Chapter 9: Problem 76
The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?
Short Answer
Step by step solution
Define the Hypotheses
Calculate the Test Statistic
Determine the Critical Value
Compare the Test Statistic to the Critical Value
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Hypothesis Testing
- **Defining Hypotheses:** We first define both the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). The null hypothesis in our case asserts that the mean cost \(\mu\) is exactly \)1.00, whereas the alternative hypothesis suggests \(\mu\) is different from $1.00.
- **Choosing a Significance Level:** In advance, we determine a significance level \(\alpha\), often set at 0.05 or 0.01, which represents the probability of rejecting the null hypothesis when it is actually true.
- **Analyzing Results:** Depending on the data and test statistics, we decide whether to accept or reject the null hypothesis. The outcome of the test provides a pathway to understand the data against the initial claim.
Importance of Critical Values
To find the critical value, we refer to a t-distribution table if the population standard deviation is unknown and the sample size is small. In our example, with a significance level of 0.01 and 11 degrees of freedom, the critical value corresponds to approximately \(\pm 3.106\).
Here’s why critical values matter:
- **Defining Boundaries:** They provide clear boundaries to decide if the null hypothesis should be rejected.
- **Risk Management:** Critical values help assess the risk of making a Type I error, which is rejecting a true null hypothesis.
- **Standardization:** They allow comparison across different studies by providing a standard benchmark for evaluating the t-statistic or z-statistic.
Calculating the T-Statistic
To calculate the t-statistic, you use this formula:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]Where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
In our example, the calculated t-statistic is approximately -0.962. This result comes from the given sample mean and standard deviation, indicating how the sample mean compares with the expected mean of $1.00.
Key aspects of the t-statistic include:
- **Sensitivity to Sample Size:** Smaller samples lead to greater variability, which the t-statistic adjusts for, making it a critical tool in small-sample tests.
- **Scale of Measurement:** The t-statistic is reported in standard deviation units, allowing for easier interpretation and comparison.
- **Decision Making:** Comparing the calculated t-statistic to the critical value helps decide whether to reject or not reject the null hypothesis.