The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate.Use a 2x2 contingency table for questions 1-3. Use the median to divide each list into 2 groups.1) Test for independence between poverty rate and gun deaths.2) Test for independence between high school graduation rate and gun deaths.3) Test for independence between unemployment and gun deaths.4) Divide the data into 3 groups by gun deaths: 0-9, 9.1-12 and 12.1-20.Use ANOVA to test the claim that the average poverty rate is the same for the three groups.Be sure to write down the mean and standard deviation for each group. 0-9 9.1-12 12.1-20n =x =s =n =x =s =n =x =s = Data for Review on Chapters 11&12STATENew JerseyVermontMinnesotaHawaiiDelawareUtahVirginiaNebraskaConnecticutMarylandIdahoAlaskaMassachusettsWashingtonWisconsinNevadaWyomingFloridaNorth DakotaPennsylvanialowaColoradoIllinoisMissouriSouth DakotaMichiganNew HampshireRhode IslandOhioKansasPOVERTY%6.80%7.60%8.10%8.60%9.20%9.20%9.20%9.50%9.70%9.70%9.90%10.00%10.10%10.20%10.20%10.60%10.60%11.10%11.20%11.20%11.30%11.40%11.50%11.60%11.80%12.00%12.00%12.10%12.30%12.50%GUN DEATHS5.79.2122.610.312.610.294.49.714.117.63.18.79.713.816.711.911.811.2-11.5126.45.31111.4H.S. GRAD's91.89191.590.487.490.486.689.888.688.288.491.48989.789.883.987.485.390.187.991.489.386.486.889.987.991.384.787.689.7UNEMPLOYMENT5.73.643.54.93.74.52.85.35.14.26.64.75.34.56.845.32.95.43.74.25.65.63.75.13.65.64.74.6 IndianaMaineNorth CarolinaCaliforniaMontanaGeorgiaNew YorkKentuckyTennesseeSouth CarolinaArizonaWest VirginiaOklahomaArkansasTexasAlabamaOregonNew MexicoLouisianaMississippiMedians12.60%12.60%13.10%13.20%13.80%14.40%14.50%14.80%15.00%15.00%15.20%15.40%15.60%15.90%16.20%16.70%16.70%17.90%18.30%20.10%11.9%137.612.17.714.412.64.213.715.415.214.114.316.519.810.617.6117.619.317.811.586.690.284.380.690.883.984.781.783.183.684.282.885.682.479.982.189.182.882.280.4888884.64.55.96.14.15.95.25.25.766.37.64.65.44.16.26.16.766.35.2

Note: “Since you have posted a question with multiple sub parts, we will provide the solution only to the first three sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately. 1) The objective is to Test for independence between poverty rate and gun deaths. State the hypotheses as follows: Null Hypothesis H0: 'poverty rate' and 'gun deaths' are independent. Alternative Hypothesis H1:'poverty rate' and 'gun deaths' are dependent. Based on the given data and with the help of given medians, the 2×2 contingency table for poverty rate and gun deaths is obtained as follows: Poverty/Gun Deaths Low High Row Total Low 15 10 25 High 10 15 25 Column Total 25 25 50 The test statistic is given by the formula: χ2=∑iOi-Ei2Ei where, Oi is the observed frequency and Ei is the expected frequency. The expected frequencies are computed using the formula: E=Row total×Column totalGrand total Calculations: Poverty/Gun Deaths Low High Row Total Low 25×2550=12.5 25×2550=12.5 25 High 25×2550=12.5 25×2550=12.5 25 Column Total 25 25 50 Substitute the values in the formula of the test statistic: χ2=∑iOi-Ei2Ei=15-12.5212.5+10-12.5212.5+10-12.5212.5+15-12.5212.5=0.5+0.5+0.5+0.5=2 Thus, the value of the test statistic is 2 Degrees of freedom: df=r-1c-1=2-12-1=1 Here, r is the number of rows and c is the number of columns. Using the Chi-square table, look for df=1 in the first column and α=0.05 in the topmost row, the intersection of both values is 3.841 Thus, the required critical value is 3.841 Since the value of the test statistic is less than the critical value i.e., (2 < 3.841) we fail to reject the null hypothesis. Therefore, it can be concluded that the variables poverty rate and gun deaths are independent. 2) The objective is to test for independence between high school graduation rate and gun deaths. State the hypotheses as follows: Null Hypothesis H0: High school graduation rate and gun deaths are independent. Alternative Hypothesis H1:High school graduation rate and gun deaths are dependent. Based on the given data and with the given medians, the 2×2 contingency table for High school graduation rate and gun deaths is obtained as follows: HS graduation/Gun Deaths Low High Row Total Low 8 17 25 High 17 8 25 Column Total 25 25 50 The test statistic is given by the formula: χ2=∑iOi-Ei2Ei where, Oi is the observed frequency and Ei is the expected frequency. The expected frequencies are computed using the formula: E=Row total×Column totalGrand total Calculations: HS graduation/Gun Deaths Low High Row Total Low 25×2550=12.5 25×2550=12.5 25 High 25×2550=12.5 25×2550=12.5 25 Column Total 25 25 50 Substitute the values in the formula of the test statistic: χ2=∑iOi-Ei2Ei=8-12.5212.5+17-12.5212.5+17-12.5212.5+8-12.5212.5=1.62+1.62+1.62+1.62=6.48 Thus, the value of the test statistic is 6.48 Degrees of freedom: df=r-1c-1=2-12-1=1 Using the Chi-square table, the critical value at degrees of freedom df=1 and significance level α=0.05 is 3.841 Since the value of the test statistic is greater than the critical value i.e., (6.48 > 3.841) reject the null hypothesis. Therefore, it can be concluded that the variables High school graduation rate and gun deaths are dependent. 3) The objective is to test for independence between unemployment and gun deaths. State the hypotheses as follows: Null Hypothesis H0: Unemployment and gun deaths are independent. Alternative Hypothesis H1:Unemployment and gun deaths are dependent. Based on the given data and with the help of given hint, the 2×2 contingency table for Unemployment and gun death is obtained as follows: Unemployment/Gun Deaths Low High Row Total Low 15 10 25 High 10 15 25 Column Total 25 25 50 The expected frequencies are computed using the formula: E=Row total×Column totalGrand total Calculations: Unemployment/Gun Deaths Low High Row Total Low 25×2550=12.5 25×2550=12.5 25 High 25×2550=12.5 25×2550=12.5 25 Column Total 25 25 50 Substitute the values in the formula of the test statistic: χ2=∑iOi-Ei2Ei=15-12.5212.5+10-12.5212.5+10-12.5212.5+15-12.5212.5=0.5+0.5+0.5+0.5=2 Thus, the value of the test statistic is 2 Degrees of freedom: df=r-1c-1=2-12-1=1 Using the Chi-square table, the critical value at degrees of freedom df=1 and significance level α=0.05 is 3.841 (shown in subpart 2). Since the value of the test statistic is less than the critical value i.e., (2 < 3.841) do not reject the null hypothesis. Therefore, it can be concluded that the variables unemployment and gun deaths are independent.

Answered: The data below lists states poverty…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.6: Summarizing Categorical Data

Problem 4BGP

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Question

The data below lists states poverty rate, gun deaths per 100,000 people, high school graduation rate and the unemployment rate. Use a 2x2 contingency table for questions 1-3. Use the median to divide each list into 2 groups. 1) Test for independence between poverty rate and gun deaths. 2) Test for independence between high school graduation rate and gun deaths. 3) Test for independence between unemployment and gun deaths. 4) Divide the data into 3 groups by gun deaths: 0-9, 9.1-12 and 12.1-20. Use ANOVA to test the claim that the average poverty rate is the same for the three groups. Be sure to write down the mean and standard deviation for each group. 0-9 9.1-12 12.1-20 n = x = s = n = x = s = n = x = s =

Data for Review on Chapters 11&12
STATE
New Jersey
Vermont
Minnesota
Hawaii
Delaware
Utah
Virginia
Nebraska
Connecticut
Maryland
Idaho
Alaska
Massachusetts
Washington
Wisconsin
Nevada
Wyoming
Florida
North Dakota
Pennsylvania
lowa
Colorado
Illinois
Missouri
South Dakota
Michigan
New Hampshire
Rhode Island
Ohio
Kansas
POVERTY%
6.80%
7.60%
8.10%
8.60%
9.20%
9.20%
9.20%
9.50%
9.70%
9.70%
9.90%
10.00%
10.10%
10.20%
10.20%
10.60%
10.60%
11.10%
11.20%
11.20%
11.30%
11.40%
11.50%
11.60%
11.80%
12.00%
12.00%
12.10%
12.30%
12.50%
GUN DEATHS
5.7
9.2
12
2.6
10.3
12.6
10.2
9
4.4
9.7
14.1
17.6
3.1
8.7
9.7
13.8
16.7
11.9
11.8
11.2
-
11.5
12
6.4
5.3
11
11.4
H.S. GRAD's
91.8
91
91.5
90.4
87.4
90.4
86.6
89.8
88.6
88.2
88.4
91.4
89
89.7
89.8
83.9
87.4
85.3
90.1
87.9
91.4
89.3
86.4
86.8
89.9
87.9
91.3
84.7
87.6
89.7
UNEMPLOYMENT
5.7
3.6
4
3.5
4.9
3.7
4.5
2.8
5.3
5.1
4.2
6.6
4.7
5.3
4.5
6.8
4
5.3
2.9
5.4
3.7
4.2
5.6
5.6
3.7
5.1
3.6
5.6
4.7
4.6

Indiana
Maine
North Carolina
California
Montana
Georgia
New York
Kentucky
Tennessee
South Carolina
Arizona
West Virginia
Oklahoma
Arkansas
Texas
Alabama
Oregon
New Mexico
Louisiana
Mississippi
Medians
12.60%
12.60%
13.10%
13.20%
13.80%
14.40%
14.50%
14.80%
15.00%
15.00%
15.20%
15.40%
15.60%
15.90%
16.20%
16.70%
16.70%
17.90%
18.30%
20.10%
11.9%
13
7.6
12.1
7.7
14.4
12.6
4.2
13.7
15.4
15.2
14.1
14.3
16.5
19.8
10.6
17.6
11
7.6
19.3
17.8
11.5
86.6
90.2
84.3
80.6
90.8
83.9
84.7
81.7
83.1
83.6
84.2
82.8
85.6
82.4
79.9
82.1
89.1
82.8
82.2
80.4
88888
4.6
4.5
5.9
6.1
4.1
5.9
5.2
5.2
5.7
6
6.3
7.6
4.6
5.4
4.1
6.2
6.1
6.7
6
6.3
5.2

Definition Definition Middle value of a data set. The median divides a data set into two halves, and it also called the 50th percentile. The median is much less affected by outliers and skewed data than the mean. If the number of elements in a dataset is odd, then the middlemost element of the data arranged in ascending or descending order is the median. If the number of elements in the dataset is even, the average of the two central elements of the arranged data is the median of the set. For example, if a dataset has five items—12, 13, 21, 27, 31—the median is the third item in ascending order, or 21. If a dataset has six items—12, 13, 21, 27, 31, 33—the median is the average of the third (21) and fourth (27) items. It is calculated as follows: (21 + 27) / 2 = 24.

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