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The manager of a service station claims that the mean amount spent on gas by its customers is \(\$ 15.90\) per visit. You want to test if the mean amount spent on gas at this station is different from \(\$ 15.90\) per visit. Briefly explain how you would conduct this test when \(\sigma\) is not known.

Short Answer

Expert verified
To find out if the mean amount spent on gas is different from $15.90, a two-tailed t-test should be conducted. First, state the null and alternative hypotheses. Then, collect sample data and compute the sample mean, standard deviation, and sample size. Calculate the test statistic using these figures. Finally, find the critical value corresponding to the chosen confidence level and compare it to the test statistic to make a decision.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis is that the mean of the amounts spent on gas is equal to $15.90. This is denoted as \(H_0: \mu = 15.90\). The alternative hypothesis is that the mean is not equal to $15.90, written as \(H_a: \mu ≠ 15.90\).
02

Collect and Analyze Sample Data

Collect a random sample from the population and calculate the sample mean \( \bar{x} \), and the sample standard deviation \(s\). Also find the sample size \(n\). These figures will be used in the next step to calculate the t-score.
03

Calculate the Test Statistic

The t-test statistic is calculated using the following formula: \( t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \) where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the population mean assumed under the null hypothesis (in this case, $15.90), \( s \) is the sample standard deviation, and \( n \) is the sample size.
04

Find the Critical Value and Make the Decision

Find the critical t-value corresponding to the confidence level and degrees of freedom, which are \( n - 1 \) for a t-test. Compare the absolute value of the test statistic to the critical value. If the test statistic is greater than the critical value, reject the null hypothesis. This means you believe the mean amount spent on gas is different than $15.90. If the test statistic is less than the critical value, you do not reject the null hypothesis, indicating the claim may be correct.

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Key Concepts

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

Understanding the Null Hypothesis
In statistics, a null hypothesis serves as the starting point for conducting a hypothesis test. It is essentially a statement, typically about a population parameter, formulated to be tested.
In our problem, the service station manager claims that the mean amount spent on gas by customers is \(15.90. Thus, the null hypothesis, denoted as \(H_0\), asserts that the mean spending is indeed \)15.90.
This can be mathematically expressed as:
\(H_0: \mu = 15.90\).
- The null hypothesis is always about population parameters, not sample statistics.- It often represents a "no effect" or "status quo" situation. - The goal of the hypothesis test is to assess the evidence against the null hypothesis.
Exploring the Alternative Hypothesis
The alternative hypothesis offers a statement that contradicts the null hypothesis. It suggests a different scenario than the one proposed under the null. In our example, you want to test whether the mean amount spent on gas is different from \(15.90. Thus, the alternative hypothesis, denoted as \(H_a\), posits that the mean is not equal to \)15.90:
\(H_a: \mu eq 15.90\).
- The alternative hypothesis reflects what we seek to support with our test.- It can be one-sided (greater than or less than a certain value) or two-sided (not equal to a specific value).

In this case, it is a two-sided hypothesis, implying we are interested in any deviation from $15.90, not just an increase or decrease.
Calculating the Test Statistic
The test statistic is a crucial component of hypothesis testing. It is calculated using sample data and compared against a critical value to decide whether to reject the null hypothesis.
When you're using a t-test, the formula for the test statistic \( t \) is:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where:- \( \bar{x} \) is the sample mean- \( \mu_0 \) is the population mean under the null hypothesis (e.g., $15.90)- \( s \) is the sample standard deviation- \( n \) is the sample size
  • A t-test is particularly useful when the population standard deviation is unknown.
  • The calculated t-score indicates how far and in what direction the sample mean deviates from the null hypothesis mean, in units of standard error.
This comparison helps determine if the observed data is significantly different from what was stated in the null hypothesis.
Determining the Critical Value
A critical value establishes the threshold for deciding whether to reject the null hypothesis. It is linked to the confidence level you choose for your test, typically 95% or 99%. In a t-test, the critical value is determined based on the degrees of freedom, which are calculated as \( n - 1 \), where \( n \) is the sample size.
- Look up the critical t-value from a t-distribution table or use statistical software.- Compare the absolute value of the calculated test statistic to this critical value.If the test statistic exceeds the critical value, the null hypothesis is rejected. This suggests the sample provides enough evidence to conclude that the mean amount spent on gas is indeed different from $15.90. On the other hand, if the statistic is less than the critical value, we do not reject the null hypothesis, indicating there is not enough evidence to support a difference. This approach forms the basis of making decisions in hypothesis testing.

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Most popular questions from this chapter

According to a 2008 survey by the Royal Society of Chemistry, \(30 \%\) of adults in Great Britain stated that they typically run the water for a period of 6 to 10 minutes while taking the shower (http://www.rsc.org/ AboutUs/News/PressReleases/2008/EuropeanShowerHabits.asp). Suppose that in a recent survey of 400 adults in Great Britain, 104 stated that they typically run the water for a period of 6 to 10 minutes when they take a shower. At the \(5 \%\) significance level, can you conclude that the proportion of all adults in Great Britain who typically run the water for a period of 6 to 10 minutes when they take a shower is less than 30 ? Use both the \(p\) -value and the critical value approaches.

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