(10:2)

2.

The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community of a village are shown below.

Age (years)	Percent of Canadian Population	Observed Number in the Village
Under 5	7.2%	46
5 to 14	13.6%	72
15 to 64	67.1%	293
65 and older	12.1%	44

Use a 5% level of significance to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village.(a) What is the level of significance? State the null and alternate hypotheses.H₀: The distributions are different. H₁: The distributions are the same.H₀: The distributions are different. H₁: The distributions are different. H₀: The distributions are the same. H₁: The distributions are the same.H₀: The distributions are the same. H₁: The distributions are different. (b) Find the value of the chi-square statistic for the sample. (Round your answer to three decimal places.) Are all the expected frequencies greater than 5?YesNo What sampling distribution will you use?uniformnormal binomialStudent'stchi-square What are the degrees of freedom? (c) Estimate theP-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 that the population fits the specified distribution of categories?Since theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, 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 insufficient to conclude that the village population does not fit the general Canadian population.At the 5% level of significance, the evidence is sufficient to conclude that the village population does not fit the general Canadian population.

3.

The type of household for the U.S. population and for a random sample of411households from a community in Montana are shown below.

Type of Household	Percent of U.S. Households	Observed Number of Households in the Community
Married with children	26%	100
Married, no children	29%	107
Single parent	9%	37
One person	25%	100
Other (e.g., roommates, siblings)	11%	67

Use a 5% level of significance to test the claim that the distribution of U.S. households fits the Dove Creek distribution.(a)What is the level of significance?State the null and alternate hypotheses.H₀: The distributions are different.H₁: The distributions are different.H₀: The distributions are the same.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 the same.(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.)Are all the expected frequencies greater than 5?YesNo What sampling distribution will you use?chi-squareStudent'st binomialuniformnormalWhat are the degrees of freedom?(c)Find or estimate theP-value of the sample test statistic. (Round your answer to three decimal places.)(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?Since theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, 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 community household distribution does not fit the general U.S. household distribution.At the 5% level of significance, the evidence is insufficient to conclude that the community household distribution does not fit the general U.S. household distribution.

4.

The types of raw materials used to construct stone tools found at an archaeological site are shown below. A random sample of 1486 stone tools were obtained from a current excavation site.

Raw Material	Regional Percent of Stone Tools	Observed Number of Tools as Current excavation Site
Basalt	61.3%	908
Obsidian	10.6%	153
Welded Tuff	11.4%	168
Pedernal chert	13.1%	209
Other	3.6%	48

Use a 1% level of significance to test the claim that the regional distribution of raw materials fits the distribution at the current excavation site.(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 different. H₀: The distributions are different. H₁: The distributions are the same.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 at least 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 Student'stuniformbinomial What are the degrees of freedom? (c) Find or estimate theP-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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application.At the 0.01 level of significance, the evidence is sufficient to conclude that the regional distribution of raw materials does not fit the distribution at the current excavation site.At the 0.01 level of significance, the evidence is insufficient to conclude that the regional distribution of raw materials does not fit the distribution at the current excavation site.

5.

The types of browse favored by deer are shown in the following table. Using binoculars, volunteers observed the feeding habits of a random sample of 320 deer.

Type of Browse	Plant Composition in Study Area	Observed Number of Deer Feeding on This Plant
Sage brush	32%	106
Rabbit brush	38.7%	118
Salt brush	12%	45
Service berry	9.3%	30
Other	8%	21

Use a 5% level of significance to test the claim that the natural distribution of browse fits the deer feeding pattern.(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 different. H₀: The distributions are different. H₁: The distributions are the same.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 at least 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-squareuniform binomialnormalStudent'st What are the degrees of freedom? (c) Estimate theP-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 that the population fits the specified distribution of categories?Since theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, 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 natural distribution of browse does not fit the feeding pattern.At the 5% level of significance, the evidence is insufficient to conclude that the natural distribution of browse does not fit the feeding pattern.

6.

The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15% pike. Five years later it sampled the lake to see if the distribution of fish had changed. It found that the500fish in the sample were distributed as follows.

Catfish	Bass	Bluegill	Pike
122	87	224	67

In the 5-year interval, did the distribution of fish change at the 0.05 level?(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 different. H₀: The distributions are the same.H₁: The distributions are different.H₀: The distributions are different.H₁: The distributions are the same.(b)Find the value of the chi-square statistic for the sample. (Round your answer to three decimal places.)Are all the expected frequencies greater than 5?YesNo What sampling distribution will you use?uniformbinomial chi-squarenormalStudent'stWhat are the degrees of freedom?(c)Estimate theP-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 that the population fits the specified distribution of categories?Since theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, 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 insufficient to conclude that current fish distribution is different than that of five years ago.At the 5% level of significance, the evidence is sufficient to conclude that current fish distribution is different than that of five years ago.

7.

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%	294
Humanities	25%	212
Natural Science	20%	203
Social Science	15%	100
All other subjects	8%	79

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 different.H₀: The distributions are different. H₁: The distributions are the same. H₀: The distributions are different. H₁: The distributions are different.H₀: The distributions are the same. H₁: The distributions are the same. (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-squarebinomial normalStudent'stuniform What are the degrees of freedom? (c) Estimate theP-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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, 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.

8.

The acuracy of a census report on a city in southern California was questioned by some government officials. A random sample of 1215 people living in the city was used to check the report, and the results are shown below.

Ethnic Origin	Census Percent	Sample Result
Black	10%	124
Asian	3%	32
Anglo	38%	472
Latino/Latina	41%	509
Native American	6%	65
All others	2%	13

Using a 1% level of significance, test the claim that the census distribution and the sample distribution agree.(a) What is the level of significance? State the null and alternate hypotheses.H₀: The distributions are different. H₁: The distributions are the same.H₀: The distributions are different. H₁: The distributions are different. H₀: The distributions are the same. H₁: The distributions are different.H₀: The distributions are the same. H₁: The distributions are the same. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least 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?binomialStudent'st uniformchi-squarenormal What are the degrees of freedom? (c) Estimate theP-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 that the population fits the specified distribution of categories?Since theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application.At the 1% level of significance, the evidence is sufficient to conclude that census distribution and the ethnic origin distribution of city residents are different.At the 1% level of significance, the evidence is insufficient to conclude that census distribution and the ethnic origin distribution of city residents are different.

9.

Snoop Incorporated is a firm that does market surveys. The Rollum Sound Company hired Snoop to study the age distribution of people who stream music. To check the Snoop report, Rollum used a random sample of 519 customers and obtained the following data.

Customer Age (yrs)	Percent of Customers from Snoop Report	Number of Customers from Sample
Less than 14	12%	98
14 - 18	29%	138
19 - 23	11%	47
24 - 28	10%	43
29 - 33	14%	76
More than 33	24%	117

Using a 1% level of significance, test the claim that the distribution of customer ages in the Snoop report agrees with that of the sample report.(a) What is the level of significance? State the null and alternate hypotheses.H₀: The distributions are different. H₁: The distributions are the same.H₀: The distributions are different. H₁: The distributions are different. H₀: The distributions are the same. H₁: The distributions are the same.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 at least 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?uniformStudent'st normalchi-square What are the degrees of freedom? (c) Estimate theP-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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application.At the 1% level of significance, the evidence is sufficient to conclude that that the distribution of customer ages and the age distribution shown in the Snoop report are different.At the 1% level of significance, the evidence is insufficient to conclude that that the distribution of customer ages and the age distribution shown in the Snoop report are different.

10.

A gambler complained about the dice. They seemed to be loaded! The dice were taken off the table and tested one at a time. One die was rolled 300 times and the following frequencies were recorded.

Outcome	1	2	3	4	5	6
Observed frequencyO	64	47	61	35	50	43

Do these data indicate that the die is unbalanced? Use a 1% level of significance.Hint: If the die is balanced, all outcomes should have the same expected frequency.(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 at least 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?uniformchi-square normalStudent'st What are the degrees of freedom? (c) Estimate theP-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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application.At the 1% level of significance, the evidence is sufficient to conclude that the distribution of observed outcomes for the die is different from the expected distribution of a fair die.At the 1% level of significance, the evidence is insufficient to conclude that the distribution of observed outcomes for the die is different from the expected distribution of a fair die.

(10;1)

5.

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.

	Personality Type
Occupation	E	I	Row Total
Clergy (all denominations)	64	43	107
M.D.	69	93	162
Lawyer	59	78	137
Column Total	192	214	406

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.(a) What is the level of significance? State the null and alternate hypotheses. H₀: Myers-Briggs preference and profession are independent H₁: Myers-Briggs preference and profession are not independent.H₀: Myers-Briggs preference and profession are independent H₁: Myers-Briggs preference and profession are independent. H₀: Myers-Briggs preference and profession are not independent H₁: Myers-Briggs preference and profession are not independent.H₀: Myers-Briggs preference and profession are not independent H₁: Myers-Briggs preference and profession are independent. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least 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?binomialchi-square normalStudent'stuniform What are the degrees of freedom? (c) Find or estimate theP-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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.

6.

The following table shows the Myers-Briggs personality preferences for a random sample of 519 people in the listed professions. T refers to thinking and F refers to feeling.

	Personality Type
Occupation	T	F	Row Total
Clergy (all denominations)	52	96	148
M.D.	78	81	159
Lawyer	110	102	212
Column Total	240	279	519

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance.(a) What is the level of significance? State the null and alternate hypotheses. H₀: Myers-Briggs preference and profession are independent. H₁: Myers-Briggs preference and profession are not independent.H₀: Myers-Briggs preference and profession are not independent. H₁: Myers-Briggs preference and profession are not independent. H₀: Myers-Briggs preference and profession are independent. H₁: Myers-Briggs preference and profession are independent.H₀: Myers-Briggs preference and profession are not independent. H₁: Myers-Briggs preference and profession are independent. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least 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?normaluniform chi-squareStudent'stbinomial What are the degrees of freedom? (c) Find or estimate theP-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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 1% level of significance, there is insufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.At the 1% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.

7.

The following table shows site type and type of pottery for a random sample of 628 sherds at an archaeological location.

	Pottery Type
Site Type	Mesa Verde Black-on-White	McElmo Black-on-White	Mancos Black-on-White	Row Total
Mesa Top	73	56	60	189
Cliff-Talus	83	70	60	213
Canyon Bench	87	70	69	226
Column Total	243	196	189	628

Use a chi-square test to determine if site type and pottery type are independent at the 0.01 level of significance.(a) What is the level of significance? State the null and alternate hypotheses.H₀: Site type and pottery are independent. H₁: Site type and pottery are independent.H₀: Site type and pottery are independent. H₁: Site type and pottery are not independent. H₀: Site type and pottery are not independent. H₁: Site type and pottery are not independent.H₀: Site type and pottery are not independent. H₁: Site type and pottery are independent. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least 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?uniformnormal binomialStudent'stchi-square What are the degrees of freedom? (c) Find or estimate theP-value of the sample test statistic. (Round your answer to three decimal places.) 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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application.At the 1% level of significance, there is sufficient evidence to conclude that site and pottery type are not independent.At the 1% level of significance, there is insufficient evidence to conclude that site and pottery type are not independent.

8.

The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.

Ceremonial Ranking	Cooking Jar Sherds	Decorated Jar Sherds (Noncooking)	Row Total
A	81	54	135
B	95	50	145
C	74	80	154
Column Total	250	184	434

Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.(a)What is the level of significance?State the null and alternate hypotheses.H₀: Ceremonial ranking and pottery type are not independent.H₁: Ceremonial ranking and pottery type are independent.H₀: Ceremonial ranking and pottery type are not independent.H₁: Ceremonial ranking and pottery type are not independent. H₀: Ceremonial ranking and pottery type are independent.H₁: Ceremonial ranking and pottery type are not independent.H₀: Ceremonial ranking and pottery type are independent.H₁: Ceremonial ranking and pottery type are independent.(b)Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least 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-squareStudent'st normalbinomialuniformWhat are the degrees of freedom?(c)Find or estimate theP-value of the sample test statistic. (Round your answer to three decimal places.)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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis.(e)Interpret your conclusion in the context of the application.At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.

9.

The following table shows age distribution and location of a random sample of 166 buffalo in a national park.

Age	Lamar District	Nez Perce District	Firehole District	Row Total
Calf	12	13	16	41
Yearling	12	11	10	33
Adult	31	28	33	92
Column Total	55	52	59	166

Use a chi-square test to determine if age distribution and location are independent at the 0.05 level of significance.(a) What is the level of significance? State the null and alternate hypotheses.H₀: Age distribution and location are independent. H₁: Age distribution and location are independent.H₀: Age distribution and location are not independent. H₁: Age distribution and location are not independent. H₀: Age distribution and location are not independent. H₁: Age distribution and location are independent.H₀: Age distribution and location are independent. H₁: Age distribution and location are not independent. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least 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?normaluniform binomialStudent'stchi-square What are the degrees of freedom? (c) Find or estimate theP-value of the sample test statistic. (Round your answer to three decimal places.) 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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application.At the 5% level of significance, there is sufficient evidence to conclude that age distribution and location are not independent.At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.

10.

Mr. Acosta, a sociologist, is doing a study to see if there is a relationship between the age of a young adult (18 to 35 years old) and the type of movie preferred. A random sample of 93 adults revealed the following data. Test whether age and type of movie preferred are independent at the 0.05 level.

	Person's Age
Movie	18-23 yr	24-29 yr	30-35 yr	Row Total
Drama	10	14	10	34
Science Fiction	11	8	11	30
Comedy	8	6	15	29
Column Total	29	28	36	93

(a) What is the level of significance? State the null and alternate hypotheses.H₀: Age and movie preference are independent. H₁: Age and movie preference are not independent.H₀: Age and movie preference are not independent. H₁: Age and movie preference are independent. H₀: Age and movie preference are not independent. H₁: Age and movie preference are not independent.H₀: Age and movie preference are independent. H₁: Age and movie preference are independent. (b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least 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?Student'stuniform chi-squarenormalbinomial What are the degrees of freedom? (c) Find or estimate theP-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 theP-value >, we fail to reject the null hypothesis.Since theP-value >, we reject the null hypothesis. Since theP-value, we reject the null hypothesis.Since theP-value, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application.At the 5% level of significance, there is sufficient evidence to conclude that age of young adult and movie preference are not independent.At the 5% level of significance, there is insufficient evidence to conclude that age of young adult and movie preference are not independent.