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Chapter 5: Problem 17

A study suggests that \(60 \%\) of college student spend 10 or more hours per week communicating with others online. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. You randomly sample 160 students from your dorm and find that \(70 \%\) spent 10 or more hours a week communicating with others online. A friend of yours, who offers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. $$ \begin{array}{l} H_{0}: \hat{p}<0.6 \\ H_{A}: \hat{p}>0.7 \end{array} $$

Short Answer

Expert verified
The null and alternative hypotheses are incorrectly formulated; correct them to \(H_0: p = 0.6\) and \(H_A: p \neq 0.6\).

Step by step solution

01

Understand the Null and Alternative Hypotheses

In hypothesis testing, we need to establish a null hypothesis \(H_0\) and an alternative hypothesis \(H_A\). The null hypothesis usually states that there is no effect or no difference, and includes a specific population parameter. The alternative hypothesis is what you want to prove and usually states that there is some effect or difference.
02

Identify Errors in the Null Hypothesis

The null hypothesis \(H_0\) is incorrectly stated as \(\hat{p} < 0.6\). Usually, \(H_0\) is about equality or no change. For this scenario, it should be \(H_0: p = 0.6\) where \(p\) is the population proportion, not \(\hat{p}\), which is the sample proportion.
03

Identify Errors in the Alternative Hypothesis

The alternative hypothesis \(H_A\) is incorrectly stated as \(\hat{p} > 0.7\). The goal is to see if there's a significant difference from the claimed \(60\%\), so \(H_A\) should be formulated about the population proportion: \(H_A: p eq 0.6\). This is looking for a difference, not just an increase.
04

Formulate the Correct Hypotheses

Given the context of the question, an appropriate set of hypotheses are:- \(H_0: p = 0.6\) (The true population proportion of students who spend 10 or more hours online is 60%.)- \(H_A: p eq 0.6\) (The true population proportion is different from 60%.) This depicts a two-tailed test, which is appropriate if the investigational aim is to identify any difference rather than a specific increase or decrease.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the **null hypothesis** is our starting point. It is usually a statement of no effect or no change that we try to test against. - For this exercise, it should reflect the **population proportion** claimed in the original study, which is 60%.- Therefore, the null hypothesis is written as: \( H_0: p = 0.6 \), where \( p \) is the true population proportion. This hypothesis asserts that the proportion of students spending 10 or more hours online is exactly 60%. It's like saying "nothing special is happening," a foundation to disprove or support by collecting sample data.
Alternative Hypothesis
The **alternative hypothesis** is what we test against the null hypothesis. It's usually a statement that there is an effect or a difference. - In this context, the alternative hypothesis challenges the original claim about the population proportion.- The initial serious error was in stating it as \( \hat{p} > 0.7 \), which incorrectly uses the **sample proportion**. Instead, the correct way is: \( H_A: p eq 0.6 \). We aim to establish that the true population proportion is different from 60%, which may be lower or higher. This represents a two-tailed test, suggesting any deviation from the original claim is significant.
Population Proportion
The **population proportion** is a key concept in understanding results from a broader group - It represents the percentage of all individuals in a population having a particular characteristic. - In our case, it is the proportion of all college students who spend 10 or more hours online weekly, claimed to be 60%. This is our focus parameter in hypothesis testing. The null hypothesis revolves around this value, and by getting appropriate sample data, we test if the true population proportion deviates significantly from this claimed figure.
Sample Proportion
The **sample proportion** provides insight when testing hypotheses against the population proportion. - It's derived from the data collected in a sample and is a snapshot of the population proportion. - For the given exercise, the sample consists of 70% from 160 students. This sample proportion of 70% allows us to conduct the test. It serves to estimate if the true population proportion might actually differ from the claimed 60%. By evaluating the difference between the sample and population proportions, we decide whether or not to reject the null hypothesis.

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Most popular questions from this chapter

A patient named Diana was diagnosed with Fibromyalgia, a long-term syndrome of body pain, and was prescribed anti-depressants. Being the skeptic that she is, Diana didn't initially believe that anti-depressants would help her symptoms. However after a couple months of being on the medication she decides that the anti-depressants are working, because she feels like her symptoms are in fact getting better. (a) Write the hypotheses in words for Diana's skeptical position when she started taking the anti-depressants. (b) What is a Type 1 Error in this context? (c) What is a Type 2 Error in this context?

Of all freshman at a large college, \(16 \%\) made the dean's list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) Calculate the variability of this distribution. (d) What is the formal name of the value you computed in (c)? (e) Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations?

Exercise 5.11 provides a \(95 \%\) confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. (a) A local newspaper claims that the average waiting time at this ER exceeds 3 hours. Is this claim supported by the confidence interval? Explain your reasoning. (b) The Dean of Medicine at this hospital claims the average wait time is 2.2 hours. Is this claim supported by the confidence interval? Explain your reasoning. (c) Without actually calculating the interval, determine if the claim of the Dean from part (b) would be supported based on a \(99 \%\) confidence interval?

In a random sample 765 adults in the United States, 322 say they could not cover a $$\$ 400$$ unexpected expense without borrowing money or going into debt. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic can we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) A cable news pundit thinks the value is actually \(50 \%\). Should she be surprised by the data? (g) Suppose the true population value was found to be \(40 \%\). If we use this proportion to recompute the value in part (e) using \(p=0.4\) instead of \(\hat{p},\) does the resulting value change much?

As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic can we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) The historical rate of defects is \(10 \%\). Should the engineer be surprised by the observed rate of defects during the current week? (g) Suppose the true population value was found to be \(10 \%\). If we use this proportion to recompute the value in part (e) using \(p=0.1\) instead of \(\hat{p},\) does the resulting value change much?
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