Transcript The table shows a random sample of 100 hikers and the area of hiking preferred.
Slide 1
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 2
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 3
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 4
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 5
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 6
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 2
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 3
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 4
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 5
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.
Slide 6
The table shows a random sample of 100 hikers
and the area of hiking preferred. Are hiking area
preference and gender independent?
Hiking Preference Area
Coastline
Lake/Stream
Mountains
Gender
Female
18
16
11
Male
16
25
14
Ho: Gender and preferred hiking area are
independent.
Ha: Gender and preferred hiking area are not
independent.
•The table contains the observed (O) frequencies.
• If the null hypothesis is true, the expected
percentages (E) are calculated by the formula
(row total)(column total) ÷ total surveyed
• A Test of Independence is right-tailed.
• The degrees of freedom (df)
= (# rows – 1)(# columns – 1) = (2 – 1)(3 – 1) = 2
Distribution for the Test: Chi-Square
Mean of the distribution = number of dfs = 2
To find the pvalue:
•
•
•
•
•
Go to MATRIX in calculator, scroll to EDIT, choose [A]
We have a 2 x 3 matrix.
Enter the values from the table into the matrix. QUIT.
Go to STAT, TESTS, scroll down to χ2-test, press Enter.
Press Enter, Enter, Enter. The test statistic and p-value are given.
Test statistic: 1.4679
p-value: 0.4800
• If the Null is true, there is a 0.4800 probability that the test statistic is
greater than 1.4679.
Decision: Assume
α = 0.05 (α < p-value)
DO NOT REJECT Ho.
Conclusion: There is NOT sufficient evidence
to conclude that gender and hiking preference
are independent.