Question

The director of library services at a 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 below.

Subject Area	Percent of Books on Subject in Circulation Library on This Subject	Number of Books in Sample on This Subject
Business	32%	265
Humanities	25%	220
Natural Science	20%	221
Social Science	15%	113
All other subjects	8%	69

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.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are the same.
H₁: The distributions are the same.H₀: The distributions are different.
H₁: The distributions are the same.    H₀: The distributions are the same.
H₁: The distributions are different.H₀: The distributions are different.
H₁: The distributions are different.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

chi-squarenormal    uniformbinomialStudent's t

What are the degrees of freedom?


(c) Estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students.At the 5% level of significance, the evidence is insufficient to conclude that the subject distribution of books in the library is different from that of books checked out by studentsThe director of library services at a 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 below.

Subject Area	Percent of Books on Subject in Circulation Library on This Subject	Number of Books in Sample on This Subject
Business	32%	265
Humanities	25%	220
Natural Science	20%	221
Social Science	15%	113
All other subjects	8%	69

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.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: The distributions are the same.
H₁: The distributions are the same.H₀: The distributions are different.
H₁: The distributions are the same.    H₀: The distributions are the same.
H₁: The distributions are different.H₀: The distributions are different.
H₁: The distributions are different.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

chi-squarenormal    uniformbinomialStudent's t

What are the degrees of freedom?


(c) Estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students.At the 5% level of significance, the evidence is insufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students

Answer 1

Solution:

Here, we have to use chi square test for goodness of fit.

(a) What is the level of significance?

We are given level of significance = α = 0.05

State the null and alternate hypotheses.

H₀: The distributions are the same.
H₁: The distributions are different.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to two decimal places. Round the test statistic to three decimal places.)

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Calculation tables for test statistic are given as below:

Subject Area	Prop.	O	E	(O - E)^2/E
Business	0.32	265	284.160	1.291897523
Humalities	0.25	220	222.000	0.018018018
Natural science	0.2	221	177.600	10.60563063
Social science	0.15	113	133.200	3.063363363
all other subjects	0.08	69	71.040	0.058581081
Total	1	888	888	15.03749062

Are all the expected frequencies greater than 5?

Answer: Yes

What sampling distribution will you use?

Answer: Chi-square

What are the degrees of freedom?

We are given

N = 5

Degrees of freedom = df = N – 1 = 5 – 1 = 4

df = 4

Chi square = ∑[(O – E)^2/E] = 15.03749062

Test statistic = 15.037

(c) Estimate the P-value of the sample test statistic.

P-value = 0.004624

(By using Chi square table or excel)

0.025 < P-value < 0.050

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

P-value < α = 0.05

So, we reject the null hypothesis

Since the P-value ≤ α, we reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, the evidence is sufficient to conclude that the subject distribution of books in the library is different from that of books checked out by students.