Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Each day I am getting better in math A "subliminal" message is below our threshold of awareness but may nonetheless influence us. Can subliminal messages help students learn math? A group of 18students who had failed the mathematics part of the City University of New York Skills Assessment Test agreed to participate in a study to find out. All received a daily subliminal message, flashed on a screen too rapidly to be consciously read. The treatment group of 10students (assigned at random) was exposed to "Each day I am getting better in math." The control group of 8students was exposed to a neutral message, "People are walking on the street." All 18students participated in a summer program designed to improve their math skills, and all took the assessment test again at the end of the program. The following table gives data on the subjects' scores before and after the program.

a. Explain why a two-sample t-test and not a paired t-test is the appropriate inference procedure in this setting.

b. The following boxplots display the differences in pretest and post-test scores for the students in the control (C) and treatment (T) groups. Write a few sentences comparing the performance of these two groups.

c. Do the data provide convincing evidence at the α=0.01,3051526=0.200=20%significance level that subliminal messages help students like the ones in this study learn math, on average?

d. Can we generalize these results to the population of all students who failed the mathematics part of the City University of New York Skills Assessment Test? Why or why not?

Short Answer

Expert verified

(a) As students are assigned to one of the samples randomly.

(b) The Control distribution is heavily skewed to the right, whereas the Treatment distribution is marginally tilted to the left. The Treatment group's center is higher than the Control group's center. The spread between the two distributions appears to be similar. The distributions do not appear to have any outliers.

(c) No, There is no convincing evidence for subliminal messages that help students like the ones in this study learn math, on average.

(d) No, we can not generalize the result of all the students failed in mathematics.

Step by step solution

01

Part (a) Step 1: Given information

We need to find the reason for the two-sample t-test.

02

Part (a) Step 2: Explanation 

The first sample contains the treatment group's ten pupils, whereas the second sample comprises the control group's eight students.

Because the students were assigned to one of the samples at random, the students in the two samples will be completely unrelated, making the two-sample t-methods suitable.

03

Part (b) Step 1: Given information

We need to interpret the given boxplot.

04

Part (b) Step 2: Explanation

We know that

  • Because the box of the boxplot is to the left between the whiskers, the Control distribution is heavily skewed to the right. Because the box of the boxplot leans somewhat to the right between the whiskers, the Treatment distribution is slightly skewed to the left.
  • Because the box of the Treatment boxplot lies further to the right than the box of the Control boxplot, the center of the Treatment group is higher than the center of the Control group.
  • Because the two boxplots are almost the same width, the spread of the two distributions seems to be about the same.
  • Unusual features: There appear to be no outliers in the distributions, as neither boxplot has independent dots.
05

Part (c) Step 1: Given information

We need to find whether there is convincing for evidence subliminal messages help students like the ones in this study learn math, on average.

06

Part (c) Step 2: Explanation

We know that

We know that

The null hypothesis asserts that the variables are unrelated, whereas the alternative hypothesis asserts that they are.

H0:μ1=μ2Hα:μ1<μ2

where μ1=true mean difference for students does not receive the treatment.

And μ2=true mean difference for students receiving the treatment.

And expected frequencies are a product of row and column total divided by table total.

And The squared differences between the actual and predicted frequencies, divided by the expected frequency, make up the chi-square subtotals.

Therefore, the data is no convincing evidence for subliminal messages that help students like the ones in this study learn math, on average.

07

Part (d) Step 1: Given information

We need to find whether the result can be generalized or not.

08

Part (d) Step 2: Explanation

We know that

  • The kids opted to participate, the sample is not random, and it is possible that some students declined to participate.
  • The students who accepted to participate may have different features than those who refused, causing us to continually overestimate or underestimate the mean difference.

This means that we can't extrapolate the findings to the entire population of students who failed the mathematics section of the City University of New York Skills Assessment Tests, because we can't be sure whether the differences are due to the treatment or to the students' different characteristics (compared to the general population).

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 Vaia!

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

Children make choices Many new products introduced into the market are

targeted toward children. The choice behavior of children with regard to new products is of particular interest to companies that design marketing strategies for these products. As part of one study, randomly selected children in different age groups were compared on their ability to sort new products into the correct product category (milk or juice). Here are some of the data:

Researchers want to know if a greater proportion of 6- to 7-year-olds can sort correctly than 4- to5-year-olds.

a. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.

b. Check if the conditions for performing the test are met.

I want red! Refer to Exercise 1.

a. Find the probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample, assuming that the company’s claim is true.

b. Suppose that the Child and Adult samples contain an equal proportion of red jelly beans. Based on your result in part (a), would this give you reason to doubt the

company’s claim? Explain your reasoning.

A quiz question gives random samples of n=10observations from each of two Normally distributed populations. Tom uses a table of t distribution critical values and 9degrees of freedom to calculate a 95%confidence interval for the difference in the two population means. Janelle uses her calculator's two-sample t Interval with 16.87degrees of freedom to compute the 95%confidence interval. Assume that both students calculate the intervals correctly. Which of the following is true?

(a) Tom's confidence interval is wider.

(b) Janelle's confidence Interval is wider.

(c) Both confidence Intervals are the same.

(d) There is insufficient information to determine which confidence interval is wider.

(e) Janelle made a mistake, degrees of freedom has to be a whole number.

Suppose the probability that a softball player gets a hit in any single at-bat is 0.300. Assuming that her chance of getting a hit on a particular time at bat is independent of her other times at bat, what is the probability that she will not get a hit until her fourth time at bat in a game?

a.(43)(0.3)1(0.7)33051526=0.200=20.0%43(0.3)1(0.7)3

b.(43)(0.3)3(0.7)13051526=0.200=20.0%43(0.3)3(0.7)1

C.(41)(0.3)3(0.7)13051526=0.200=20.0%41(0.3)3(0.7)1

d.(0.3)3(0.7)13051526=0.200=20.0%(0.3)3(0.7)1

e.(0.3)1(0.7)33051526=0.200=20.0%(0.3)1(0.7)3

Better barley Does drying barley seeds in a kiln increase the yield of barley? A famous experiment by William S. Gosset (who discovered the t distributions) investigated this question. Eleven pairs of adjacent plots were marked out in a large field. For each pair, regular barley seeds were planted in one plot and kiln-dried seeds were planted in the other. A coin flip was used to determine which plot in each pair got the regular barley seed and which got the kiln-dried seed. The following table displays the data on barley yield (pound per acre) for each plot.

Do these data provide convincing evidence at the α=0.05level that drying barley seeds in a kiln increases the yield of barley, on average?

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.

Sign-up for free