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Chapter 4: Problem 7

State the null and alternative hypotheses for the statistical test described. Testing to see if there is evidence that a mean is less than 50 .

Short Answer

Expert verified
The null hypothesis is \( H_0: \mu = 50 \) and the alternative hypothesis is \( H_1: \mu < 50 \).

Step by step solution

01

Formulate the Null Hypothesis

The Null hypothesis, often denoted as \( H_0 \), is a statement that indicates no effect or difference from an assumed standard or normal condition. Here, it is assumed that the mean is 50. Therefore, the null hypothesis would be \( H_0: \mu = 50 \), where \( \mu \) represents the mean.
02

Formulate the Alternative Hypothesis

The Alternative Hypothesis, usually denoted as \( H_1 \) or \( H_a \), indicates the existence of an effect or difference. In this context, the claim is that the mean is less than 50. Therefore, the alternative hypothesis would be \( H_1: \mu < 50 \).

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Key Concepts

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

Null Hypothesis
The null hypothesis, often represented as \( H_0 \), plays a crucial role in hypothesis testing. It is a formal statement suggesting that no effect or significant difference exists in a population parameter, based on the sample data. In simpler terms, it assumes the "status quo" or the standard against which any potential variations are measured.
Consider it as the default position, which in this scenario implies that the average or mean is equal to 50. Hence, in our given exercise, the null hypothesis is expressed as \( H_0: \mu = 50 \), where \( \mu \) denotes the population mean.
This hypothesis provides a baseline for statistical tests, allowing researchers to determine if the evidence collected is strong enough to reject this initial assumption.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \) or \( H_a \), represents what you aim to prove or investigate through statistical testing. It challenges the null hypothesis by suggesting that there is an effect or a difference in the population parameter of interest.
In our case, it questions whether the mean is indeed less than 50. Thus, the alternative hypothesis for the exercise is \( H_1: \mu < 50 \).
This hypothesis indicates a one-tailed test, specifically a left-tailed test, since it focuses on determining if the mean is significantly lower than the stated value. It is crucial in hypothesis testing as it provides a direction for what the researcher expects to find based on the sample data.
Mean Comparison
Mean comparison involves analyzing whether there is a statistically significant difference between a sample mean and a known value, such as the mean assumed in the null hypothesis. This type of analysis is pivotal in making informed decisions about population parameters based on sample data.
Assessing differences in means often involves the use of tests such as the t-test or z-test, depending on the sample size and distribution characteristics. These tests calculate the probability of observing the sample data if the null hypothesis is true.
  • If this probability, or p-value, is sufficiently low, typically less than 0.05, researchers may reject the null hypothesis in favor of the alternative.
  • Conversely, a higher p-value suggests insufficient evidence to support the alternative hypothesis.

In our specific problem, the focus is on comparing the sample mean to 50, determining if it is indeed lower than this value, which would support the alternative hypothesis.

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Most popular questions from this chapter

Give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=28 / 40=0.70\) with \(n=40\)

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?

Pesticides and ADHD Are children with higher exposure to pesticides more likely to develop ADHD (attention-deficit/hyperactivity disorder)? In one study, authors measured levels of urinary dialkyl phosphate (DAP, a common pesticide) concentrations and ascertained ADHD diagnostic status (Yes/No) for 1139 children who were representative of the general US population. \(^{7}\) The subjects were divided into two groups based on high or low pesticide concentrations, and we compare the proportion with ADHD in each group. (a) Define the relevant parameter(s) and state the null and alternative hypotheses. (b) In the sample, children with high pesticide levels were more likely to be diagnosed with ADHD. Can we necessarily conclude that, in the population, children with high pesticide levels are more likely to be diagnosed with ADHD? (Whether or not we can make this generalization is, in fact, the statistical question of interest.) (c) In the study, evidence was found to support the alternative hypothesis. Explain what that means in the context of pesticide exposure and ADHD?

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer. \(^{52}\) The study reviewed the records of all 1,050 skin cancer patients referred to the St. Louis University Cancer Center in 2004\. Of the 42 patients with melanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11 . (a) Is this an experiment or an observational study? (b) Of the patients with melanoma, what proportion had the cancer on the left side? (c) A bootstrap \(95 \%\) confidence interval for the proportion of melanomas occurring on the left is 0.579 to \(0.861 .\) Clearly interpret the confidence interval in the context of the problem. (d) Suppose the question of interest is whether melanomas are more likely to occur on the left side than on the right. State the null and alternative hypotheses. (e) Is this a one-tailed or two-tailed test? (f) Use the confidence interval given in part (c) to predict the results of the hypothesis test in part (d). Explain your reasoning. (g) A randomization distribution gives the p-value as 0.003 for testing the hypotheses given in part (d). What is the conclusion of the test in the context of this study? (h) The authors hypothesize that skin cancers are more prevalent on the left because of the sunlight coming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do the data in this study support a conclusion that more melanomas occur on the left side because of increased exposure to sunlight on that side for drivers?

In Exercises 4.40 to 4.44 , null and alternative hypotheses for a test are given. Give the notation \((\bar{x},\) for example) for a sample statistic we might record for each simulated sample to create the randomization distribution. \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\)
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