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91Ó°ÊÓ

The examinations in a large multisection statistics class are scaled after grading so that the mean score is 75 . The professor thinks that students in the 8:00 A.M. class have trouble paying attention because they are sleepy and suspects that these students have a lower mean score than the class as a whole. The students in the 8:00 A.M. class this semester can be considered a sample from the population of all students in the course, so the professor compares their mean score with 75 . State the hypotheses \(H_{0}\) and \(H_{a}\).

Short Answer

Expert verified
\(H_0: \mu = 75\), \(H_a: \mu < 75\).

Step by step solution

01

Define the Null Hypothesis

The null hypothesis (denoted as \(H_0\)) is a statement of no effect or no difference. In this question, the professor suspects that students in the 8:00 A.M. class score lower than the average class score. Therefore, the null hypothesis would state that the mean score of the 8:00 A.M. class is equal to the mean score of the entire class. Formally, this is expressed as \(H_0: \mu = 75\), where \(\mu\) represents the mean score of the 8:00 A.M. class.
02

Define the Alternative Hypothesis

The alternative hypothesis (denoted as \(H_a\)) is what you believe to be true if the null hypothesis is rejected. The professor believes that students in the 8:00 A.M. class score lower than average due to being sleepy. Therefore, the alternative hypothesis would state that the mean score of the 8:00 A.M. class is less than the mean score of the entire class. Formally, this is expressed as \(H_a: \mu < 75\).

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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 (denoted as \(H_0\)) serves as a default statement that suggests no change or effect is present. When it comes to statistics, this hypothesis proposes that any kind of observed difference or effect in the data is purely due to random chance, and not because of a specific cause.
For the case involving the 8:00 A.M. class in a statistics course, the null hypothesis claims that there is no difference between the mean scores of students attending this early class and the established mean score of the whole class. This is formalized by stating \( H_0: \mu = 75 \). Here, \( \mu \) represents the mean of the 8:00 A.M. class.
By testing this hypothesis, it is usually understood that until there is sufficient statistical evidence gathered, the null hypothesis remains valid. The purpose is to either provide strong evidence against this hypothesis (leading to its rejection), or to find insufficient evidence, allowing it to stand.
Alternative Hypothesis
The alternative hypothesis (symbolized as \(H_a\) or \(H_1\)) comes into play when there is a belief that something beyond chance is affecting results. In other words, it's what you might believe to be true if your null hypothesis turns out to be unlikely, according to your data.
In the example of the 8:00 A.M. statistics class, the professor suspects that students score less well in the morning due to sleepiness. Consequently, the alternative hypothesis is that the mean score for the early class is actually lower than the others. This can be expressed as \(H_a: \mu < 75\), indicating the belief that \(\mu\), or average score of the 8:00 A.M. class, is less than the total class mean of 75.
  • It represents a significant shift away from the null hypothesis.
  • It usually indicates the existence of an affect, difference, or relationship.
  • While the null assumes the status quo, the alternative suggests something new at play.

The goal of hypothesis testing is to see if the data supports this alternative proposition. If so, the null hypothesis can be rejected in favor of the alternative.
Mean Score Comparison
At its core, hypothesis testing often revolves around comparing means. It's a straightforward yet powerful way to understand whether there's a meaningful difference between groups. In this context, we are comparing the mean score of the entire class against the mean score of one particular 8:00 A.M. cohort.
The established class mean score is 75. The question we aim to answer is whether the students' mean score from the earlier class is significantly different from 75. In statistics, circumstances like these often employ a mean score comparison method.
  • First, you calculate the mean of the 8:00 A.M. class.
  • Next, you compare it against the known class mean of 75.
  • The statistical analysis determines if any observed difference is significant or likely due to random chance.

A considerable difference, along with a sufficiently large sample size, might indicate that students indeed find it challenging to focus in the early hours, which adversely affects their scores. Through this comparison, educators can identify differentiating factors in student performance.

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Most popular questions from this chapter

You are testing \(H_{0}: \mu=0\) against \(H_{\mathrm{a}}: \mu>0\) based on an SRS of 20 observations from a Normal population. What values of the \(z\) statistic are statistically significant at the \(\alpha=0.005\) level? (a) All values for which \(z>2.576\) (b) All values for which \(z>2.807\) (c) All values for which \(|z|>2.807\)

A test of \(H_{0}: \mu=0\) against \(H_{a}: \mu \neq 0\) has test statistic \(z=1.65\). Is this test statistically significant at the \(5 \%\) level \((\alpha=0.05)\) ? Is it statistically significant at the \(1 \%\) level \((\alpha=0.01)\) ?

The National Institute of Standards and Technology (NIST) supplies "standard copper samples" whose melting point is supposed to be exactly \(1084.80^{\circ} \mathrm{C}\). To do so, NIST must check that samples that they intend to supply meet this condition. Is there reason to think that the true melting point of a new copper sample is not \(1084.80^{\circ} \mathrm{C}\) ? To find out, NIST measures the melting point of this sample six times. Repeated measurements of the same thing vary, which is why NIST makes six measurements. These measurements are an SRS from the population of all possible measurements. This population has Normal distribution with mean \(\mu\) equal to the true melting point and standard deviation \(\sigma=0.25^{\circ} \mathrm{C}\). (a) We seek evidence against the claim that \(\mu=1084.80\). What is the sampling distribution of the mean \(x^{-} \bar{x}\) in many samples of six measurements of one sample if the claim is true? Make a sketch of the Normal curve for this distribution. (Draw a Normal curve, then mark on the axis the values of the mean and 1, 2, and 3 standard deviations on either side of the mean.) (b) Suppose that the sample mean is \(\mathrm{x}^{-} \bar{x}=1084.90\). Mark this value on the axis of your sketch. Another copper sample has \(\mathrm{x}^{-} \bar{x}=1084.50\) for six measurements. Mark this value on the axis as well. Explain in simple language why one result is good evidence that the true melting point differs from \(1084.80\) and why the other result gives no reason to doubt that \(1084.80\) is correct.

Family caregivers of patients with chronic illness can experience anxiety. Do regular support-group meetings affect these feelings of anxiety? It is possible that they reduce anxiety, perhaps through sharing experiences with other caregivers in similar situations, or increase anxiety, perhaps by reinforcing painful experiences by recounting them to others. To explore the effect of support-group meetings, several familiy caregivers were enrolled in a support group. After three months, researchers administered a test to measure anxiety, with larger scores indicating greater anxiety. Assume these caregivers are a random sample from the population of all family caregivers. A \(95 \%\) confidence interval for the population mean anxiety score \(\mu\) after participating in a support group is \(7.2 \pm 0.7 .{ }^{13}\) Use the method described in the previous exercise to answer these questions. (a) Suppose we know that the mean anxiety score for the population of all family caregivers is \(6.4\). With a two-sided alternative, can you reject the null hypothesis that \(\mu=6.4\) at the \(5 \%(\alpha=0.05)\) significance level? Why? (b) Suppose we know that the mean anxiety score for the population of all family caregivers is \(6.6\). With a two-sided alternative, can you reject the null hypothesis that \(\mu=6.6\) at the \(5 \%(\alpha=0.05)\) significance level? Why?

Every society has its own marks of wealth and prestige. In ancient China, it appears that owning pigs was such a mark. Evidence comes from examining burial sites. The skulls of sacrificed pigs tend to appear along with expensive ornaments, which suggests that the pigs, like the ornaments, signal the wealth and prestige of the person buried. A study of burials from around 3500 B.C. concluded that "there are striking differences in grave goods between burials with pig skulls and burials without them . . . A test indicates that the two samples of total artifacts are statistically significantly different at the \(0.01\) level."8 Explain clearly why "statistically significantly different at the \(0.01\) level" gives good reason to think that there really is a systematic difference between burials that contain pig skulls and those that lack them.

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