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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

Scientists studying lion attacks on humans in Tanzania \(^{32}\) found that 95 lion attacks happened between \(6 \mathrm{pm}\) and \(10 \mathrm{pm}\) within either five days before a full moon or five days after a full moon. Of these, 71 happened during the five days after the full moon while the other 24 happened during the five days before the full moon. Does this sample of lion attacks provide evidence that attacks are more likely after a full moon? In other words, is there evidence that attacks are not equally split between the two five-day periods? Use StatKey or other technology to find the p-value, and be sure to show all details of the test. (Note that this is a test for a single proportion since the data come from one sample.)

Interpreting a P-value In each case, indicate whether the statement is a proper interpretation of what a p-value measures. (a) The probability the null hypothesis \(H_{0}\) is true. (b) The probability that the alternative hypothesis \(H_{a}\) is true. (c) The probability of seeing data as extreme as the sample, when the null hypothesis \(H_{0}\) is true. (d) The probability of making a Type I error if the null hypothesis \(H_{0}\) is true. (e) The probability of making a Type II error if the alternative hypothesis \(H_{a}\) is true.

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 \(\alpha=0.01\) ). Testing a new drug with potentially dangerous side effects to see if it is significantly better than the drug currently in use. If it is found to be more effective, it will be prescribed to millions of people.

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 \(H_{a}: p<0.5\) Sample data: \(\hat{p}=38 / 100=0.38\) with \(n=100\)

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 \(\alpha=0.01\) ). Using a sample of 10 games each to see if your average score at Wii bowling is significantly more than your friend's average score.
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