A.) The average (mean) annual income was greater than $45,000
Null Hypothesis: The average (mean) annual income is greater than or equal to $45,000.
Ho: u > $45,000
Alternative Hypothesis: The average (mean) annual income was less than $45,000
Ha: u < $45,000
I will use a = .05 as the significance level, and observing the sample size of n < 30 which tells me I need to use a Z-Test to find the mean of this test and the hypothesis. As the alternative hypothesis is Ha: u < $45,000, the given test is a one-tailed Z-Test.
The critical value for the significance level, α=0.05 for a one-tailed z-test is -1.645.
Decision Rule: We must reject the hypothesis if z >1.645
Test Statistics from Minitab:
One-Sample Z: Income ($1000)
Test of mean = 45, < 45
The proposed standard deviation = 14.64
95% Upper Variable N Mean StDev SE Mean Bound Z P Income ($1000) 50 43.48 14.55
2.06 46.86 0.49 0.311 Confidence Intervals from MiniTab:
One-Sample Z
The assumed standard deviation = 14.64
N Mean SE Mean 95% CI
50 42.61 2.06 (39.45, 47.51)
Summary
Since the value of the test statistic is not in the critical I will fail to reject the null hypothesis. Therefore I can conclude that the data gives enough evidence to show the null hypothesis is true i.e. based on the sample data we can say with 95% confidence or 5% significance the claim that the average mean annual income was greater than $45,000, is false.
The Essay on Elements of a Test Hypothesis
Each of a sample of 49 unemployed men was given a mental health examination using the General Health Questionnaire (GHQ). The GHQ is widely recognized measure of present mental health , with lower values indicating better mental health. The mean and standard deviation of the GHQ scores were x = 10. 94 and s = 5. 10, ? respectively. (a). Specify the appropriate null and alternative hypothesis if we ...
I am also able to test this hypothesis using a p-value approach.
B.) The true population proportion of customers who live in an urban area exceeds 45%.
Null Hypothesis: The true population proportion of customers who live in an urban area is greater than or equal to 45%.
Ho: p > .45
Alternate Hypothesis: The true population proportion of customers who live in an urban area is less than 45%. Ha: p < .45
I will use a significance level of, α=0.05. I will use a Z-Test of proportions to test the given hypothesis. And with the alternative hypothesis being Ho: p > .45, the given test is a one-tailed Z-Test.
The critical value for significance level, α=0.05 for a one-tailed Z-Test is given as 1.645.
Decision Rule: To Reject the null hypothesis if z 0.45
95% Lower Sample X N Sample p Bound Z-Value P-Value 1 21 50 0.420000 0.406599 -2.13 0.017
Confidence Interval from MiniTab:
Test and CI for One Proportion
Sample X N Sample p 95% CI 1 21 50 0.406599 (0.283195, 0.556805)
Summary
Since the value of the test statistic is in the critical region I am going to reject the null hypothesis. Therefore I can determine the data gives me enough evidence to show the null hypothesis is false. I can say with 95% confidence or 5% significance, the true population proportion of customers who live in a suburban area is less than 45%, is true.
So, here the p-value for this test is 0.017. Which is smaller than the significance level 0.05, thus I am rejecting the null hypothesis based on the p-value approach as well.
The 95% Upper Confidence Bound in this case is 0.406599; which tells me, with a probability of 0.95, the true population proportion of customers who live in a suburban area is less than 40.6599%.
The Essay on Interventions in order to increase the National Achievement Test of 4th Year Students of Malabon National High School
The primordial purpose of this study is to provide the students with a complete and balanced education and to develop their performance in National Achievement Test (Nat) . Thus, the results of this study will benefit the students, the teachers and the school administrators. This study is significant because it will provide the indispensable facts for the students of Malabon National High School ...
C.) The average (mean) number of years lived in the current home is less than 13 years.
Null Hypothesis: The average (mean) number of years lived in the current home is less than or equal to 8 years. Ho: u < 8
Alternate Hypothesis: The average (mean) number of years lived in the current home is greater than 8 years.
Ha: u > 8
I will utilize, α=0.05. Because the sample size is less than 30, a Z-Test will be utilized to find the mean of the hypothesis. As the alternative hypothesis is Ha: u > 8, the given test is a one-tailed Z-Test.
The critical value for significance level, α=0.05 for a lower-tailed z-test is given as 1.645. Decision Rule: Reject null hypothesis z >1.645
Test Statistic from Minitab:
One-Sample Z: Years
Test of mu = < 8, > 8
The proposed standard deviation = 4.4855
95% Upper Variable N Mean StDev SE Mean Bound Z P Years 50 12.380 4.4855 0.722 8.557 2.52 0.006
Summary
Since the value of the test statistic is in the critical region, I am rejecting the null hypothesis. Therefore, I can conclude the data gives enough evidence to show the null hypothesis is false. I can say with 95% confidence or 5% significance the claim, the average (mean) number of years lived in the current home is greater than 8 years, is true.
So the p-value for this test is 0.006. Which is smaller than the significance level 0.05, as a result I am rejecting the null hypothesis based on the p-value approach as well.
The 95% Lower Confidence Bound in this case is 8.557; which tells me, with a probability of 0.95, the average (mean) number of years lived in the current home greater than 8.557.
D.) The average (mean) credit balance for suburban customers is more than $4300.
Null Hypothesis: The average (mean) credit balance for suburban customers is greater than or equal to $3200
Ho: u > 3200
Alternative Hypothesis: The average (mean) credit balance for suburban customers is less than $3200.
Ha: u < 3200
I will use the Significance Level from the case, α=0.05.
The Coursework on Students Social Lifestyle and First Year Average Exam Grade
... relationship between a Students social lifestyle and their first year average exam grades Null Hypothesis: There is no relationship between a Students social lifestyle ... larger than 0.05 at the 5% significance level which means the null hypothesis is accepted. Finally a heteroskedasticity test showed the value of 0.9975 ...
Because the sample size is 13 a t-test will be used for mean to test the hypothesis. As the alternative hypothesis is Ha: u < 3200, the test is a one-tailed t-test.
The critical value for given alpha, α=0.05 for a one-tailed z-test is 1.645.
Decision Rule: Reject the null hypothesis if z
Now the test statistic can be computed by the following formula which gives, Z =
Z = -1.35 with corresponding p-value 0.1.
And the 95% Upper Bound = 3251.
Summary
Since the value of the test statistic is not in the critical region I am going to fail to reject the null hypothesis. Therefore I can conclude the data gives enough evidence to show the null hypothesis is true. I can say, with 95% confidence or 5% significance, the claim, the average (mean) credit balance for rural customers is less than $3200, is false.
So the p-value for this test is 0.1. Which is larger than the significance level 0.05, so I am failing to reject the null hypothesis based on the p-value approach as well.
The 95% Upper Confidence Bound in this case is 3251; which tells me, with probability 0.95, I can claim the average (mean) credit balance for rural customers is less than 3251.
