Question
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
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.
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