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A consumer organization wants to know whether there is a difference in the price of a particular toy at three different types of stores. The price of the toy was checked in a sample of five discount stores, five variety stores, and five department stores. The results are shown below. Use the .05 significance level. $$ \begin{array}{|ccc|} \hline \text { Discount } & \text { Variety } & \text { Department } \\ \hline \$ 12 & \$ 15 & \$ 19 \\ 13 & 17 & 17 \\ 14 & 14 & 16 \\ 12 & 18 & 20 \\ 15 & 17 & 19 \\ \hline \end{array} $$

Short Answer

Expert verified
Perform an ANOVA test to determine if there are significant price differences; compare F-statistic with critical value.

Step by step solution

01

State the Hypotheses

We begin by stating the null and alternative hypotheses for the test. The null hypothesis (\(H_0\)) is that there is no difference in the mean prices of the toy across the three store types. The alternative hypothesis (\(H_a\)) is that there is a difference in at least two groups.- \(H_0: \mu_1 = \mu_2 = \mu_3\)- \(H_a:\) At least one mean is different.
02

Choose the Significance Level

The significance level (\(\alpha\)) is given as 0.05. This is the criteria we will use to decide whether or not to reject the null hypothesis.
03

Perform ANOVA Test

To check for differences between group means, we will perform an ANOVA test:- Calculate the group means for each store type.- Compute the overall mean.- Find the sum of squares between groups (SSB) and sum of squares within groups (SSW).- Calculate the degrees of freedom for between groups (dfb = 2) and within groups (dfw = 12).- Compute the mean squares between (MSB = SSB/dfb) and within (MSW = SSW/dfw).- Calculate the F-ratio: \( F = \frac{MSB}{MSW} \).For this dataset, use statistical software or an ANOVA table to determine the F-statistic.
04

Decision Rule

We will compare the calculated F-statistic to the critical F-value from the F-distribution table at \( \alpha = 0.05 \) for dfb = 2 and dfw = 12. If the F-statistic is greater than the critical value, we reject the null hypothesis.
05

Conclusion

Based on the value of the F-statistic and the critical F-value, determine whether to reject or fail to reject the null hypothesis. Conclude if there is a significant difference in toy prices among the store types.

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Key Concepts

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

Hypotheses Testing
Hypotheses testing involves making an assumption about a parameter and finding evidence to support or refute this assumption with data. We start by formulating a **null hypothesis** (\(H_0\)) and an **alternative hypothesis** (\(H_a\)). In the context of an ANOVA test, the null hypothesis states that all group means are equal, implying no effect or no difference. For the toy store price example, this means:
\(H_0: \mu_1 = \mu_2 = \mu_3\), where \(\mu_1, \mu_2,\) and \(\mu_3\) represent the average prices of toys in discount, variety, and department stores respectively.
The alternative hypothesis suggests that at least one group mean is different, indicating a potential difference:
\(H_a:\) at least one \(\mu\) is different.
This method helps in decision-making, allowing us to evaluate if observed data is consistent with the null hypothesis or if there is enough evidence to support the alternative.
Significance Level
The **significance level** (\(\alpha\)) in hypothesis testing determines the threshold for rejecting the null hypothesis. It is the probability of falsely rejecting a true null hypothesis (Type I error). In most statistical analysis, a common significance level is 0.05, as used in this toy price example.
Using \(\alpha = 0.05\) means there is a 5% risk of concluding that a difference exists when in fact, it does not. It's like setting a bar for the evidence needed to reject \(H_0\). If the probability that the data supporting \(H_a\) happening by random chance is lower than 5%, we reject \(H_0\).
This threshold ensures that critical decisions are based on a strong likelihood of correctness rather than arbitrary results.
Statistical Analysis
**Statistical analysis** of data involves organizing, interpreting, and drawing conclusions. The ANOVA test is a powerful statistical technique used to compare means among three or more groups. ANOVA helps assess whether the observed differences between group means are more than what we might expect by chance.
For this test, we calculate:
  • Group means: the average price of toys for each store type.
  • Overall mean: the average price of toys across all stores.
  • Sum of squares between groups (SSB): measures variability between different groups.
  • Sum of squares within groups (SSW): measures variability within each group.

The resulting calculations break down into variance components:
  • Mean Square Between (MSB) = SSB divided by its degrees of freedom.
  • Mean Square Within (MSW) = SSW divided by its degrees of freedom.

This analysis prepares the data for the calculation of the F-ratio, determining if group means are statistically different.
F-ratio
The **F-ratio** is a critical part of the ANOVA test and represents the ratio of variation between group means to the variation within the groups. This ratio helps in determining if the group means differences are statistically significant.
The computation involves:
  • F-ratio = MSB/MSW
The higher the F-ratio, the greater the likelihood that the observed data groups have different means.
We assess the F-ratio against a critical value from the F-distribution table, which considers the degrees of freedom for between groups (dfb) and within groups (dfw). In our example: dfb=2 and dfw=12.
If the F-ratio exceeds the critical value from the F-table (at \(\alpha = 0.05\)), the null hypothesis is rejected, concluding a statistically significant difference in toy prices among different store types. This approach remains foundational for comparing multiple group means efficiently.

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Most popular questions from this chapter

The following are four observations collected from each of three treatments. Test the hypothesis that the treatment means are equal. Use the . 05 significance level. $$ \begin{array}{|rcc|} \hline \text { Treatment } 1 & \text { Treatment } 2 & \text { Treatment } 3 \\\ \hline 8 & 3 & 3 \\ 6 & 2 & 4 \\ 10 & 4 & 5 \\ 9 & 3 & 4 \\ \hline \end{array} $$ a. State the null and the alternate hypotheses. b. What is the decision rule? c. Compute SST, SSE, and SS total. d. Complete an ANOVA table. e. State your decision reqarding the null hypothesis.

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