Part I. Chi-Square Goodness of Fit Test (equal frequencies) For a recent year the number of homicides that occurred in New York City are given in the table below. Use a 0.05 level of significance to test the claim that the homicides in NYC are equally

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Home >Mathematics homework help >Statistics homework help >Part I. Chi-Square Goodness of Fit Test (equal frequencies) For a recent year the number of homicides that occurred in New York City are given in the table below. Use a 0.05 level of significance to test the claim that the homicides in NYC are equally

STAT 3001

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This assignment is worth a total of 60 points.

Part I. Chi-Square Goodness of Fit Test (equal frequencies)

For a recent year the number of homicides that occurred in New York City are given in the table below. Use a 0.05 level of significance to test the claim that the homicides in NYC are equally likely to occur in each of the 12 months.

Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
38	30	46	40	46	49	47	50	50	42	37	37

Instructions for performing this test in Stat Disk can be found in the Stat Disk User’s Manual under Goodness of Fit, Equal Frequencies.

Instructions	Answers
1. Use the Chi-Square Goodness-of-Fit test to see if there is a difference between the frequency of homicides in different months of the year. Use a significance level of .05. Paste results here.	Num. categories: 12 0.05 alpha 11 Df 0.497004209 p-value 19.67513757 critical value 10.375 test-statistic
2. What are we trying to show here?	We are trying to show that there is an equal likelihood of homicides occurring in all 12 months from January to December.
3. What is the p-value and what does it represent in the context of this problem?	The p-value was 0.4970. The (p-value = 0.4970 > α= 0.05) meaning that there is not enough evidence to reject the null hypothesis.
4. State in your own words what the results of this Goodness-of-Fit test tells us.	The (critical value = 19.6752 > the test statistic value = 10.3750), hence it is not possible to reject the null hypothesis which states that there is an equal likeliness of homicides to occur in each of the 12 months from January to December.
5. Repeat the above procedure using only the summer months of Jun through Sep. Paste results here. Did you get different results? Would these results support the police commissioner’s claim that more homicides occur in the summer when the weather is nicer.	4: num. categories 49 expected value 0.05 alpha 3 Df 0.989013641 p-value 7.814727903 critical value 0.12244898 test statistic The results were different as evidenced by different expected values, degrees of freedom, p-value, critical value and test statistic. Even so, the critical value was still larger than the test statistic, meaning that there is still no enough evidence to reject the hypothesis that more homicides happen in summer.

Part II. Chi-Square Goodness of Fit Test (unequal frequencies)

Acme Toy Company prints baseball cards. The company claims that 30% of the cards are rookies, 60% veterans, and 10% are All-Stars. Suppose a random sample of 100 cards has 50 rookies, 45 veterans, and 5 All-Stars. Is this consistent with Acme's claim? Use a 0.05 level of significance. Instructions for performing this test in Stat Disk can be found in the Stat Disk User’s Manual under Goodness of Fit, Unequal Frequencies.

Instructions	Answers
6. Complete the table as necessary. [Hint: You will need to compute the observed frequencies based on the percentages for the 100 samples.		Rookies	Veterans	All-Stars
1.	OBSERVED	50	45	5
2.	EXPECTED	0.3	0.6	0.1
7. Use the Chi-Square Goodness-of-Fit test for Unequal Frequencies to see if there is a difference between the observed frequencies and the expected frequencies Use a significance level of .05. Paste results here.	Num Categories: 3 Degrees of freedom: 2 Test Statistic, X^2: 19.5833 Critical X^2: 5.991471 P-Value: 0.0001
8. State the null and alternative hypothesis.	Null Hypothesis: of all the baseball cards printed by Ame Toy company, 30% are rookies, 60% are veterans and 10% are All-stars. Alternative Hypothesis: at least one of the proportions in the null hypothesis would be different.
9. What conclusion would you reach, given the result of your Goodness-of-Fit test? Does the data support the company’s claim?[State in your own words following the guidelines for a conclusion statement learned last week.]	There is enough evidence to reject the null hypothesis as depicted by the (p-value = .0001< α = 0.05). the small p-value provides us with substantial evidence to reject the null hypothesis that the proportions will be 30%, 60% and 10 percent for rookies, veterans and All-stars respectively.

Part III. Chi-Square Test of Independence

A randomized controlled trial was designed to compare the effectiveness of splinting versus surgery in the treatment of carpal tunnel syndrome. Results are given in the table below. Use a significance level of 0.01 to test the claim that success is independent of the type of treatment.

	Successful	Unsuccessful
Splint treatment	60	23
Surgery treatment	67	6

Hint: Instructions for performing this test in Stat Disk can be found in the Stat Disk User’s Manual under the heading Chi Square Test of Independence (Contingency Tables).

Instructions	Answers
10. Just looking at the numbers in the table, what is your best guess about the relationship between type of treatment and success? Are they independent or is there a relationship?	Looking at the numbers, I would guess that both the split and surgery treatments have more successful treatments compared to successful treatments with surgery producing more success. They are therefore independent.
11. Compute a Chi-Square Test of Independence on this data using a 0.01 level of significance. Paste your results here.	Degrees of freedom: 1 Test Statistic, X^2: 9.7504 Critical X^2: 6.634903 P-Value: 0.0018
12. What are the null and alternative hypothesis for this test?	Null hypothesis: there is no association between surgery and splinting for carpal tunnel syndrome. Alternative hypothesis: there is an association between surgery and splinting for carpal tunnel syndrome.
13. What is the p-value for this result? What does this represent?	The (p-value = .0018 < α = .01) shows that there is enough evidence to reject the null hypothesis.
14. State your conclusion related to the context of this problem.	Since there is enough evidence to reject the null hypothesis, the alternative is accepted, meaning that there is an association between surgery and splinting for carpal tunnel syndrome. The two variables are therefore not independent.

Part IV. Apply this to your own situation

Using one of the above statistical tests, compose and SOLVE an actual problem from the context of your own personal or professional life. You will need to make up some data and describe which test you will use to analyze the situation. Here’s an example:

	Example: Do not use this problem!!
State the problem that you are analyzing.	Last year I asked my friends what sport they preferred. 50% reported soccer, 30% reported rugby and 20% reported cricket. I wanted to know if this had changed and whether my friends want to keep watching these games.
Make up some data for the new situation.	I asked all my 30 friends the kind of game they preferred. These friends are those that are on my phonebook. This is what they said. •8 said cricket •7 said rugby •15 said soccer
Determine which type of Chi-Square test you will perform.	Since these are unequal frequencies, I will perform a Chi-Square Goodness-of-Fit Test (Unequal Frequencies).
Specify your null and alternative hypotheses.	H0: There is no difference this year in the preferences of sports among my friends H1: my friends have changed their preferences for sports.
Setup the test		cricket	rugby	Soccer
3.	OBSERVED	8	7	15
4.	EXPECTED	0.3	0.2	0.5
Perform the test	Num Categories: 3 Degrees of freedom: 2 Test Statistic, X^2: 0.2778 Critical X^2: 5.991471 P-Value: 0.8703
State your conclusion	We have evidence to believe we should reject the null hypothesis with the test statistic being less than the critical value at the given level of significance of 0.05.

Submit your final draft of your Word file by going to Week 5, Project, and follow the directions under Week 5 Assignment 2. Please use the naming convention "WK5Assgn2+first initial+last name" as the Submission Title.