/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 98 A study \(^{14}\) compared perso... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A study \(^{14}\) compared personality characteristics between 49 children of alcoholics and a control group of 49 children of nonalcoholics who were matched on age and gender. On a measure of well-being, the 49 children of alcoholics had a mean of \(26.1(s=7.2)\) and the 49 subjects in the control group had a mean of \(28.8(s=6.4) .\) The difference scores between the matched subjects from the two groups had a mean of \(2.7(s=9.7)\). a. Are the groups to be compared independent samples or dependent samples? Why? b. Show all steps of a test of equality of the two population means for a two- sided alternative hypothesis. Report the P-value and interpret. c. What assumptions must you make for the inference in part b to be valid?

Short Answer

Expert verified
a. The samples are dependent. b. Test statistic \( t = 1.95 \), \( p \approx 0.057 \), no significant difference. c. Differences should be normally distributed.

Step by step solution

01

Identify Sample Type

The samples are dependent because the groups are matched by age and gender. The difference scores are used, indicating a paired t-test is appropriate.
02

State Hypotheses

The null hypothesis (H_0) is that there is no difference between the means: \( H_0: \mu_d = 0 \), where \( \mu_d \) represents the mean difference. The alternative hypothesis (H_a) is the means are different: \( H_a: \mu_d eq 0 \).
03

Calculate Test Statistic

For a paired t-test, the t-statistic is calculated as follows:\[t = \frac{\bar{x}_d - \mu_d}{s_d/\sqrt{n}}\]where \( \bar{x}_d = 2.7 \), \( s_d = 9.7 \), and \( n = 49 \). Plugging in these values:\[t = \frac{2.7 - 0}{9.7/\sqrt{49}} = \frac{2.7}{1.3857} \approx 1.95\]
04

Determine Degrees of Freedom and Critical Value

The degrees of freedom for a paired sample t-test is \( n - 1 \), so here it is \( 48 \). Using a t-table or calculator, look up the critical t-value for \( 48 \) degrees of freedom at \( \alpha = 0.05 \) for a two-tailed test, which is approximately \( 2.01 \).
05

Compare Test Statistic to Critical Value

Since \( t = 1.95 \) is less than \( 2.01 \), we do not reject the null hypothesis. Thus, there is no statistically significant difference between the means at the \( \alpha = 0.05 \) level.
06

Find the P-value and Interpret

Using the t-distribution table or software, the p-value corresponding to \( t = 1.95 \) with \( 48 \) degrees of freedom is around \( 0.057 \). This p-value is greater than \( 0.05 \), supporting the decision not to reject \( H_0 \). This suggests no significant difference between the two groups.
07

List Assumptions

For the paired t-test to be valid, the differences between pairs should be approximately normally distributed, and the data should be interval or ratio scale.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Paired t-test
The paired t-test is a statistical method used when you have two related samples. Typically, these samples are matched or are from the same group before and after a treatment. The process is to see if there is a significant difference in the means of these paired observations.

In the context of the exercise, the groups being compared are considered related because they're matched by age and gender. This makes them dependent samples since similar characteristics imply a closer relationship than having randomly selected groups.

Using a paired t-test involves calculating the difference for each matched pair, which essentially converts your dataset into a single column of differences. This mean difference can then be tested to check if it significantly deviates from zero (or another hypothesized value). The statistic is calculated to understand these changes quantitatively and conclusively.
Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics that allows us to make inferences or educated guesses about a population parameter. The process involves formulating two hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)).

The null hypothesis represents the status quo, or no effect scenario, in our study. For the paired t-test in our exercise, it states that there is no difference in the mean well-being scores between children of alcoholics and the control group. The alternative hypothesis contradicts this, suggesting there is a difference in these means.

After crafting these hypotheses, statistical tests are conducted to determine whether the null hypothesis can be rejected or not, based on a pre-determined significance level (often \(\alpha = 0.05\)). In our study, the calculated \(t\)-statistic and the corresponding p-value help us decide the fate of our null hypothesis.

If the p-value is less than \(\alpha\), we reject \(H_0\) and accept that there is a significant difference; otherwise, we fail to reject \(H_0\). In our example, since the p-value is greater than 0.05, there isn't enough evidence to suggest the means are significantly different.
Dependent Samples
Dependent samples are groups of paired individuals or the same individuals observed under two different conditions. In the exercise at hand, the groups are considered dependent due to age and gender-matched pairing. This connection implies that observations are not completely independent of each other and that changes in one group might reflect changes in the other.

When dealing with dependent samples, paired sample analysis like the paired t-test is the go-to method. This method accounts for the matching and often results in more powerful tests since it reduces variation attributable to individual differences.

Using dependent samples allows researchers to detect differences with greater precision. It's crucial, however, to ensure that the differences, not individual raw scores, are normally distributed, as this assumption affects the validity of paired statistical methods. Therefore, analyzing dependent samples through a paired t-test offers a robust comparison platform when these conditions are met.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The percentage of women who get breast cancer sometime during their lifetime is higher now than in 1900 . Suppose that breast cancer incidence tends to increase with age, and suppose that women tend to live longer now than in \(1900 .\) Explain why a comparison of breast cancer rates now with the rate in 1900 could show different results if we control for the age of the woman.

Kidnapping in southern and eastern European countries The following data on kidnapping offences in countries of east and south Europe in 2014 were obtained from https://data.unodc.org. (Crime and Criminal Justice \(\rightarrow>\) Crime \(\rightarrow>\) Kidnapping \(\rightarrow\) Filter by Region and Sub Region as appropriate) \(\begin{array}{ll}\text { Eastern Europe: } & 31,95,12,3,292,88,369,10\end{array}\) Southern Europe: \(\quad 2,1,3,1,58,297,22,376,11,5,99,8\) Using statistical software, a. Construct and interpret a plot comparing responses by region. Kidnapping in southern and eastern European countries The following data on kidnapping offences in countries of east and south Europe in 2014 were obtained from https://data.unodc.org. (Crime and Criminal Justice \(->\) Crime \(\rightarrow>\) Kidnapping \(->\) Filter by Region and Sub Region as appropriate) Eastern Europe: 31,95,12,3,292,88,369,10 Southern Europe: 2,1,3,1,58,297,22,376,11,5,99,8 Using statistical software, a. Construct and interpret a plot comparing responses by region.

A USA Today story (May 22, 2010 ) about the medical benefits of moderate drinking of alcohol stated that a major French study links those who drink moderately to a lower risk for cardiovascular disease but challenges the idea that moderate drinking is the cause. "Instead, the researchers say, people who drink moderately tend to have a higher social status, exercise more, suffer less depression and enjoy superior health overall compared to heavy drinkers and lifetime abstainers. A causal relationship between cardiovascular risk and moderate drinking is not at all established." The study looked at the health status and drinking habits of 149,773 French adults. a. Explain how this story refers to an analysis of three types of variables. Identify those variables. b. Suppose socioeconomic status is treated as a control variable when we compare moderate drinkers to abstainers in their heart attack rates. Explain how this analysis shows that an effect of an explanatory variable on a response variable can change at different values of a control variable.

The Centers for Disease Control (www.cdc.gov) periodically administers large randomized surveys to track health of Americans. In a survey of 4431 adults in \(2003 / 2004,66 \%\) were overweight (body mass index \(\mathrm{BMI} \geq 25\) ). In the most recently available survey of 5181 adults in \(2011 / 2012,69 \%\) were overweight. a. Estimate the change in the population proportion who are overweight and interpret. b. The standard error for estimating this difference equals \(0.0096 .\) What is the main factor that causes se to be so small? c. The \(95 \%\) confidence interval comparing the population proportions in \(2011 / 2012\) to the one in \(2003 / 2004\) is (0.011,0.049) . Interpret, taking into account whether 0 is in this interval.

Energy drinks: health risks and toxicity A study was carried out in Saudi Arabia in which 31 male university students (18 overweight/obese and 13 having normal weight) were enrolled from December 2013 to December 2014 (www.annsaudimed.net). The heart rate variability was significantly less in obese subjects as compared to subjects with normal weight at 60 minutes after consuming an energy drink as indicated by the mean heart rate range MHRR (P-value \(=0.012\) ). a. The conclusion was based on a significance test comparing means. Define notation in context, identify the groups and the population means and state the null hypothesis for the test. b. What information you are not able to obtain from the P-value approach which you could learn if the confidence interval comparing the means was provided?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.