Chapter 13: Problem 5
An information retrieval system has ten storage locations. Information has been stored with the expectation that the long-run proportion of requests for location \(i\) is given by the expression \(p_{i}=(5.5-|i-5.5|) / 30\). A sample of 200 retrieval requests gave the following frequencies for locations 1-10, respectively: \(4,15,23,25,38,31,32\), 14,10 , and 8 . Use a chi-squared test at significance level .10 to decide whether the data is consistent with the a priori proportions (use the \(P\)-value approach).
Short Answer
Step by step solution
Determine Null and Alternative Hypotheses
Calculate Expected Frequencies
Compute the Chi-Squared Statistic
Determine Degrees of Freedom
Find the P-Value
Compare the P-Value with Significance Level
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Statistical Hypothesis Testing
Null and Alternative Hypotheses
- If the evidence suggests that the data does not fit the expected pattern under \(H_0\), \(H_1\) is considered more plausible.
- Conversely, if the data aligns closely with the expected model, we do not reject \(H_0\).
Degrees of Freedom
P-Value Approach
- A lower \(P\)-value indicates stronger evidence against the null hypothesis.
- If the \(P\)-value is less than or equal to the significance level (commonly set at 0.05 or 0.10), you reject \(H_0\).
- If the \(P\)-value is greater than the significance level, there is not enough evidence to reject \(H_0\).