1.

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)	67	40	107
M.D.	72	90	162
Lawyer	57	80	137
Column Total	196	210	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 not independent H₁: Myers-Briggs preference and profession are 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 independent H₁: Myers-Briggs preference and profession 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?chi-squareStudent'st normalbinomialuniform 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.

2.

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	77	56	56	189
Cliff-Talus	81	69	63	213
Canyon Bench	92	69	65	226
Column Total	250	194	184	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 not independent. H₁: Site type and pottery are independent.H₀: Site type and pottery are independent. H₁: Site type and pottery are independent. H₀: Site type and pottery are not independent. H₁: Site type and pottery are not independent.H₀: Site type and pottery are independent. H₁: Site type and pottery 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?uniformStudent'st normalbinomialchi-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 independant.

3.

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	16	12	13	41
Yearling	11	9	13	33
Adult	32	29	31	92
Column Total	59	50	57	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?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 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.

4.

The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students. In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing.

Myers-Briggs Preference	Arts & Science	Business	Allied Health	Row Total
IN	63	13	20	96
EN	78	38	38	154
IS	59	38	18	115
ES	70	44	40	154
Column Total	270	133	116	519

Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.(a) What is the level of significance? State the null and alternate hypotheses.H₀: Myers-Briggs type and area of study are independent. H₁: Myers-Briggs type and area of study are not independent.H₀: Myers-Briggs type and area of study are independent. H₁: Myers-Briggs type and area of study are independent. H₀: Myers-Briggs type and area of study are not independent. H₁: Myers-Briggs type and area of study are independent.H₀: Myers-Briggs type and area of study are not independent. H₁: Myers-Briggs type and area of study 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?binomialStudent'st uniformchi-squarenormal 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 Myers-Briggs type and area of study are not independent.At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs type and area of study are not independent.

5.

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	6	16	12	34
Science Fiction	10	9	11	30
Comedy	10	6	13	29
Column Total	26	31	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?uniformStudent'st chi-squarebinomialnormal 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.

6.

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%	50
5 to 14	13.6%	85
15 to 64	67.1%	278
65 and older	12.1%	42

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 the same. H₁: The distributions are different.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 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?chi-squarenormal Student'stbinomialuniform

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.

7.

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%	108
Married, no children	29%	119
Single parent	9%	31
One person	25%	88
Other (e.g., roommates, siblings)	11%	65

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 different. H₀: The distributions are the same.H₁: The distributions are the same.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-squareuniform Student'stbinomialnormal

What 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.

8.

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%	910
Obsidian	10.6%	160
Welded Tuff	11.4%	164
Pedernal chert	13.1%	199
Other	3.6%	53

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 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 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-squareStudent'st normaluniformbinomial 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.

9.

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%	92
Rabbit brush	38.7%	133
Salt brush	12%	44
Service berry	9.3%	27
Other	8%	24

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 different. H₁: The distributions are the same.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 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?uniformbinomial Student'stchi-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 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.

10.

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
118	83	214	85

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 different.H₁: The distributions are the same.H₀: The distributions are the same.H₁: The distributions are different. H₀: The distributions are the same.H₁: The distributions are the same.H₀: The distributions are different.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?binomialnormal uniformStudent'stchi-squareWhat 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.