an accident. Its value is equal to 290.

\({\bar x_1}\)denotes the sample mean time length of children in the ICU who were wearing seat belts and have a value equal to 0.83.

\({\bar x_2}\)denotes the sample mean time length of children in the ICU who were not wearing seat belts and have a value equal to 1.39.

\({s_1}\)is the sample standard deviation of the time length of children in the ICU who were wearing seat belts. It has a value equal to 1.77.

\({s_1}\)is the sample standard deviation of the time length of children in the ICU who were not wearing seat belts. It has a value equal to 3.06.

Substitute the respective values in the above formula to obtain the test statistic:

\(\begin{array}{c}t = \frac{{\left( {0.83 - 1.39} \right) - 0}}{{\sqrt {\frac{{{{\left( {1.77} \right)}^2}}}{{123}} + \frac{{{{\left( {3.06} \right)}^2}}}{{290}}} }}\\ = - 2.33\end{array}\)

State the critical value

The degrees of freedom is the smaller of the two values\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\).

The values are computed below:

\(\begin{array}{c}\left( {{n_1} - 1} \right) = 123 - 1\\ = 122\end{array}\)

\(\begin{array}{c}\left( {{n_2} - 1} \right) = 290 - 1\\ = 289\end{array}\)

Thus, the value of the degrees of freedom is equal to the smaller of 122 and 289, which is 122.

The critical value can be obtained using the t distribution table with degrees of freedom equal to 122 and the significance level equal to 0.05 for a left-tailed test.

Thus, the critical value is equal to -1.6574.

The p-value for t equal to -2.33 is equal to 0.0108.

Conclusion of the test

Since the test statistic value is less than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence tosupport the claim that thechildren wearing seat belts have a lower mean length of time in an ICU than the mean for children not wearing seat belts.

Construct a confidence interval

The confidence interval has the following expression:

\(CI = \left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\)

If a one-tailed hypothesis test is conducted at a 0.05 level of significance, then the confidence level to construct the confidence interval is equal to 90%.

Thus, the level of significance to construct the confidence interval becomes\(\alpha = 0.10\).

The margin of error is given by the following formula:
\(E = {t_{\frac{\alpha }{2}}}\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \).

Substitute the respective values in the above formula to compute the margin of error:

\(\begin{array}{c}E = 1.6574 \times \sqrt {\frac{{{{\left( {1.77} \right)}^2}}}{{123}} + \frac{{{{\left( {3.06} \right)}^2}}}{{290}}} \\ = 0.3983\end{array}\)

Substituting the value of E and the sample means, the following confidence interval is obtained:

\(\begin{array}{c}\left( {0.83 - 1.39} \right) - 0.3983 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {0.83 - 1.39} \right) + 0.3983\\ - 0.9583 < \left( {{\mu _1} - {\mu _2}} \right) < - 0.1617\end{array}\)

Thus, the 90% confidence interval of the difference in the two population means is equal to (-0.9583, -0.1617).

Conclusion based on the confidence interval

Therefore, 95 per cent of the time, the value of the difference in the population means would lie within the values of-0.9583 and -0.1617.

It can be observed that 0 does not lie within the interval. This implies that the two population means cannot be the same.

Thus, there is enough evidence tosupport the claim that thechildren wearing seat belts have a lower mean length of time in an ICU than the mean for children not wearing seat belts.

Final Conclusion

It can be concluded that the mean time length of children in the ICU who were wearing seat belts is less than the mean time length of children in the ICU who were not wearing seat belts.