Assume that the two samples are independent simple random...

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 $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ ) 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 Joumal 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?

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 $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ )
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 Joumal 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?

Key Concepts

Significance Level

This is the probability threshold used in hypothesis testing to determine whether to reject the null hypothesis. Typically denoted by alpha (?), it represents the risk of making a Type I error, which is falsely rejecting a true null hypothesis. A common significance level is 0.05, meaning there is a 5% risk of error in rejecting the null hypothesis.

Confidence Interval

A confidence interval is a range of values, derived from sample statistics, that is likely to contain the true population parameter with a specified level of confidence. When comparing two means, the confidence interval for the difference between the means provides an estimated range in which the true difference lies, and it is constructed to be consistent with the hypothesis test conducted.

Welch's t-test

Welch's t-test is a variation of the two-sample t-test that does not assume equal population variances. It is used to compare the means of two independent samples drawn from normally distributed populations. This test adjusts the degrees of freedom based on the sample variances and sizes, providing more reliable results when the assumption of equal variances is violated.

Independent Samples

This concept refers to samples that are collected in such a way that the observations in one sample have no relationship with the observations in the other sample. It is a fundamental assumption in many statistical tests where the comparison of parameters, such as means, between two groups is necessary without interference from pairing or matching between individuals in different groups.

Hypothesis Testing

Hypothesis testing involves setting up a null hypothesis, which represents a statement of no effect or no difference, and an alternative hypothesis, which represents the claim being tested. A statistical test is then used to determine whether there is sufficient evidence to reject the null hypothesis in favor of the alternative, based on the observed sample data and a predetermined significance level.

Transcript

00:01 So we're looking at people wearing seatbelts and not wearing seatbelts children and finding out how many days they stay in the icu.

00:07 And so we're going to assume that those students with a seatbelt have equal stay in the hospital as those who do not wear a seatbelt.

00:18 And alternately, that these seat -beled children have a mean stay less than those who are not seat -beled.

00:24 And so we're assuming that the difference between these two seatbelt minus not seatbelt is zero.

00:36 And we're actually getting something that's negative.

00:39 And this will be our p value.

00:40 So let's look at the data and let's get our test statistic.

00:44 And we had sample sizes, two sample sizes.

00:47 One sample size was 123 and the other was 290.

00:51 So that was for the seatbelt.

00:54 And this is for the non -seat belt.

00:57 And so we're going to use degrees of freedom of 122 to be conservative.

01:02 And there are relatively large sample sizes anyway.

01:05 And so let's find our test statistic.

01:08 And we have our 0 .83 minus our 1 .39.

01:13 And then we're going to divide that by the standard deviation of the seat belted group squared, divided by the sample size.

01:22 And then the standard deviation of the non -seat belted children divided by the sample size of it.

01:32 And when we do that calculation, we get the test statistic comes out to be negative 2 .330.

01:40 And so now we want to find, and that's what this value is, we want to find a likelihood if the difference is actually zero or the means are equal, how likely is this number or more extreme to come up.

01:54 And so i'm going to use my tcdf to find this, my tcdf, and i'm going to use negative.

02:04 Well, mine already has negative 1 times 10 to the 99th hour in it.

02:08 And then our upper value is going to be that negative 2 .330.

02:14 My degrees of freedom is again the 122.

02:17 And i will paste and let that do the calculation.

02:21 And i have a p value of about 1%.

02:24 And that is less than my 5 % significance level.

02:28 So i definitely have sufficient evidence to reject the null, meaning that we believe that the children that had seatbelts on have a smaller stay in icu than those that are.

02:48 Our non -seat belt.

02:49 And again, you'd write a nice sentence for that.

02:51 So now we want to find the confidence interval...

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