Run a one-way ANOVA to compare three or more groups of people (specified by a categorical variable) on one continuous variable.
Run a one-way ANOVA to compare three or more groups of people (specified by a categorical variable) on one continuous variable.
Write the null and alternate hypotheses.
What is the calculated mean for each group?
What is the calculated F value?
What is the p and critical F value?
Is the p value significant? Is calculated F value greater than Critical F value? Conclusion?
Run a chi-squared test to determine whether one group of people has higher rates of a given outcome relative to the other groups of people in the data set.
Write the null and alternate hypotheses.
What is the calculated chi-square value?
What is the p and critical value?
Is the p value significant? Is calculated chi-square value greater than Critical F value? Conclusion?
Requirements:
Hypothesis Testing Continued
Cynthia R. Glover
West Coast University
PH 521 Biostatistics
Dr. Gagandeep Gill
August 6, 2023
Number of Medications
Age
Language
Fracture
Ability
Smoking History
3.1 Ho: The proportion of English speaking patients is equal to 50%
Ho: p= 50%
Ha: The proportion of English speaking patients is greater than 50%?”
Ha: p>50%
What is the proportion of English speaking patients
p= 29/66 = 0.4394
3.3 What is the standard error and z statistics
Standard error = ) = = 0.0611
z statistic = (p-hat –p)/ standard error = (0.4394-0.5)/0.0611 = -0.99
Find critical z and p value. Is the p value significant? Conclusion?
Z critical = 1.645
p value = p(z>-0.99) = 0.8389
The p value is not significant as it is greater than the significance value (0.05). We thus fail to reject the null and conclude that there is insufficient evidence to support the claim that the proportion of English speaking patients is greater than
4.1
Ho: The number of medications used regularly by smokers is similar to the number used by nonsmokers
Ha: The number of medications used regularly by smokers is different from the number used by nonsmokers
What is the calculated mean for each group?
What is the confidence interval for each group?
Confidence interval = mean ± t critical * s/√n
Non smokers
n= 42
t critical at 5% significance level and 41 df = 2.02
Confidence interval = 2.9762 ± 2.02* 3.1273/√42
= 2.9762 ± 0.9745
= (2.0017, 3.9507)
Smokers
n= 24
t critical at 5% significance level and 23 df =2.069
Confidence interval = 3.7917± 2.069* 4.2219/√24
= 3.7917 ± 1.7827
= (2.0089, 5.5744)
What is the critical t value?
df = N-2 = 66-2 = 64
t critical = 1.998
Is the p value significant? Conclusion?
Standard error = Sp
Sp =
Standard error = 0.9108
T statistic = (mean 1 – mean 2)/standard error
T statistic = (2.9762- 3.7917)/0.9108 = -0.895
p value = 0.3740
The p value is not significant as it is greater than the significance value (0.05). This fails to reject the null hypothesis and I can say that there is not enough evidence to support the claim that the number of medications used regularly by smokers is different from the number used by nonsmokers.
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