/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A new weight-watching company, W... [FREE SOLUTION] | 91Ó°ÊÓ

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A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds with a standard deviation of 2.8 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the \(p\) -value.

Short Answer

Expert verified
At the 0.05 significance level, we conclude that the average weight loss is less than 10 pounds.

Step by step solution

01

Define Hypotheses

We start by setting up the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). - The null hypothesis \(H_0\) is: \( \mu = 10 \) pounds (mean weight loss is 10 pounds).- The alternative hypothesis \(H_a\) is: \( \mu < 10 \) pounds (mean weight loss is less than 10 pounds).
02

Collect Sample Data

You have the following data: - Sample size \(n = 50\)- Sample mean \(\bar{x} = 9\) pounds- Sample standard deviation \(s = 2.8\) pounds.
03

Calculate Test Statistic

The formula to calculate the test statistic for the sample mean is:\[z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]Substituting the values:\[z = \frac{9 - 10}{\frac{2.8}{\sqrt{50}}} \approx -2.525\]
04

Determine Critical Value and Significance Level

The significance level is given as \( \alpha = 0.05 \). Since the test is one-tailed (we are checking if the mean is less than 10), we look up the critical value for \(z\) in a standard normal distribution table for \( \alpha = 0.05 \), which is \(-1.645\).
05

Compare Test Statistic to Critical Value

Compare the calculated test statistic \(z = -2.525\) with the critical value \(-1.645\). Since \(-2.525\) is less than \(-1.645\), we reject the null hypothesis.
06

Calculate p-value

To find the p-value, locate \(z = -2.525\) in the standard normal distribution table. The p-value corresponds to the probability of obtaining a \(z\) value less than \(-2.525\), which is approximately 0.0058.
07

Make a Decision

Since the p-value (0.0058) is less than the significance level (0.05), we reject the null hypothesis. This indicates there is enough evidence to support the claim that the average weight loss is less than 10 pounds.

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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 denoted as \( H_0 \), serves as the starting point for hypothesis testing in statistics. It represents a baseline assumption, often implying "no effect" or "no difference." In the context of our weight loss study, the null hypothesis is that the average weight loss is exactly 10 pounds. More formally, we state it as \( \mu = 10 \).

Why do we establish a null hypothesis? Essentially, we do this to have a reference point that we can test against. If evidence strongly contradicts this hypothesis, we then consider other alternatives. It's like starting any investigation by asking, "What if nothing happened?"

To test the null hypothesis, we gather data and perform statistical tests to see if our sample data sufficiently challenges \( H_0 \). If it does, we may reject the null hypothesis in favor of an alternative. However, if the data doesn't provide strong evidence against it, we keep the null hypothesis and conclude there is no strong evidence of a change.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \) or \( H_1 \), is what researchers typically want to prove. It is a statement contrary to the null hypothesis. In our particular case on weight loss, the alternative hypothesis asserts \( \mu < 10 \), meaning the average weight loss is actually less than 10 pounds.

This hypothesis doesn't merely oppose the null; it suggests the presence of an effect or a difference. When crafting an alternative hypothesis, researchers know they'll need substantial evidence to shift the null hypothesis. It's akin to suggesting that a new theory or observation better explains the data.

The alternative hypothesis gains support if the collected data show significant deviation from what the null hypothesis predicts. This is why it's critical to ensure that your data are robust and your statistical testing method is accurate, as these elements provide the foundational support for drawing conclusions.
p-value
The \( p \)-value is a crucial component of hypothesis testing. It represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed statistic, assuming \( H_0 \) is true. Essentially, it provides a metric to gauge the strength of evidence against the null hypothesis.

In our weight loss example, the calculated \( p \)-value is approximately 0.0058. This means there is a 0.58% chance that we'd observe such a significant difference in average weight loss (or something more extreme) purely by random chance, assuming \( H_0: \mu = 10 \) is valid.

A small \( p \)-value suggests that the sample results are unlikely under the null hypothesis, which leads us to consider the alternative hypothesis as more plausible. Researchers often predefine a threshold, known as the significance level, to help decide when a \( p \)-value is small enough to reject \( H_0 \).
Significance Level
The significance level, represented by \( \alpha \), is a threshold set by the researcher that helps determine when to reject the null hypothesis. It essentially dictates how much risk we are willing to take in making a Type I error, where we incorrectly reject \( H_0 \) when it's actually true.

In many fields, including our weight loss case, the significance level is often set at 0.05. This tells us there is a 5% risk of concluding that the average weight loss is less than 10 pounds when, in fact, it isn't.

The significance level corresponds to the critical value in the test. For a one-tailed test like ours (where we're only concerned if the mean falls below a certain value), we can compare our test statistic against a critical value defined by this threshold. If our \( p \)-value is less than \( \alpha \), we reject \( H_0 \).
  • A lower \( \alpha \) leads to stricter requirements to reject \( H_0 \).
  • A higher \( \alpha \) makes it easier to find something statistically significant, risking false positives.
Bringing it all together, we note that in our example, since our \( p \)-value of 0.0058 is less than 0.05, we reject the null hypothesis, concluding that participants likely lose less than 10 pounds.

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Most popular questions from this chapter

The following hypotheses are given. $$\begin{array}{l}H_{0}: \pi \leq .70 \\\H_{1}: \pi>.70\end{array}$$ A sample of 100 observations revealed that \(p=.75 .\) At the .05 significance level, can the null hypothesis be rejected? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

The following hypotheses are given. $$\begin{array}{l}H_{0}: \pi=.40 \\\H_{1}: \pi \neq .40\end{array}$$ A sample of 120 observations revealed that \(p=.30 .\) At the .05 significance level, can the null hypothesis be rejected? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

The following information is available. $$\begin{array}{l}H_{0}: \mu=50 \\\H_{1}: \mu \neq 50\end{array}$$ The sample mean is \(49,\) and the sample size is \(36 .\) The population follows the normal distribution and the standard deviation is \(5 .\) Use the .05 significance level.

According to a recent survey, Americans get a mean of 7 hours of sleep per night. A random sample of 50 students at West Virginia University revealed the mean number of hours slept last night was 6 hours and 48 minutes \((6.8\) hours \() .\) The standard deviation of the sample was 0.9 hours. Is it reasonable to conclude that students at West Virginia sleep less than the typical American? Compute the \(p\) -value.

Many grocery stores and large retailers such as Wal-Mart and K-Mart have installed selfcheckout systems so shoppers can scan their own items and cash out themselves. How do customers like this service and how often do they use it? Listed below is the number of customers using the service for a sample of 15 days at the Wal-Mart on Highway 544 in Surfside, South Carolina. $$\begin{array}{|rrrrrrrrrr|}\hline 120 & 108 & 120 & 114 & 118 & 91 & 118 & 92 & 104 & 104 \\\112 & 97 & 118 & 108 & 117 & & & & & \\\\\hline\end{array}$$ Is it reasonable to conclude that the mean number of customers using the self- checkout system is more than 100 per day? Use the .05 significance level.

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