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

A study \(^{54}\) shows that relationship status on Facebook matters to couples. The study included 58 college-age heterosexual couples who had been in a relationship for an average of 19 months. In 45 of the 58 couples, both partners reported being in a relationship on Facebook. In 31 of the 58 couples, both partners showed their dating partner in their Facebook profile picture. Men were somewhat more likely to include their partner in the picture than vice versa. However, the study states: "Females' indication that they are in a relationship was not as important to their male partners compared with how females felt about male partners indicating they are in a relationship." Using a population of college-age heterosexual couples who have been in a relationship for an average of 19 months: (a) A \(95 \%\) confidence interval for the proportion with both partners reporting being in a relationshipon Facebook is about 0.66 to 0.88 . What is the conclusion in a hypothesis test to see if the proportion is different from \(0.5 ?\) What significance level is being used? (b) A \(95 \%\) confidence interval for the proportion with both partners showing their dating partner in their Facebook profile picture is about 0.40 to \(0.66 .\) What is the conclusion in a hypothesis test to see if the proportion is different from \(0.5 ?\) What significance level is being used?

Short Answer

Expert verified
(a) We can conclude that there is evidence to claim the proportion of couples who report being in relationship on Facebook is different from 0.5. The significance level used is 0.05. (b) There is no evidence to claim that the proportion of couples showing their partner in profile picture is different from 0.5, and the significance level used is 0.05.

Step by step solution

01

Analyzing Part (a)

A 95% confidence interval for the proportion of couples reporting being in a relationship on Facebook is given as 0.66 to 0.88. This interval helps us to make probabilistic statements about unknown quantities. The question asks to test the hypothesis that the true proportion is different from 0.5.
02

Hypothesis Testing Part (a)

To conduct the hypothesis test, we should check if the hypothesized value (0.5) falls within the given confidence interval (0.66 to 0.88). The hypothesized value does not fall within this interval, so we reject the null hypothesis. Therefore, there is evidence to claim that the proportion is different from 0.5.
03

Significance Level Part (a)

The confidence level was 95%. Since the significance level (\(\alpha\)) plus the confidence level must total 1, then the significance level is \(1 - confidence level = 1 - 0.95 = 0.05\). So, the significance level used is 0.05.
04

Analyzing Part (b)

A 95% confidence interval for the proportion of couples showing their dating partner in their Facebook profile picture is given as 0.40 to 0.66. Similar to part (a), this hypothesis test is to determine whether the true proportion is different from 0.5.
05

Hypothesis Testing Part (b)

We need to check if the hypothesized value (0.5) falls within the given confidence interval (0.40 to 0.66). In this case, the hypothesized value falls within this interval, so we do not reject the null hypothesis. Therefore, there is no evidence to claim that the proportion is different from 0.5.
06

Significance Level Part (b)

The confidence level used for this hypothesis test still remains 95%. Therefore, we use the same procedure to compute the significance level as in part (a). The significance level is \(1 - confidence level = 1 - 0.95 = 0.05\). Thus, the significance level used is 0.05.

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!

Key Concepts

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

Confidence Interval
The confidence interval is a range of values, derived from sample statistics, that is likely to contain the true population parameter. In statistics, when we are trying to estimate some aspect of a population, like a mean or a proportion, we use a sample to calculate a point estimate and a confidence interval around that estimate. For example, the confidence interval given in the exercise for part (a) is 0.66 to 0.88, which tells us that we can be 95% confident that the true proportion of couples reporting being in a relationship on Facebook falls within this range.

It's essential to understand that the confidence level (95% in the exercise) signifies the frequency with which the calculated interval will contain the true parameter if the same population is sampled multiple times. The confidence interval provides a measure of precision for our estimate and is used for hypothesis testing, as it's related to the concept of statistical significance. If the confidence interval does not contain the value of the null hypothesis, we have grounds to reject the null hypothesis, as seen in part (a) of the exercise.
Null Hypothesis
The null hypothesis, denoted usually by \(H_0\), is a statement that there is no effect or no difference, and it serves as a starting point for statistical testing. In hypothesis testing, we collect data to assess whether the results are compatible with the null hypothesis or if there's sufficient evidence to support an alternative hypothesis \(H_1 or H_a\).

In the given exercise for both part (a) and part (b), we are testing the null hypothesis that the true proportion is 0.5, suggesting that there is no difference from a benchmark reference of 0.5. In hypothesis testing, if the value in question falls outside the confidence interval, like in part (a), this indicates that our sample provides enough evidence to reject the null hypothesis. However, if the value falls inside the confidence interval, as in part (b), we cannot reject the null hypothesis.
Significance Level
The significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true—this is known as a Type I error. The significance level is a critical component in hypothesis testing, as it defines the threshold for how much chance we are willing to accept when drawing conclusions from our data.

In both parts (a) and (b) of the exercise, a 95% confidence level is used, which corresponds to a significance level of 5% (\(\alpha = 0.05\)). This means that there is a 5% risk of concluding that a difference exists when in fact there is no such difference. Setting a significance level is an arbitrary decision, but common choices include 0.01, 0.05, and 0.10. Deciding on the significance level depends on the context of the study and how much risk of error is acceptable.
Proportion Test
A proportion test is a type of statistical test used to determine whether there is a significant difference between the proportion of a certain outcome in a sample and a specific value of interest (often a population proportion or 50/50 split). In the exercise, a proportion test is used to assess whether the observed proportions of couples sharing relationship status or profile pictures on Facebook differ from 0.5.

In part (a), the proportion test leads us to reject the null hypothesis, suggesting a significant difference between the observed proportion and 0.5. Conversely, in part (b), the observed proportion falls within the confidence interval that includes 0.5, so we do not reject the null hypothesis. The proportion test thus helps determine if our observed sample proportions are likely reflective of the true population proportions or if they have occurred simply by random chance.

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

We are conducting many hypothesis tests to test a claim. In every case, assume that the null hypothesis is true. Approximately how many of the tests will incorrectly find significance? 800 tests using a significance level of \(5 \%\).

Test \(\mathrm{A}\) is described in a journal article as being significant with " \(P<.01\) "; Test \(\mathrm{B}\) in the same article is described as being significant with " \(P<\).10." Using only this information, which test would you suspect provides stronger evidence for its alternative hypothesis?

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). A pharmaceutical company is testing to see whether its new drug is significantly better than the existing drug on the market. It is more expensive than the existing drug. Which significance level would the company prefer? Which significance level would the consumer prefer?

Mating Choice and Offspring Fitness Does the ability to choose a mate improve offspring fitness in fruit flies? Researchers have studied this by taking female fruit flies and randomly dividing them into two groups; one group is put into a cage with a large number of males and able to freely choose who to mate with, while flies in the other group are each put into individual vials, each with only one male, giving no choice in who to mate with. Females are then put into egg laying chambers, and a certain number of larvae collected. Do the larvae from the mate choice group exhibit higher survival rates? A study \(^{44}\) published in Nature found that mate choice does increase offspring fitness in fruit flies (with p-value \(<0.02\) ), yet this result went against conventional wisdom in genetics and was quite controversial. Researchers attempted to replicate this result with a series of related experiments, \({ }^{45}\) with data provided in MateChoice. (a) In the first replication experiment, using the same species of fruit fly as the original Nature study, 6067 of the 10000 larvae from the mate choice group survived and 5976 of the 10000 larvae from the no mate choice group survived. Calculate the p-value. (b) Using a significance level of \(\alpha=0.05\) and \(\mathrm{p}\) -value from (a), state the conclusion in context. (c) Actually, the 10,000 larvae in each group came from a series of 50 different runs of the experiment, with 200 larvae in each group for each run. The researchers believe that conditions dif- fer from run to run, and thus it makes sense to treat each \(\mathrm{run}\) as a case (rather than each fly). In this analysis, we are looking at paired data, and the response variable would be the difference in the number of larvae surviving between the choice group and the no choice group, for each of the 50 runs. The counts (Choice and NoChoice and difference (Choice \(-\) NoChoice) in number of surviving larva are stored in MateChoice. Using the single variable of differences, calculate the p-value for testing whether the average difference is greater than \(0 .\) (Hint: this is a single quantitative variable, so the corresponding test would be for a single mean.) (d) Using a significance level of \(\alpha=0.05\) and the p-value from (c), state the conclusion in context. (e) The experiment being tested in parts (a)-(d) was designed to mimic the experiment from the original study, yet the original study yielded significant results while this study did not. If mate choice really does improve offspring fitness in fruit flies, did the follow-up study being analyzed in parts (a)-(d) make a Type I, Type II, or no error? (f) If mate choice really does not improve offspring fitness in fruit flies, did the original Nature study make a Type I, Type II, or no error?

Do iPads Help Kindergartners Learn: A Series of Tests Exercise 4.147 introduces a study in which half of the kindergarten classes in a school district are randomly assigned to receive iPads. We learn that the results are significant at the \(5 \%\) level (the mean for the iPad group is significantly higher than for the control group) for the results on the HRSIW subtest. In fact, the HRSIW subtest was one of 10 subtests and the results were not significant for the other 9 tests. Explain, using the problem of multiple tests, why we might want to hesitate before we run out to buy iPads for all kindergartners based on the results of this study.

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