/*! 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 27 In astudy of memory recall, 8 st... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In astudy of memory recall, 8 students from a large psychology class were selected at random and given 10 minutes to memorize a list of 20 nonsense words. Each was asked to list as many of the words as he or she could remember both 1 hour and 24 hours later. The data are given in the accompanying table. Is there convincing evidence to suggest that the mean number of words recalled after 1 hour is greater than the mean recall after 24 hours by more than 3 ? Use a significance level of \(\alpha=0.01\).

Short Answer

Expert verified
After conducting a hypothesis test for the difference in means, we compared the test statistic with the critical value and p-value at a significance level of 0.01. If the test statistic was greater than the critical value or the p-value was less than 0.01, we would reject the null hypothesis and conclude that there is convincing evidence to suggest that the mean number of words recalled after 1 hour is greater than the mean recall after 24 hours by more than 3. If not, we do not have convincing evidence for this claim.

Step by step solution

01

State the hypotheses

The null hypothesis (H0) states that there is no significant difference in mean recall after 1 hour and 24 hours, while the alternative hypothesis (Ha) states that the mean recall after 1 hour is greater than the mean recall after 24 hours by more than 3. \(H_0: \mu_1 - \mu_2 = 3\) \(H_a: \mu_1 - \mu_2 > 3\) where \(\mu_1\) is the mean recall after 1 hour, and \(\mu_2\) is the mean recall after 24 hours.
02

Calculate the sample means and standard deviations

We need to calculate the means and standard deviations for both the 1-hour and 24-hour recall data. From the table, we can compute them. Let \(M_1\) and \(M_2\) be the sample means, and \(S_1\) and \(S_2\) be the standard deviations for the 1-hour and 24-hour recalls, respectively.
03

Calculate the test statistic

Since we have two dependent samples, we will use the following test statistic: \(t = \frac{(M_1 - M_2) - 3}{\sqrt{\frac{S_1^2}{n} + \frac{S_2^2}{n}}}\) where \(n\) is the sample size (8 students). Plug in the values from Step 2 and calculate the test statistic.
04

Determine the critical value and p-value

Using a significance level of 0.01 and the test statistic calculated in Step 3, we can determine the critical value and p-value by consulting the t-distribution table or using a calculator or statistical software. Since we are performing a one-tailed test, we will use a one-tailed critical value.
05

Compare the test statistic with the critical value and p-value

Compare the calculated test statistic with the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis. Alternatively, if the p-value is less than the significance level (0.01), we also reject the null hypothesis.
06

Draw the conclusion

Based on the comparison in Step 5, we make our decision: - If we rejected the null hypothesis, then we have convincing evidence to suggest that the mean number of words recalled after 1 hour is greater than the mean recall after 24 hours by more than 3. - If we failed to reject the null hypothesis, then we do not have convincing evidence to suggest the mean number of words recalled after 1 hour is greater than the mean recall after 24 hours by more than 3.

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.

t-distribution
When dealing with smaller sample sizes, like in our memory recall study with just 8 students, using the normal distribution isn't always accurate. This is where the t-distribution shines. It's similar to the normal distribution but has thicker tails. This means it accounts for more variability, which is common with tiny sample sets.
The t-distribution becomes critical when computing confidence intervals and conducting hypothesis tests. The slightly altered shape allows us to make accurate inferences about the population, even with limited data, by providing more conservative estimates.
  • As the sample size increases, the t-distribution gradually resembles the normal distribution.
  • The degrees of freedom, which is the sample size minus one ( -1), influences the thickness of the tails.
  • We use tables or statistical software to find critical values in the t-distribution that correspond to our desired significance level.
Understanding how the t-distribution works is vital for conducting accurate statistical tests, especially with smaller sample sizes or unknown population variances.
significance level
The significance level, \(\alpha\), tells us the threshold for rejecting the null hypothesis. In our example, the specified \(\alpha\) is 0.01, which means we're aiming for a very strict criterion.
A lower significance level, like 0.01 instead of the more common 0.05, indicates that we're allowing a smaller probability of incorrectly rejecting the null hypothesis. This lower threshold decreases the risk of a Type I error, where we wrongly conclude that there is an effect when there isn't one.
  • A significance level of 0.01 implies only a 1% chance of making such a mistake.
  • This is particularly important in studies where results need to be very reliable.
  • In practice, selecting a significance level involves balancing the need for confidence with the risk of missing a real effect.
The selection of an appropriate significance level is key in research studies and affects the conclusions drawn from hypothesis tests.
p-value
The p-value represents the probability of observing test results at least as extreme as the results actually observed. This is provided that the null hypothesis is true.
In hypothesis testing, it's a critical number: it allows us to measure the evidence against the null hypothesis. In our study on memory recall, once we calculate our test statistic, we find the corresponding p-value to determine the strength of our results.
  • If the p-value is less than the significance level (0.01 in our case), we reject the null hypothesis.
  • If it's greater, we fail to reject the null hypothesis, indicating insufficient evidence to support the alternative.
  • The p-value helps us quantify how "extreme" our findings are in relation to what we would expect by random chance.
Interpreting p-values correctly is fundamental for drawing conclusions in statistical tests and understanding the strength of results.
dependent samples analysis
When conducting hypothesis tests with dependent samples, we are looking at two sets of data that are related in some way. In the memory recall study, each student's results after 1 hour and 24 hours are paired, meaning they are not independent.
This pairing requires a specific analytical approach, often involving a t-test for dependent samples. This is because the differences between paired observations generally exhibit more correlation.
  • The dependent t-test takes into account the natural pairing between observations in the two data sets.
  • This specific test improves sensitivity and reduces variability by focusing on the differences within each pair.
  • The calculation involves using the difference between each pair's observations as a single metric.
Dependent samples analysis is essential when the observations are not independent, ensuring accurate statistical conclusions by leveraging their inherent connection.

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 article "Dieters Should Use a Bigger Fork" (Food Network Magazine, January/February 2012) described an experiment conducted by researchers at the University of Utah. The article reported that when people were randomly assigned to cither cat with a small fork or to eat with a large fork, the mean amount of food consumed was significantly less for the group that ate with the large fork. a. What are the two treatments in this experiment? b. In the context of this experiment, explain what it means to say that the mean amount of food consumed was significantly less for the group that ate with the large fork.

The paper referenced in the previous exercise also gave information on calorie content. For the sample of Burger King meal purchases, the mean number of calories was 1008 , and the standard deviation was 483 . For the sample of McDonald's meal purchases, the mean number of calories was \(908,\) and the standard deviation was 624 . Based on these samples, is there convincing evidence that the mean number of calories in McDonald's meal purchases is less than the mean number of calories in Burger King meal purchases? Use \(\alpha=0.01\).

The article "A Shovel with a Perforated Blade Reduces Energy Expenditure Required for Digging Wet Clay" (Human Factors, \(2010: 492-502\) ) described a study in which each of 13 workers performed a task using a conventional shovel and using a shovel with a blade that was perforated with small holes. The authors of the cited article provided the following data on energy expenditure (in \(\mathrm{Kcal} / \mathrm{kg}\) (subject) \(/ \mathrm{lb}\) (clay):

Research has shown that for baseball players, good hip range of motion results in improved performance and decreased body stress. The article "Functional Hip Characteristics of Baseball Pitchers and Position Players" (The American Journal of Sports Medicine, \(2010: 383-388\) ) reported on a study involving independent samples of 40 professional pitchers and 40 professional position players. For the sample of pitchers, the mean hip range of motion was 75.6 degrees and the standard deviation was 5.9 degrees, whereas the mean and standard deviation for the sample of position players were 79.6 degrees and 7.6 degrees, respectively, Assuming that these two samples are representative of professional baseball pitchers and position players, estimate the difference in mean hip range of motion for pitchers and position players using a \(90 \%\) confidence interval.

The report referenced in the previous exercise also gave data for representative samples of 250 adults in Canada and 250 adults in England. The sample mean amount of sleep on a work night was 423 minutes for the Canada sample and 409 minutes for the England sample. Suppose that the sample standard deviations were 35 minutes for the Canada sample and 42 minutes for the England sample. a. Construct and interpret a \(95 \%\) confidence interval estimate of the difference in the mean amount of sleep on a work night for adults in Canada and adults in England. b. Based on the confidence interval from Part (a), would you conclude that there is evidence of a diflerence in the mean amount of sleep on a work night for the two countries? Explain why or why not.

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.