In: Statistics and Probability
1a Test the claim that the mean GPA of Orange Coast students is
smaller than the mean GPA of Coastline students at the 0.10
significance level.
The null and alternative hypothesis would be:
H0:pO≥pCH0:pO≥pC
H1:pO<pCH1:pO<pC
H0:pO=pCH0:pO=pC
H1:pO≠pCH1:pO≠pC
H0:μO≤μCH0:μO≤μC
H1:μO>μCH1:μO>μC
H0:pO≤pCH0:pO≤pC
H1:pO>pCH1:pO>pC
H0:μO=μCH0:μO=μC
H1:μO≠μCH1:μO≠μC
H0:μO≥μCH0:μO≥μC
H1:μO<μCH1:μO<μC
The test is:
right-tailed
two-tailed
left-tailed
The sample consisted of 35 Orange Coast students, with a sample
mean GPA of 3.35 and a standard deviation of 0.06, and 35 Coastline
students, with a sample mean GPA of 3.38 and a standard deviation
of 0.02.
The test statistic is: (to 2 decimals)
The p-value is: (to 2 decimals)
Based on this we:
1b)
You are conducting a test of the claim that the row variable and the column variable are dependent in the following contingency table.
X | Y | Z | |
---|---|---|---|
A | 41 | 28 | 13 |
B | 35 | 57 | 17 |
Give all answers rounded to 3 places after the decimal point, if necessary.
(a) Enter the expected frequencies below:
X | Y | Z | |
---|---|---|---|
A | |||
B |
(b) What is the chi-square test-statistic for
this data?
Test Statistic: χ2=χ2=
(c) What is the critical value for this test of
independence when using a significance level of αα = 0.10?
Critical Value:
χ2=χ2=
(d) What is the correct conclusion of this hypothesis test at the 0.10 significance level?
Remember to give all answers rounded to 3 places after the decimal point, if necessary.
please if you can show work from ti 84 please
1a
The null and alternative hypothesis would be:
H0:pO≥pC
H1:pO<pC
The test is: left-tailed
using excel>addin>phstat>two variance '
we have
Separate-Variances t Test for the Difference Between Two Means | |
(assumes unequal population variances) | |
Data | |
Hypothesized Difference | 0 |
Level of Significance | 0.05 |
Population 1 Sample | |
Sample Size | 35 |
Sample Mean | 3.35 |
Sample Standard Deviation | 0.06 |
Population 2 Sample | |
Sample Size | 35 |
Sample Mean | 3.38 |
Sample Standard Deviation | 0.02 |
Intermediate Calculations | |
Numerator of Degrees of Freedom | 0.0000 |
Denominator of Degrees of Freedom | 0.0000 |
Total Degrees of Freedom | 41.4634 |
Degrees of Freedom | 41 |
Separate Variance Denominator | 0.0107 |
Difference in Sample Means | -0.03 |
Separate-Variance t Test Statistic | -2.8062 |
Lower-Tail Test | |
Lower Critical Value | -1.6829 |
p-Value | 0.0038 |
The test statistic is: -2.81
The p-value is: 0.00
Based on this we:
1b)
You are conducting a test of the claim that the row variable and the column variable are dependent in the following contingency table.
using excel>addin>phstat>chi square test
we have
Chi-Square Test | |||||||||
Observed Frequencies | |||||||||
Column variable | Calculations | ||||||||
X | Y | Z | Total | fo-fe | |||||
A | 41 | 28 | 13 | 82 | 8.371728 | -8.49215 | 0.120419 | ||
B | 35 | 57 | 17 | 109 | -8.37173 | 8.492147 | -0.12042 | ||
Total | 76 | 85 | 30 | 191 | |||||
Expected Frequencies | |||||||||
Column variable | |||||||||
X | Y | Z | Total | (fo-fe)^2/fe | |||||
A | 32.62827 | 36.49215 | 12.87958 | 82 | 2.148009 | 1.976221 | 0.001126 | ||
B | 43.37173 | 48.50785 | 17.12042 | 109 | 1.615933 | 1.486699 | 0.000847 | ||
Total | 76 | 85 | 30 | 191 | |||||
Data | |||||||||
Level of Significance | 0.1 | ||||||||
Number of Rows | 2 | ||||||||
Number of Columns | 3 | ||||||||
Degrees of Freedom | 2 | ||||||||
Results | |||||||||
Critical Value | 4.60517 | ||||||||
Chi-Square Test Statistic | 7.228835 | ||||||||
p-Value | 0.026933 | ||||||||
Reject the null hypothesis | |||||||||
Reject the null hypothesis |
(a) Enter the expected frequencies below:Give all answers rounded to 3 places after the decimal point, if necessary.
Column variable | |||
X | Y | Z | |
A | 32.628 | 36.492 | 12.880 |
B | 43.372 | 48.508 | 17.120 |
(c) What is the critical value for this test of
independence when using a significance level of αα = 0.10?(b) What
is the chi-square test-statistic for this
data?
Test Statistic: χ2=7.223
Critical Value: χ2=4.605
(d) What is the correct conclusion of this hypothesis test at the 0.10 significance level?