Please provide the following information. (a) What is the...

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Library: Book Circulation The director of library services at Fairmont College did a survey of types of books (by subject) in the circulation library. Then she used library records to take a random sample of 888 books checked out last term and classified the books in the sample by subject. The results are shown next. Using a $5 \%$ level of significance, test the claim that the subject distribution of books in the library fits the distribution of books checked out by students.

Please provide the following information.
(a) What is the level of significance? State the null and alternate hypotheses.
(b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom?
(c) Find or estimate the $P$ -value of the sample test statistic.
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories?
(e) Interpret your conclusion in the context of the application.
Library: Book Circulation The director of library services at Fairmont College did a survey of types of books (by subject) in the circulation library. Then she used library records to take a random sample of 888 books checked out last term and classified the books in the sample by subject. The results are shown next.
Using a $5 \%$ level of significance, test the claim that the subject distribution of books in the library fits the distribution of books checked out by students.

Key Concepts

Decision Rule and Test Conclusion

This concept involves comparing the p-value to the level of significance to decide whether to reject or fail to reject the null hypothesis. Additionally, the conclusion must be interpreted in the context of the specific application, linking the statistical results back to the practical implications of the study.

Level of Significance

This concept refers to the probability threshold set by the researcher (commonly noted as alpha, such as 0.05) to determine whether the observed data are statistically unlikely under the assumption of the null hypothesis. It serves as a cutoff for deciding if the results of the test can be deemed statistically significant.

Null and Alternative Hypotheses

These are formal statements used to test a claim in statistical hypothesis testing. The null hypothesis typically represents a position of no effect or no difference (for example, that the observed distribution matches the expected distribution), while the alternative hypothesis represents the claim that there is an effect or a difference. Formulating these hypotheses is a critical step in the testing process.

Chi-Square Goodness-of-Fit Test

This statistical test is used to determine whether there is a significant difference between the observed frequencies and the expected frequencies in one or more categories. It is commonly applied when assessing the fit of an observed distribution to a theoretical model, making it fundamental in categorical data analysis.

Expected Frequencies

Expected frequencies are the counts that would be predicted by the theoretical model or distribution, assuming the null hypothesis is true. A common assumption in the chi-square goodness-of-fit test is that each expected frequency should be greater than a specified threshold (often 5) to ensure the validity of the test.

Sampling Distribution and Degrees of Freedom

In the context of the chi-square test, the sampling distribution refers to the chi-square distribution, which is used to assess the probability of the observed statistic under the null hypothesis. Degrees of freedom, typically calculated as the number of categories minus one, adjust the shape of the chi-square distribution and are crucial for determining critical values and p-values.

P-Value Estimation

The P-value is the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. It quantifies the strength of the evidence against the null hypothesis, with smaller p-values indicating stronger evidence.

Transcript

00:01 What do we have in this question? now in this question, the director of library services at fairmont college did a survey of the types of books in the circulation library.

00:11 Then she used the library records to take a random sample of 888 books checked out last term and classified the books in the sample by their subjects.

00:21 Now what are the subject areas? so we are going to need a table.

00:25 So this column is subject area.

00:35 Okay after that we have the percent of books in the circulation library so let me just write this as the percent of books in the library in the library on that particular subject after that the observed values the number of books in the sample on this subject so let me just write this as the observed values okay this is what we have so far so what are the business area the subject area the first one is business.

01:14 The next is humanities.

01:20 Humanities.

01:21 After that we have natural sciences.

01:22 Let me just write this as an s.

01:24 Then we have social sciences or social studies.

01:27 So let me just write this as sst.

01:30 And the last category is that of the other subjects.

01:35 Okay? the other subjects.

01:36 Now what are the percentage of books in the entire library on these subjects? it is 32 % then it is 25 % then it is 20 % then it is 15 % and then for the last one this is 8 % now what are the observed values from that 888 books how are the books divided into these subjects it was 268 then it was 214 then it was 215 15 150 15 and 76 okay now this is a table that is given to us now what are they asking they are saying use a 5 % level of significance which means our alpha is 0 .05 use 5 % level of significance and test the claim that the subject distribution of the books in the library fits the distribution of the books checked out by students so we need a null and the alternative hypothesis right a null and an alternative hypothesis null hypothesis null hypothesis is there is subject distribution.

02:53 Null hypothesis for a study will be that the subject distribution of books in the library, of books in the library, fits the distribution of books checked out by students, checked out by students, checked out by students.

03:44 What will be the alternative hypothesis? the alternative hypothesis will be that the subject distribution of books in the library, in the library doesn't fit.

04:09 Doesn't fit the distribution of books in the sample, in the sample.

04:27 Okay, so what we are using is a kai square statistic.

04:30 Now, what is the first step in performing a kai square statistic? we need to find the expected values for all the categories, the expected values for all the categories.

04:41 And how exactly do i find that? it is going to be sample size multiplied by the probability of the proportion for each category, multiplied by the probability for each category.

05:08 All right, so let us just go here.

05:10 What was our sample size? our sample size was 888.

05:14 Was 888.

05:15 Now, this is the column for expected values, for expected values.

05:27 All right.

05:30 So let us just apply the formula that we just learned.

05:33 How exactly do i apply it in this question? my total sample size is 888.

05:41 My total sample size is 888 and i want 32 % of that so 0 .32.

05:46 So this is going to be 284 .16, 284 .16.

05:54 Then we want 25 % of 888.

05:57 So 888 divided by 4.

06:00 This is 22.

06:02 22.

06:04 Then we want 20.

06:06 Of 888 888 into 0 .2 this is 177 .6 177 .6 then we want 15 % of 888 888 in 2 .15 133 .23 .2 133 .2 and then we want 8 % of 888 so this is 0 .08 into 888 and this is 71 .04 now i have the expected values for all the categories in question.

06:37 Now, what is going to be the next step? the next step is we are going to find the individual kai square values for all of the categories.

06:45 How do we do that? they are given by the formula.

06:50 Difference of observed and expected values, square them, and divide that by e...

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