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(a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The waiting time for customers at MacBurger Restaurants follows a normal distribution with a population standard deviation of 1 minute. At the Warren Road MacBurger, the quality assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. At the . 05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

Short Answer

Expert verified
Reject the null hypothesis; mean waiting time is less than 3 minutes.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_{0}\)) is that the mean waiting time is 3 minutes. The alternate hypothesis (\(H_{a}\)) is that the mean waiting time is less than 3 minutes.\[H_{0}: \mu = 3 \H_{a}: \mu < 3\]
02

Decision Rule

We will use a Z-test since the population standard deviation is known. The significance level (\(\alpha\)) is 0.05. For a left-tailed test, look up the critical value in the Z-table for \(\alpha = 0.05\), which is -1.645. Reject \(H_{0}\) if the Z-test statistic is less than -1.645.
03

Calculate Test Statistic

Use the formula for the Z-test statistic: \[Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Substitute the values:\[\bar{X} = 2.75, \mu = 3, \sigma = 1, n = 50 \]\[Z = \frac{2.75 - 3}{\frac{1}{\sqrt{50}}} = \frac{-0.25}{0.1414} \approx -1.77 \]
04

Make Decision Regarding H0

The calculated Z-value is -1.77. Since -1.77 is less than -1.645, we reject \(H_{0}\).
05

Calculate and Interpret p-value

The p-value is the probability of observing a test statistic as extreme as the one observed. Using the standard normal distribution table, the p-value for Z = -1.77 is approximately 0.038. Since 0.038 is less than the significance level of 0.05, we reject \(H_{0}\). This indicates that there is sufficient evidence to conclude that the mean waiting time is less than 3 minutes.

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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 is a foundational concept in hypothesis testing. It represents a default or initial assumption that there is no effect or no difference, and that any observed deviation from this is due to random chance. In the scenario of the MacBurger waiting times, the null hypothesis \(H_{0}\) states that the mean waiting time is 3 minutes. Essentially, \(H_{0}: \mu = 3\).
By setting a null hypothesis, you have a baseline that you compare your sample data against. It's like saying, "Let's assume nothing has changed. Can we prove otherwise?" In research and statistics, supporting a null hypothesis means acknowledging that any observed effect might just be random, not a significant pattern.
When testing a hypothesis, if the evidence strongly contradicts the null hypothesis, it may be rejected. However, rejecting the null hypothesis does not prove it false; it merely provides enough statistical evidence against it.
Alternate Hypothesis
The alternate hypothesis (H_{a}) is what researchers aim to support. It represents a new or different reality from the one posed by the null hypothesis. In this case, the alternate hypothesis suggests that the mean waiting time is less than 3 minutes: \(H_{a}: \mu < 3\).
This hypothesis signals a change or effect that the researchers suspect and wish to explore further. If statistical evidence favors the alternate hypothesis, it can lead to new insights or changes in practice.
Laying out an alternate hypothesis clearly can guide the direction of an analysis. Setting it opposite to the null ensures a comprehensive assessment that can either reinforce the existing understanding or point towards new possibilities.
p-value
The p-value is a crucial indicator in hypothesis testing. It measures the probability of obtaining the observed results, or something more extreme, assuming the null hypothesis is true. Lower p-values indicate stronger evidence against the null hypothesis.
The p-value for the MacBurger example is calculated to be approximately 0.038. This tells us that if the population mean waiting time really is 3 minutes, there’s only a 3.8% chance of seeing a sample mean of 2.75 or less.
When dealing with p-values, it's essential to compare them to a predefined significance level (often 0.05). If the p-value is smaller, as it is here, it suggests rejecting the null hypothesis, which supports a conclusion that the mean waiting time truly is less than 3 minutes.
z-test
A z-test is used in hypothesis testing to determine if there is a significant difference between sample and population means. It is suitable when the standard deviation of the population is known and the sample size is sufficiently large.
  • The calculation involves the formula:
    \[ Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
  • Here, \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
The test statistic in the MacBurger scenario was calculated as -1.77, which compares to a critical value to decide on the null hypothesis. Since -1.77 is more extreme than -1.645, the threshold for a 0.05 significance level, it led to rejecting the null. The z-test thus confirms a meaningful deviation from what's expected under the null hypothesis.
Statistical Decision-Making
Statistical decision-making is about using data and statistical methods to determine a course of action. It relies heavily on concepts like the null hypothesis, p-value, and decision rules.
In hypothesis testing, making decisions involves evaluating evidence in light of these established criteria. Here, the decision rule stated to reject the null hypothesis if the z-test statistic was less than -1.645.
The ultimate decision regarding \(H_{0}\) is an integration of all the statistical outcomes. Given that the calculated z-value (-1.77) and the p-value (0.038) both support rejecting \(H_{0}\), the decision is that there is enough evidence to conclude that the mean waiting time is indeed less than 3 minutes at the specified significance level.

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